8-th math - karnataka textbook societyktbs.kar.nic.in/new/website...

TEXT BOOK COMMITTEE

ChairmanDr. B. J. Venkatachala, Professor, H.B.C.S.E(T.I.F.R, Mumbai),

Department of Mathematics, Indian Institute of Science, Bengaluru.

Members1. Dr. G. Sheela, Asst. Professor, Department of Education, Manasagangotri,

Mysore University, Mysuru.

2. Sri T. K. Raghavendra, Lecturer, D.I.E.T., Chickballapur.

3. Sri A. Ramaswamy, Asst. Master, Govt. Empress High School, Tumkuru.

4. Sri Vinay A. Joseph, Asst. Master and P.R.O., St. Xavier’s High School,

Shivajinagar, Bengaluru.

5. Smt. Vasanthi Rao, Retired Teacher, Rajajinagar, Bengaluru.

6. Sri G. M. Jangi, Artist, D.S.E.R.T., Bengaluru.

Scrutinizers1. Dr. Ashok M. Limkar, Subject Inspector of Mathematics, D.D.P.I., Office,

Vijayapura.

2. Sri A. S. Hanuman, Subject Inspector of Mathematics, D.D.P.I., Office,

Shivamogga.

Editorial Committee Members1. Dr. K. S. Sameera Simha, Joint Secretary, Vijaya Educational Institutions,

Jayanagar, Bengaluru.

2.Dr. S. Shiva Kumar, Professor, R.V. Engineering College, Bengaluru.

Chief advisors1. Sri Nagendra Kumar, Managing Director, Karnataka Text Book Soci-ety, Bengaluru-85.

2. Sri Panduranga, Deputy Direcor(in-charge), Karnataka Text Book Society, Bengaluru-85.

Chief Co-ordinatorProf. G. S. Mudambadithaya, Curriculum review and Text book preparation,

Karnataka Text Book Society, Bengaluru-85.

Programme Co-ordinatorSmt. R. N. Shashikala, Asst. Director, Karnataka Text Book Society,

Bengaluru-85.

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Foreword

The Government of India through NCERT have brought out NCF-2005to revise the curriculum of schools and suggested all the states to intro-duce revised textbooks in the schools based on the new curriculum. Ac-cordingly state Governments took up the work and requested respectiveDSERTs to start introducing new curriculum and texts. Karnataka Gov-ernment has suggested to its DSERT to take up the challenge to fulfil thevision of NCF-2005. DSERT, Karnataka started the process: constitutedcommittees to revise the syllabi, identified the writers and requested thesepeople to write texts books based on the new syllabi incorporating the ex-pectations of NCF-2005. Karnataka Text Book Society, took the initiativeand coordinated the whole programme of writing these text books.

The current work, a text book in mathematics for 8-th standard, is astep taken in this direction. An effort has been made here to look at themathematics needed at 8-th standard through a different lens. At firstglance, this may look a totally unconventional approach. Some may feelthat it is hard on the part of 8-th standard students. On the other handthat is the correct age for the students to learn new concepts and ideas.Students are receptive to new intellectual challenges. It is the onus ofthe teachers to teach new things to the students and prepare them to thechallenges of the ever changing world. This text book is also an effortto integrate our students with the national mainstream where CBSE hassurged forward and parents think that their wards will be better off bylearning CBSE texts.

We have tried here to tell something new about numbers and numbersystem. Similarly, some thing new about graphs, postulates of geometryand congruency of triangles are also introduced with more expectations.Quadrilaterals have been introduced now itself. There are optional prob-lems at the end to challenge the students.

It is my earnest request to all my teacher friends to take up the newchallenge. Let the parents of our students not feel that their wards arealways in the back seats.

B. J. VenkatachalaHomi Bhabha Centre for

Science Education, TIFR, Mumbai

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Table of ContentsIndian Mathematics - A brief introduction (i)

Chapter 1.

Unit 1. Playing with numbers 1

Unit 2. Squares, square-roots, cubesand cube-roots 27

Unit 3. Rational numbers 53

Unit 4. Commercial arithmetic 80

Unit 5. Statistics 102

Chapter 2.

Unit 1. Algebraic Expressions 132

Unit 2. Factorisation 148

Unit 3. Linear equations in one variable 156

Unit 4. Exponents 172

Unit 5. Introduction to graphs 192

Chapter 3.

Unit 1. Axioms, postulates and theorems 219

Unit 2. Theorems on triangles 250

Unit 3. Congruency of triangles 266

Unit 4. Construction of triangles 291

Unit 5. Quadrilaterals 306

Chapter 4.

Unit 1. Mensuration 330

Optional problems 341

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INDIAN MATHEMATICS - A BRIEF INTRODUCTION

Indian Mathematics dates back to the Vedic times. The firstsignificant mathematical texts of Vedic times are Shulva Sutras.Shulva is a sanskrit word for chord. These contain the details ofconstruction of sacrificial altars. These ancient texts introducesurds of the type

√2,√3, etc. (In fact most of the ancient mathe-

matics was developed because of the interest in Yajna and Yagaand astrology.) Baudhayana Sutra and Apastamba Sutra give avery good approximation to

√2 in the form

1 +1

3+

1

3× 4− 1

3× 4× 34,

which is correct up to 5 decimal places.

The classical Pythagoras’ theorem is stated in the above su-

tras, far earlier than the Greeks discovered it. Another ancientunsolved problem known as squaring a circle finds its place inShulva sutras. One is required to construct, using only a rulerand a compass, a square whose area is equal to that of the givencircle. Shulva sutras give approximate methods for construct-ing such a square. This remained unsolved over two thousandyears, and only in 18th century it was proved that such a con-struction is impossible.

Indian mathematicians are credited with being the first togive an approximate value for π. Aryabhata I (476 AD) gave an

approximate value for π as 3.1416; he mentions that a circle ofdiameter 20000 units has circumference approximately equalto 62832 units. It is interesting to note that Aryabhata I clearlymentions, in the fifth century itself, that π is not rational and heis using its approximate value. Only in 1761, Lambart provedthat π is an irrational number, and in 1882, Lindeman provedthat π is, in fact, a transcendental number.

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(ii) Indian Mathematics

The most remarkable contribution for which the entire worldstill salutes India is the invention of the decimal system by in-troducing zero and infinity. The use of decimal system is soeasy that even children can grasp it and use it in their calcula-tions. If you really want to appreciate the simplicity of the deci-mal system and the concept of place value, you must first study

the prevalent Roman system of representing numbers. Accord-ing to Florian Cajori, an eminent historian of great repute “ ofall the mathematical discoveries, no one has contributedmore to the general progress of intelligence than Zero.” Us-ing base 10, Indians were able to grasp very large numbers(SeeChapter 2, Unit 4, Exponents for more details).

Ancient Jain contribution to mathematics is another impor-tant milestone in the history of Indian Mathematics. Their find-

ings are recorded in famous Jain texts, dating back to 500 BCto 200 BC. Here again, you see an approximate value for π as√10 and it is calculated up to 13 decimal places.Ancient India has vastly contributed in the areas: the meth-

ods of Arithmetic called Vyakthaganita; The method of Algebracalled Avyakthaganita. Ancient Indian Mathematicians had in-troduced all the four operations: addition, multiplication, sub-

traction and division. They also knew how to operate with frac-tions, solving simple equations, finding square and square-root,finding cube and cube-root, and also knew about permutationand combination.

Mahavira(9th century AD), a great Jain mathematician fromKarnataka, gave the well known formula

n C r =n!

(n− r)!r!

for the first time in the history of mathematics, in his Ganita

Sara Sangraha. Aryabhata I is one of our greatest mathemati-cians and astronomers of all times. He is the one who system-atically developed mathematics and is called, justifiably, the fa-ther of Algebra. He gave tables to trigonometric ratio Sine,

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Indian Mathematics (iii)

called jya in Sanskrit, for angles from 0 to 90 degrees at inter-

vals of 33

4degrees. He was the first Indian to declare: the Earth

is round and that stars appear to move from East to West for astationary observer on the Earth.

Aryabhata I was followed by Bhaskara I (6th century AD) whoprovided interesting geometrical treatment for many algebraicformulae and gave a very good rational approximation for sin θ,even for large values of angle θ. Brahmagupta(628 AD) wasthe first to give a formula for the area of a cyclic quadrilateral inthe form

√(s− a)(s− b)(s− c)(s− d), where a, b, c, d are the sides

of the quadrilateral and s, the semi-perimeter. He is also thefirst mathematician to obtain cyclic quadrilaterals with rationalsides.

Another important area, where our ancient mathematiciansmade significant contributions, is the solution of equations ofthe form ax + by = c, and x2 − Ny2 = 1 in integers, wherea, b, c, N are given integers, and N is generally a square-free pos-itive integer. Such equations are called Diophantine equationsin modern terminology. The second equation is wrongly calledPell’s equation by the great Euler, but the name still contin-

ues. These days, many authors call the equation of the typex2 − Ny2 = 1 as Brahmagupta-Pell’s equation. Aryabhata I dis-cusses ax + by = c, whereas Brahmagupta made a significantcontribution in the understanding of the equation x2 −Ny2 = 1.However, later, Bhaskara II(popularly called Bhaskaracharya)developed a new method, called Chakravala method, for solv-ing x2 − Ny2 = 1 in integers. Here is an interesting thing about

the equation x2 − Ny2 = 1. In 1657, Fermat (who is famousfor his contribution to number theory) proposed the problem ofsolving x2 − 61y2 = 1 for integers x, y, to European mathemati-cians. But none was able to solve this problem. In 1732, Eulergave a complete solution to this equation. But surprisingly, asa matter of divine coincidence, the same equation x2 − 61y2 = 1was solved by Bhaskara II(1150 AD)in his Bijaganitam by his

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(iv) Indian Mathematics

chakravala method more than 5 centuries before. The smallestsolution of x2 − 61y2 = 1 given by Bhaskara II is x = 226153980and y = 1766319049. Bhaskara II also introduced calculus, theconcept of derivative, though not rigorously. He clearly men-tions equivalent of d(sinx) = cosx dx.

We also see in the works of Bhaskara II and Mahavira very

beautiful and interesting problems, rich in poetic imagination.Incidentally, Bhaskara II was born in the current Bijapur dis-trict of Karnataka state (the exact place of his birth is still apoint of debate among the scholars) and moved to the currentMaharashtra state. Here we give two problems, originally statedin verses, whose translation runs as follows:

1. (From Lilavati of Bhaskaracharya)

A beautiful maiden asks me which is the number when mul-tiplied by 3, then increased by three-fourths of the product,divided by 7, diminished by one-third of the quotient, mul-tiplied by itself, diminished by 52, the square-root found,addition of 8, division by 10 gives the number 2?

2. (From Ganita Sara Sangraha of Mahavira)

Three merchants find a purse lying on a road. One mer-chant says, “ If I keep the purse, I shall have twice as muchmoney as the two of you together.” Give me the purse and Ishall have three times as much" said the second merchant.

The third merchant said, “ I shall be better off than eitherof you if I keep the purse; I shall have five times as much asthe two of you together.” How much money is in the purse?How much money does each merchant have?

The achievements of Indian Mathematicians are remarkablein certain specific areas: arithmetic; theory of equations; spher-ical trigonometry and astronomy; geometrical treatment of al-gebraic equations; plane trigonometry; and mensuration. Un-fortunately, after Bhaskara II, because of the invasion, IndianMathematics went into hibernation, except for a brief period

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Indian Mathematics (v)

during which the Kerala school by Nilakanta and Madhava madesome important contributions in the area of series approxima-tion to tan function. With Ramanujan, at the end of 19th cen-tury, it recovered its earlier glory. Ramanujan was a prodigiousmathematician. In his brief span of 32 years, he made wonder-

ful contributions to number theory, hyper-geometric series, di-vergent series, elliptic functions and integrals, and mock-thetafunction. Even today, world-wide mathematicians are trying tounderstand the depth of his mathematics and trying to provethe conjectures he made. It is also worthwhile to mention thatChadrashekhara Samantha of Orissa, at the end of 19th century,made some important contributions to astronomy.

(For more details about the Indian mathematics see “Indian

Mathematics and Astronomy: Some landmarks” by Dr S Bal-achandra Rao, published by Bharatiya Vidya Bhavan, Bengaluru)

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CHAPTER 1 UNIT 1

PLAYING WITH NUMBERS

After studying this unit you learn:

• to write a given natural number in its general form ( in base 10);

• to formulate some games and puzzles involving numbers;

• given two positive integers, how to divide one by the other to get the

quotient and the remainder;

• construction of a 3× 3 magic square;

• divisibility tests for 4,3,9,5,11;

• about some unsolved problems involving numbers.

1.1.1 Introduction

Numbers have played an important role in the intellectual development

of mankind. This still forms a perfect play-ground for activities of chil-

dren. One can create simple puzzles which are brain-teasers. They can be

used to play (mental) games among children. We shall explore some nice

properties of numbers which help in engaging children with puzzles and

these puzzles will help in arousing the curiosity of children. On the other

hand the numbers can help us in placing some of the hitherto unproved

conjectures and perhaps those will help the children to further explore the

wonderful universe of numbers.

You write 76 or 315 and say these are natural numbers. For example,

you say that 6 is the digit in the unit’s place of 76, and 7 is the digit in

ten’s place. Similarly, looking at 315, you say that 5 is in the unit’s place,

1 is in the ten’s place and 3 is in the hundred’s place. You understand the

place value of each digit, given a natural number. We explore a little bit

more about these concepts and create puzzles using them.

Looking at the numbers 2, 24, 46, 88 or 122, you immediately say that

these are even numbers and each is divisible by 2. Can you say whether a

number is divisible by 3 by just looking at the number? Can you say that

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2 Unit 1

a number is divisible by 4,5, 9 or 11 without actually dividing it by these?

Can you device simple rules which help us to decide whether a number is

divisible by 3,4,5,9,11?

1.1.2 Numbers in general form

Consider the number 45. We write this as

45 = 40 + 5 = (4× 10) + (5× 1).

Similarly, 34 = 30+4 = (3×10)+(4×1). What can we do with 354? Observe

354 = 300 + 50 + 4 = (3× 100) + (5× 10) + (4× 1).

Activity 1: Write the following numbers in the form described as above:

75, 88, 121, 361, 1024, 2011, 4444, 2345.

Can you see that any natural number can be written in the above form? It

is immaterial how many digits are there in the number. Suppose you have

123456789, a 9-digit number. You may write it as

123456789 = 100000000 + 20000000 + 3000000 + 400000

+50000 + 6000 + 700 + 80 + 9

= (1× 100000000) + (2× 10000000) + (3× 1000000)

+(4× 100000) + (5× 10000) + (6× 1000)

+(7× 100) + (8× 10) + (9× 1).

You will learn later that this can be written in a compact form as

123456789 = (1× 108) + (2× 107) + (3× 106) + (4× 105) + (5× 104)

+ (6× 103) + (7× 102) + (8× 101) + (9× 100).

This is called the base 10 representation of the given natural numbers

or the generalised form of the number. This system of representing a

number using base 10 was invented by early Indian mathematicians.

Consider, for example, 136. You write this in the generalised form as:

136 = (1× 100) + (3× 10) + (6× 1).

Can you see that 6 is associated with 1; 3 is associated with 10; and 1 is

associated with 100? This is the reason, 6 is called the digit in the unit’s

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Playing with numbers 3

place; 3 is the digit in the ten’s place; and 1 is the digit in the hundred’s

place.

Suppose you have a number abcd, with the digits in the unit’s place,

ten’s place’ hundred’s place and thousand’s place respectively as d,c,b and

a. Then its generalised form is

abcd = (a× 1000) + (b× 100) + (c× 10) + (d× 1).

To avoid the confusion that abcd may represent the product of a, b, c and

d, the number is written in the form abcd. Thus

abcd = (a× 1000) + (b× 100) + (c× 10) + (d× 1).

Indian mathematics emerged in the Indian subcontinent from 1200 BC

until the end of the 18th century and after that the modern era dawned.

In the classical period of Indian mathematics (400 AD to 1200 AD), impor-

tant contributions were made by scholars like Aryabhata, Brahmagupta,

and Bhaskara II. The decimal number system in use today and the binary

number system were first recorded in Indian mathematics. Indian mathe-

maticians made early contributions to the study of the concept of zero as a

number, negative numbers, arithmetic, and algebra. These mathematical

concepts were transmitted to the Middle East, China, and Europe and led

to further developments that now form the foundations of many areas of

mathematics.

All mathematical works were orally transmitted until approximately 500

BC; thereafter, they were transmitted both orally and in manuscript form.

The oldest mathematical document produced on the Indian subcontinent

is the birch bark Bakhshali Manuscript, discovered in 1881 in the village

of Bakhshali, near Peshawar (modern day Pakistan).

The representation using base 10 is only a convenient thing. One can use

different bases and represent numbers. For example computers use base

2 representation(called binary codes) and base 16 representation(hexa-

decimal codes). However, in daily life the use of decimal system(base 10

representation) is the most useful thing and the Indian contribution is for-

ever remembered.

Exercise 1.1.2

1. Write the following numbers in generalised form:

39, 52, 106, 359, 628, 3458, 9502, 7000.

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4 Unit 1

2. Write the following in the normal form:

(i) (5× 10) + (6× 1); (ii) (7× 100) + (5 × 10) + (8× 1); (iii) (6 × 1000) +

(5× 10) + (8× 1); (iv) (7× 1000) + (6× 1); (v) (1× 1000) + (1× 10).

3. Recalling your earlier knowledge, represent 555 in base 5.

4. What is the representation of 1024 in base 2?

1.1.3. Some games and puzzles involving digits

Here we describe some properties of numbers which will help you to

evolve a game to amuse your friends.

Game 1.

You can play a trick with your friend. You do this in several steps.

Step 1. Ask your friend to choose a 2-digit number in his mind and

not to reveal it to you.

Step 2. Tell him to reverse the digits of the number he chose and get

another number.

Step 3. Now tell him to add both the numbers and divide the sum by

11.

Step 4. Surprise him by telling him that the remainder is 0.

At no stage he reveals you the number or its reversal or their sum. Still

you can conclude that the remainder of the sum when divided by 11 is

zero.

For example suppose your friend chooses 41. The number obtained by

reversing its digits is 14. Their sum is 41 + 14 = 55. When 55 is divided by

11, the remainder is 0.

Are you not curious how it works?

Suppose the two digit number is ab. Then you know that ab = (a× 10) +

(b× 1). The reversed number is ba = (b× 10)+ (a× 1). Thus you get the sum

of a number and its reversal as:

ab+ ba = (a× 10) + (b× 1) + (b× 10) + (a× 1) = 11(a+ b).

Now you see why the remainder when divided by 11 is zero.

Activity 2: You can create a game of your own. Instead of taking the

sum of a 2-digit number and its reversal, suppose you take their positive

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difference. Take several examples, say, 21, 34, 86, 79, 95. Which divisor

is common to all the differences: 21 − 12, 43 − 34, 86 − 68, 97 − 79, 95 − 59?

What is the game you can formulate?

Game 2.

This time, tell your friend to choose a 3-digit number and to keep it in

his mind. Let him get the number obtained by reversing the digits of the

original number and tell him to find the difference between the original

number and the reversed number. Ask him to divide this difference by 99.

You may surprise him by telling the remainder is zero, even if you do not

know any thing about his choice. For example, if your friend chooses 891,

the reversed number is 198 and their difference is 891− 198 = 693 = 99× 7.

Hence the remainder is zero after dividing the difference by 99.

Activity 3: Take several 3 digit numbers, say, 263, 394, 512, 765, 681,

898, 926. Find the difference between each number and the number ob-

tained by its reversal. Find the remainder when the difference is divided

by 99 in each case.

Why does this work in general? If the number chosen is abc, then the

reversed number is cba. Thus

abc− cba = (a× 100) + (b× 10) + (c× 1)− (c× 100)− (b× 10)− (a× 1)

= (99× a)− (99× c)

= 99(a− c).

Hence you see that the difference is always divisible by 99.

Game 3.

Now you start with a 3-digit number, say 132. You can get two more

numbers 213 and 321, by cyclically permuting the digits of 132. Add all

of them; you get

132 + 213 + 321 = 666 = 18× 37.

Repeat this with numbers 196, 225, 308, 446, 589, 678, 846. Do you

observe that in each case the resulting number is divisible by 37? Can

you now formulate a game using this property?

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6 Unit 1

Statement 1.

Given a 3-digit number abc, consider two more numbers obtained

by cyclical permutation of its digits, namely, bca and cab. Then 37

divides the sum abc + bca+ cab.

The proof is not hard. Let us see how it works for 132. We have

132 = (1× 100) + (3× 10) + (2× 1),

321 = (3× 100) + (2× 10) + (1× 1),

213 = (2× 100) + (1× 10) + (3× 1).

Thus we get

132 + 321 + 213 = 1× (10 + 100 + 1) + 3× (100 + 10 + 1) + 2× (100 + 1 + 10)

= (1 + 3 + 2)× 111

= 6× 3× 37.

The same method works for for any number abc. Using the general form,

you get

abc = (a× 100) + (b× 10) + (c× 1),

bca = (b× 100) + (c× 10) + (a× 1),

cab = (c× 100) + (a× 10) + (b× 1).

Hence

abc + bca + cab = (a× 100) + (b× 10) + (c× 1) + (b× 100) + (c× 10) + (a× 1)

+(c× 100) + (a× 10) + (b× 1)

= 111(a+ b+ c).

But 111 = 37× 3 and hence the number on the right side is divisible by 37.

We may conclude that abc + bca + cab is divisible by 37.

Alpha numerals and puzzles

You can create puzzles involving numbers and letters of an alphabet.

Look at the following examples.

Example 1. Find the digit represented by P in the following addition.

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Solution:

4 1 P

+ Q 1 5

5 2 6

You see that P , being a digit, cannot exceed 9. The

only way you can arrive to 6 from 5 is adding 1.

Hence P = 1. Similarly, you get Q = 1. You may

check that 411 + 115 = 526.

Example 2. Find the digits A and C in the following multiplication.

Solution:

3 A

× 1 2

C 8 4

Here the last digit of 2 × A is 4. Hence either A = 2

or A = 7. Which one to choose?

Suppose A = 2. Then you get the product as 32 ×12 = 384. This shows that C = 3. On the other hand,

if A = 7, then the product is 37× 12 = 444. However,

the digit in the ten’s place of the result must be

8 and not 4. We may therefore reject A = 7. We

conclude A = 2 and C = 3.

Example 3. In the following addition, A, B, C represent different digits.

Find them and the sum.

Solution:

A A

+ B B

+ C C

B A C

Here you observe that the last digit of A + B + C is

C, so that A + B = 10. (Why is that A + B = 0 not

possible?) Since C is a digit, C ≤ 9. Hence the carry

from unit’s place to ten’s place is 1.

Since we are adding only 3 digits, the carry in ten’s

place of the sum cannot exceed 2. Hence B cannot

be more than 2. Thus you observe that B = 1 or 2.

The addition of digits in ten’s place gives(along with

carry 1 from unit’s place) A+B + C + 1 = 10 + C + 1

and this must leave remainder A when divided by

10.If B = 1, then A = 9 and hence C+1 = 9 giving C = 8. We get 99+11+88 =

198, which is a correct answer. If B = 2, you get A = 8 and C + 1 = 8 giving

C = 7. But then 88 + 22 + 77 = 187. This does not fit in as hundred’s place

in the sum is 1 but not 2.

The correct answer is 99 + 11 + 88 = 198.

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8 Unit 1

Exercise 1.1.31. In the following, find the digits represented by the letters:

(i)

3

+ B

7

(ii)

1 6

+ 2 A

B 1

(iii)

2 A

× A

12A

(iv)1 A A

+ 1 A A

2 A A

(v)1 A

× 1 A

1 B A

(vi)3 A

× A

2 B A

2. In the adjacent sum, A,B,C are consecutive dig-

its. In the third row, A,B,C appear in some order.

Find A,B,C.

A B C

+ C B A

+ – – –

1 2 4 2

3. Find A, B, C in the adjacent multiplication table.A B C

× A A

A C 6 C

4. Is it possible to have the adjacent multiplication

table? Give reasons.

A A

× B B

C C C

1.1.4. Divisibility and remainders

One of the important property related to integers is the concept of di-

visibility. In your earlier class, you have studied how to divide a natural

number by another natural number to get a quotient and a remainder.

If you divide 91 by 13, you see that 13 completely divides 91 and you do

not get any remainder. On the other hand, dividing 85 by 15, you see that

15× 5 = 75 and 15× 6 = 90, so that you cannot divide 85 completely by 15.

On actual division, you get 5 as quotient and 10 as remainder.

13)

91(

7

91

—–

00

15)

85(

5

75

—–

10

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Suppose you divide 304 by 12. The quotient is 25 and the remainder is

4. If you divide 887 by 17, you obtain the quotient 52 and the remainder

3.

12)

304(

25

24

——-

64

60

——-

4

17)

887(

52

85

——-

37

34

——-

3

We write these divisions in the following form:

91 = (13× 7) + 0,

85 = (15× 5) + 10,

304 = (12× 25) + 4,

887 = (17× 52) + 3.

Do you observe that 0 < 13, 10 < 15, 4 < 12, 3 < 17? Can you conclude

that the remainder does not exceed the number from which you divide?

Activity 4: Find the quotient and the remainder in each of the following

cases:

(i) 100 divided by 2,3,5,7,11,13,17,23, 29 and 31.

(ii) 300 divided by 37, 41,43,47,53,59,61,67.

Our observation can be put in a formal way:

Given a non-negative integer a and another integer b > 0, there

exist unique integers q and r such that a = (b × q) + r, where 0 ≤ r < b.

We say b divides a if the remainder is zero, that is, r = 0.

A similar statement can be made when a number is divided by another

non-zero number. You will learn more about these in higher classes. The

above statement can be used as a basis for a nice game which you can

play with your friends.

Game 4.

Ask your friend to choose a number smaller than 1000. Tell him to divide

this number by 7, 11, 13 respectively and ask for the remainders obtained

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10 Unit 1

by these divisions. Using these remainders, you can construct the number

chosen by your friend.

Suppose your friend has chosen 128. Then the remainder when divided

by 7 is 2; the remainder when divided by 11 is 7 and the remainder when

divided by 13 is 11. Now form the sum

(2× 715) + (7× 364) + (11× 924).

If you simplify this, you get 14142. Divide this by 1001. You see that

14142 = (1001× 14) + 128,

so that the remainder is 128. This is the number chosen by your friend.

Are you not thrilled?

These are the steps in this game.

Step 1. Tell your friend to choose a number less than 1000, in his

mind.

Step 2. Tell him to divide the number by 7, 11, 13 and ask him to

give you three remainders.

Step 3. Now you construct the number he thought of using the three

remainders as follows. Suppose the remainders he gives you are

r1(remainder when divided by 7), r2(remainder when divided by 11),

and r3(remainder when divided by 13); multiply r1 by 715, r2 by 364

and r3 by 924; take care that you are doing the correct multiplication.

Add all three numbers so obtained and divide the resulting number

by 1001. The remainder you obtain is the number chosen by your

friend.

Take another example, say 212. Observe that

212 = (7× 30) + 2; 212 = (11× 19) + 3; 212 = (13× 16) + 4.

Thus r1 = 2, r2 = 3 and r3 = 4. We obtain

(r1 × 715) + (r2 × 364) + (r3 × 924) = (2× 715) + (3× 364) + (4× 924) = 6218.

Divide 6218 by 1001. The remainder is 212, the number you started with.

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You may be wondering how such a game works. Suppose you start with an

arbitrary number a < 1000. Let r1, r2, r3 be the remainders when a is divided

by 7, 11, 13 respectively. Then you can write

a = 7q1 + r1, a = 11q2 + r2, a = 13q3 + r3,

for some integers q1, q2, q3. This shows that

r1 = a− 7q1, r2 = a− 11q2, r3 = a− 13q3.

Hence

715r1 + 364r2 + 924r3 = 715(a − 7q1) + 364(a − 11q2) + 924(a − 13q3)

= a(715 + 364 + 924) − (7 × 715)q1 − (11 × 364)q2 − (13 × 924)q3.

However you may notice that

7× 715 = 7× 11× 13× 5,

11× 364 = 11× 7× 13× 4,

13× 924 = 13× 7× 11× 12.

And 1001 = 7× 11× 13. Do you now understand why we consider the remain-

ders when divided by 7, 11 and 13? Hence you get

715r1 + 364r2 + 924r3 = a× 2003 − 1001(5q1 + 4q2 + 12q3).

You may also observe that a × 2003 = (a × 1001 × 2) + a. When you divide

715r1+364r2+924r3 by 1001, you are left with a as all other terms are divisible

by 1001. Since a < 1000, a is indeed the remainder. But that is the number

you have started with.

Activity 5:

Check the game 4 with some more numbers: 804, 515, 676, 938, 97, 181.

Exercise 1.1.41. Find the quotient and the remainder when each of the following num-

ber is divided by 13:

8,31, 44, 85, 1220, 2011.

2. Find the quotient and the remainder when each of the following num-

ber is divided by 304:

128, 636,785,1038,2236,8858,13765,58876.

3. Find the least natural number larger than 100 which leaves the re-

mainder 12 when divided by 19.

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4. What is the least natural number you have to add to 1024 to get a

multiple of 181?

5. What is the smallest number you have to add to 100000 to get a

multiple of 1234?

1.1.5 Magic squares

Can you arrange the numbers from 1 to 9 in 3 rows and 3 columns

such that the sum of the numbers in each row, each column and each

diagonal are all identical? Look at the following arrangement.(Fig. 1)

1

5

8 6

73

4 9 2

Fig. 1

6 1 8

3

49

57

2

Fig. 2

You observe that the sum of the numbers in each row is 15, the column

sum is 15 and the diagonal sum is 15. You can also arrange the numbers

as in Fig. 2. Can you see that there is a resemblance between these two

magic squares? The middle column is (1, 5, 9) in both the squares. The

right most column (8, 3, 4) in the first square is the left most column in

the second. Similarly, the left most column (6, 7, 2) in the first square is

the right most column in the second. Thus the second magic square is

obtained by flipping the right most column and the left most column. The

sum 15 is called the magic sum.

Is there a way of constructing such a magic square? Start with the

middle cell of the topmost row and put 1 there. Thus we begin with the

following:

1

Now we follow the following rules:

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Rule 1. If there is a vacant cell

along the diagonal from left to

right, you fill it with the next num-

ber. (Here there is a vacant cell

after 4 along the diagonal. Hence

we fill it with 5.)

4

−→ 5

4

Rule 2. If there is no vacant cell

along the diagonal and if there are

columns further, you fill the bot-

tom cell of the next column with

1

−→1

2

the next number. Follow Rule 1. (Here there is no vacant cell along

the diagonal after 1. Hence we go to the bottom most cell in the next

column and fill with 2.)

Rule 3. If there is no vacant cell

along the diagonal and if there are

no columns further, go to the row

above the cell you have reached

1

2

−→1

3

2

and fill the left most cell of this row with the next number and follow

Rule 1. (Here there is no vacant cell along the diagonal to move from

2 and we are already in the last column. Hence we move to the row

above and fill the left most cell with 3.)

Rule 4. If at any stage you en-

counter a cell which has already

been filled, go to the cell below the

cell you have reached and then

1

3 −→1 6

3 5

4

continue with Rule 1. (Here you cannot move from 3 to the next cell

along the diagonal since you have 1 already there. Hence we go to the

cell below 3 and fill with this 4.)

Rule 5. If you are at the end of

the main diagonal, fill the number

below the last cell in the diagonal

with the next number. And follow

1 6

3 5

4

−→1 6

3 5 7

4

the appropriate rule further. (Here you have 6 at the end of the main

diagonal. Hence we go to the cell below it and fill it with 7.)

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14 Unit 1

Let us see how it works for a 3 × 3 magic square. We start with the

central cell of the first row and put 1 there. Now apply Rule 2, as we

cannot move diagonally. We fill the bottom cell of the next column, that

is column 3, with number 2. Again, we cannot move diagonally, nor do

we have column further. We apply rule 3 and move to the row above and

fill the left most cell with 3. Now you see that you cannot move along the

diagonal as the cell there is already filled. Hence we use Rule 4 and go to

the cell below the cell we are in. We fill this with 4 and move diagonally

to fill 5 and 6. Again, we cannot move further and we are on the main

diagonal. We use Rule 5 and fill the cell below the last cell on the diagonal

with 7. Now we use Rule 3 and fill the left most cell of the row above with

8. Using Rule 2, we now fill the bottom most cell of the next column with

9. And you have the magic square!

The sequence of operations are shown below

1

→ Rule 2 →1

2

→ Rule 3 →1

3

2

→ Rule 4 →

1

3

4 2

→ Rule 1 →1

3 5

4 2

→ Rule 1 →1 6

3 5

4 2

→ Rule 5 →

1 6

3 5 7

4 2

→ Rule 3 →8 1 6

3 5 7

4 2

→ Rule 2 →8 1 6

3 5 7

4 9 2

Activity 6: Using the central cell of the first column as the starting point,

construct a 3 × 3 magic square with numbers from 1 to 9. How does this

compare with the magic square in Fig. 1? What relation is there between

the magic sum and the number in the central cell of the magic square?

Using the numbers from 3 to 11, construct a magic square.

Solution: We use the same sequence of operations as we have used earlier,

but the starting number is 3 instead of 1.

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3

→ Rule 2 →3

4

→ Rule 3 →3

5

4

→ Rule 4 →

3

5

6 4

→ Rule 1 →3

5 7

6 4

→ Rule 1 →3 8

5 7

6 4

→ Rule 5 →

3 8

5 7 9

6 4

→ Rule 3 →10 3 8

5 7 9

6 4

→ Rule 2 →10 3 8

5 7 9

6 11 4

Here the magic sum is 21.

Activity 7: Construct a 5×5 magic square using the above rules and with

numbers from 1 to 25. What relation is there between the magic sum and

the number in the central cell of the magic square?

Using the five rules, it is possible to construct an m ×m magic

square with the numbers from 1 to m2, for any odd natural num-

ber m > 1.

Exercise 1.1.5

1. Using the numbers from 5 to 13, construct a 3×3 magic square. What

is the magic sum here? What relation is there between the magic sum

and the number in the central cell?

2. Using the numbers from 9 to 17, construct a 3×3 magic square. What

is the magic sum here? What relation is there between the magic sum

and the number in the central cell?

3. Starting with the middle cell in the bottom row of the square and

using numbers from 1 to 9, construct a 3× 3 magic square.

4. Construct a 3× 3 magic square using all odd numbers from 1 to 17.

5. Construct a 5× 5 magic square using all even numbers from 1 to 50.

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1.1.6 Divisibility tests

If a number ends with any of the digits 0, 2,4,6 or 8, you immediately

say the number is divisible by 2. What is your reasoning? You write any

such number a as a = 10k+r, where r is the remainder when divided by 10.

Hence r is one of the numbers 0,2,4,6,8. You now see that 10 is divisible

by 2 and r is also divisible by 2. You conclude that 2 divides a.

It is natural to think whether such simple tests are available for divisi-

bility by other numbers. We explore some of them here.

1. Divisibility by 4

If a number is divisible by 4, it has to be divisible by 2(why?). Hence the

digit in the unit’s place must be one of 0,2,4,6,8. But look at the following

numbers: 10,22,34,46, 58. You see the last digit in each of these numbers

are as required, yet none of them is divisible by 4? Thus you may conclude

that it is not possible to decide the divisibility on just reading the last digit.

Perhaps, the last two digits may help.

If a number has two digits, you may decide the divisibility by actually

dividing it by 4. All you need is to remember the multiplication table for 4.

Suppose the given number is large, say it has more then 2 digits. Consider

the numbers, for example, 112 and 122. You see that 112 = 100 + 12, and

both 100 and 12 are divisible by 4. You may conclude that 112 is divisible

by 4. But 122 = 100 + 22; here 100 is divisible by 4, but 22 is not. Hence

122 is not divisible by 4. We invoke the following fundamental principle on

divisibility:

If a and b are integers which are divisible by an integer m 6= 0,

then m divides a + b, a− b and ab.

How does this help us to decide the divisibility of a large number by 4?

Suppose you have a number a with more than 2 digits. Divide this by 100

to get a quotient q and remainder r: a = 100q + r, where 0 ≤ r < 100. Since

4 divides 100, you will immediately see that a is divisible by 4 if and only

if r is divisible by 4. But r is the number formed by the last two digits of

a. Thus you may arrive at the following test:

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Statement 2

A number a (having more than one digit) is divisible by 4 if and only

if the 2-digit number formed by the last two digits of a is divisible by

4.

Example 4. Check whether 12456 is divisible by 4

Solution: Here, the number formed by the last two digits is 56. This is

divisible by 4 and hence so is 12456.

Example 5. Is the number 12345678 divisible by 4?

Solution: The number formed by the last 2 digits is 78, which is not

divisible by 4. Hence the given number is not divisible by 4.

Activity 8:

Ask your friend sitting adjacent to you to give several 4, 5 and 6 digit

numbers. Test divisibility by 4 for them.

Activity 9:

By dividing several 4 and 5 digit numbers by 8, formulate a divisibility test

by 8.

2. Divisibility by 3 and 9

Consider the numbers 2,23,234,2345, 23456, 234567. We observe that

among these 6 numbers, only 234 and 234567 are divisible by 3. Here,

we cannot think of the number formed by the last two digits, or for that

matter even three digits. Note that 3 divides 234, but it does not divide

34. Similarly, 3 divides 456, but 3 does not divide 23456.

Activity 10: Write down numbers 1, 11, 21, 31, 41, . . . , 141, 151. (All numbers

from 1 to 151 which ends in 1.) Form the sum of the digits of each number

and tabulate them. Check which numbers are divisible by 3 and whether

the digital sum of that number is also divisible by 3? What do you observe?

Consider the numbers 234 and 234567. The sum of the digits of the

first number is 9 and that of the second is 27. You see that both 9 and 27

are divisible by 3. (In fact, divisible by 9.) Let us explore this in a general

case for a 2-digit, 3-digit and 4-digit numbers. If n = ab is 2-digit number,

thenn = ab = (10× a) + b = 9a+ (a + b).

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This shows that n is divisible by 3 if and only if a + b is divisible by 3.

Similarly for m = pqr, we have

m = pqr = 100p+ 10q + r = (99p+ 9q) + (p+ q + r),

and you may observe that m is divisible by 3 if and only if p + q + r is

divisible by 3. Do you observe that you can deduce divisibility test for 9 as

well: 9 divides m if and only if 9 divides p+ q+ r? Now you do not have any

problem to extend the test for a 4-digit number or a number with more

digits. Observe that the sum of the digits of 234567 is 27. You can check

that 9 divides 234567.

Statement 3

An integer a is divisible by 3 if and only if the sum of the digits of a

is divisible by 3. An integer b is divisible by 9 if and only if the sum

of the digits of b is divisible by 9.

Example 6. Check whether the number 12345321 is divisible by 3. Is it

divisible by 9?

Solution: The sum of the digits is 1+2+3+4+5+3+2+1 = 21. Hence the

number is divisible by 3, but not by 9. In fact 12345321 = (9× 1371702) + 3.

Example 7. Is 444445 divisible by 3?

Solution: The sum of the digits is 25, which is not divisible by 3. Hence

444445 is not divisible by 3. Here the remainder is 1.

3. Divisibility by 5 and 10

Activity 9: Take all multiples of 5 from 51 to 100. Tabulate the last digits

of each multiple of 5.

Do you see that 0 or 5 appears as the digit in the unit’s place for every

multiple of 5? Does this observation help you to formulate a divisibility

test for 5 and also for 10?

Statement 4

An integer a is divisible by 5 if and only if it ends with 0 or 5. A

number is divisible by 10 if and only if ends with 0.

Example 8. How many numbers from 101 to 200 are divisible by 5?

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Solution: Write down all numbers from 101 to 200 which end with 0 or

5: 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170,

175, 180, 185, 190, 195, 200. There are 20 such numbers.


Solution: Note that 15 = 3 × 5. Hence the given number must be divisible

by both 3 and 5.(This is also sufficient to prove that the given number is

divisible by 15. However, a general rule is false. For example, 4 divides

12 and 6 divides 12, but their product 24 does not divide 12. Can you

formulate some rule?) It is obviously divisible by 5, as its last digit is 5.

The sum of the digits is 1 + 2 + 3 + 4 + 5 = 15 and it is divisible by 3. Hence

3 also divides 12345. We conclude that 15 divides 12345.

Example 10. How many numbers from 201 to 250 are divisible by 5, but

not by 3?

Solution: Here again, the numbers divisible by 5 are 205, 210, 215, 220,

225, 230, 235, 240, 245, 250. Now you compute the digital sum of these

numbers: you get 7,3,8,4,9,5,10,6,11,7. Among these, the only numbers

divisible by 3 are 3,6,9. Thus among the 10 numbers divisible by 5, only

three numbers are also divisible by 3. The remaining 7 numbers are not

divisible by 3.

3. Divisibility by 11

Consider the number 4587. You may check that it is divisible by 11.

(In fact, 4587 = 11× 417.) We may also write

4587 = (4× 1000) + (5× 100) + (8× 10) + 7

= (4× 1001) + (5× 99) + (8× 11) + (−4 + 5− 8 + 7)

= (11× 91× 4) + (11× 9× 5) + (11× 8)− (4− 5 + 8− 7).

Each of the numbers in the first three brackets is divisible by 11. Hence

the divisibility of 4587 by 11 is now related to the divisibility of 4−5+8−7,

which involves only the digits of the given number. The important point to

note here is that the sign alternates with + and −. We also observe that

4− 5 + 8− 7 = 0, which is divisible by 11.

Consider a 3 digit number 429. You may easily check hat 429 is divisi-

ble by 11: 429 = 11× 39. On the other hand,

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429 = (4× 100) + (2× 10) + 9

= (4× 99) + (2× 11) + (4− 2 + 9).

Since 4 − 2 + 9 = 11 is divisible by 11, you may conclude that 11 divides

429, without actually dividing it by 11.

How do you test a general 3-digit or 4-digit number? Suppose n = abc

is a 3-digit number. Then

n = 100a+ 10b+ c

= 99a+ 11b+ (a− b+ c).

Thus n is divisible by 11 if and only if 11 divides a− b + c. If m = pqrs is a

4-digit number, then

n = 1000p+ 100q + 10r + s

= 1001p+ 99q + 11r − (p− q + r − s).

Hence 11 divides n if and only if 11 divides p − q + r − s. This may be

extended to numbers with any number of digits, with due care.

Statement 5

Given a number n in decimal form, put alternatively − and + signs

between the digits and compute the sum. The number is divisible

by 11 if and only if this sum is divisible by 11. Thus a number

is divisible by 11 if and only if the difference between the sum of

the digits in odd places and the sum of the digits in even places is

divisible by 11.


Solution: Observe that 2−3+4−5+6 = 4 and hence not divisible by 11. The

test indicates that 23456 is not divisible by 11. In fact 23456 = (11×2123)+4.

A palindrome is a number which reads the same from left to right or

right to left. Thus a palindrome is a number n such that by reversing the

digits of n, you get back n. For example 232 is a 3-digit palindrome; 5445 is

a 4-digit palindrome.

Example 12. Find all 3-digit palindromes which are divisible by 11.

Solution: A 3-digit palindrome must be of the form aba, where a 6= 0 and

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b are digits. This is divisible by 11 if and only if 2a − b is divisible by 11.

This is possible only if 2a− b = 0 or 2a− b = 11 or 2a− b = −11. Since a ≥ 1

and b ≤ 9, we see that 2a− b ≥ 2 − 9 = −7 > −11. Hence 2a− b = −11 is not

possible. Suppose 2a − b = 0. Then 2a = b; thus a = 1, b = 2; a = 2, b = 4;

a = 3, b = 6; and a = 4, b = 8 are possible. We get the numbers 121, 242, 363,

484. For a = 6, b = 1, we see that 2a− b = 12− 1 = 11, and hence divisible by

11. Similarly a = 7, b = 3; a = 8, b = 5; and a = 9, b = 7 give the combinations

for which 2a− b is divisible by 11. We get four more numbers: 616, 737, 858,

and 979.

Thus the required numbers are: 121, 242, 363, 484, 616, 737, 858, 979.

Example 13. Prove that 12456 is divisible by 36 without actually dividing

it.

Solution: First notice that 36 = 4 × 9. Hence it is enough to prove that

the given number is separately divisible by 4 and 9(then it will be divisible

by their LCM which is 36). Consider the number formed by the last two

digits: 56. It is divisible by 4. Hence 12456 is divisible by 4. On the

other hand the sum of the digits is 18 and divisible by 9. Hence 12456 is

divisible by 9 as well. Combining, we get the result.

Exercise 1.1.6

1. How many numbers from 1001 to 2000 are divisible by 4?

2. Suppose a 3-digit number abc is divisible by 3. Prove that abc+bca+cab

is divisible by 9.

3. If 4a3b is divisible by 11, find all possible values of a + b.

4. Prove that a 4-digit palindrome is always divisible by 11.

5. Using the digits 4,5,6,7,8, each once, construct a 5-digit number

which is divisible by 132.

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22 Unit 1

You have seen the first few primes are:

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73. . .

There are infinitely many primes. Among these look at the pairs:

(3,5), (5,7), (11,13), (17,19), (29,31), (41,43), (59,61), (71,73).

These are primes which differ by 2. That is: the difference between them is

2. Such a pair of primes is called a twin prime. It is an outstanding unsolved

problem: Are there infinitely many twin primes?

Another classical unsolved problem involving numbers is Goldbach’s conjecture.

It dates back to 1742. Consider even numbers greater than 2: observe 4 = 2+2,

6 = 3 + 3, 8 = 3 + 5, 10 = 3 + 7, 12 = 5 + 7, 14 = 3 + 11, 16 = 5 + 11, and so on. It

was conjectured by a German mathematician Goldbach that every even number

greater than 2 is a sum of two primes. It has been extensively verified using

computers and there is a strong suspicion that the conjecture is true. But no

mathematical proof is available till today.

A natural number n is called a perfect number if it is equal to the sum of all

its proper positive divisors.(That is the sum of all its positive divisors excluding

the number itself.) The first perfect number is 6. It has three proper positive

divisors, 1,2,3; and 1 + 2 + 3 = 6. Similarly, 28 has proper divisors, 1,2,4,7,14;

and 1 + 2 + 4 + 7 + 14 = 28. The next numbers are 496 and 8128. All these were

discovered by Euclid. Euclid also proved that 2p−1(2p − 1) , with p a prime, is a

perfect number whenever 2p − 1 is a prime. Prime numbers of the form 2p − 1

are called Mersenne primes(after a seventh-century mathematician called Marin

Mersenne). It is easy to prove that 2n − 1 is prime only if n is prime. But not all

numbers of the form 2p − 1, where p a prime, are prime numbers. For example,

211 − 1 = 2047 = 23 × 89 and hence not a prime. Using this idea and computers,

one can generate some perfect numbers. The numbers 2p−1(2p − 1), where p is

equal to

2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281

are the first few perfect numbers. There are some unsolved problems related to

perfect numbers.

1. Are there infinitely many perfect numbers?(Equivalently, are there in-

finitely many Mersenne primes?)

2. All the perfect numbers so far known are even perfect numbers. This nat-

urally raises the question: Is there any odd perfect number?

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Additional Problems on “Playing with numbers”

1. Choose the correct option:

(a) The general form of 456 is

A. (4× 100) + (5× 10) + (6× 1) B. (4× 100) + (6× 10) + (5× 1)

C. (5× 100) + (4× 10) + (6× 1) D. (6× 100) + (5× 10) + (4× 1)

(b) Computers use

A. decimal system B. binary system C. base 5 system D. base

6 system

(c) If abc is a 3-digit number, then the number

n = abc + acb+ bac + bca + cab+ cba

is always divisible by

A. 8 B. 7 C. 6 D. 5

(d) If abc is a 3-digit number, then

n = abc− acb+ bac− bca + cab− cba

is always divisible by

A. 12 B. 15 C. 18 D. 21

(e) If 1K ×K1 = K2K, the letter K stands for the digit

A. 1 B. 2 C. 3 D. 4

(f) The numbers 345111 is divisible by

A. 15 B. 12 C. 9 D. 3

(g) The number of integers of the form 3AB4, where A,B denote some

digits, which are divisible by 11 is

A. 0 B. 4 C. 7 D. 9

2. What is the smallest 5-digit number divisible by 11 and containing

each of the digits 2,3,4,5,6?

3. How many 5-digit numbers divisible by 11 are there containing each

of the digits 2,3,4,5,6?

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24 Unit 1

4. If 49A and A49, where A > 0, have a common factor, find all possible

values of A.

5. Write 1 to 10 using 3 and 5, each at least once, and using addition

and subtraction. (For example, 7 = 5 + 5− 3.)

6. Find all 2-digit numbers each of which is divisible by the sum of its

digits.

7. The page numbers of a book written in a row gives a 216 digit number.

How many pages are there in the book?

8. Look at the following pattern:

1

1 1

11

1 1

2

3 3

1 14 46

This is called Pascal’s triangle. What is the middle number in the 9-th

row?

9. Complete the adjoining magic

square.(Hint: In a 3×3 magic square,

the magic sum is three times the

central number.)

8

73

10. Find all 3-digit natural numbers which are 12 times as large as the

sum of their digits.

11. Find all digits x, y such that 34x5y is divisible by 36.

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12. Can you divide the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 into two groups such

that the product of numbers in one group divides the product of num-

bers in the other group and the quotient is minimum?

13. Find all 8-digit numbers 273A49B5 which are divisible by 11 as well

as 25.

14. Suppose a, b are integers such that 2+ a and 35− b are divisible by 11.

Prove that a+ b is divisible by 11.

15. A list of numbers and corresponding

codes are given in the adjacent table.

Find the number L.

589 724 1306 L

524 386 9701 2011

16. In the multiplication table A8× 3B = 2730, A and B represent distinct

digits different from 0. Find A+B.

17. Find the least natural number which leaves the remainders 6 and 8

when divided by 7 and 9 respectively.

18. Prove that the sum of cubes of three consecutive natural numbers is

always divisible by 3.

Glossary

Divisibility: an integer a is said to be divisible by another non-zero integer

b, if a = qb for some integer q.

Quotient: if a = qb, for some integers a, b 6= 0 and q, then q is the quotient

upon division of a by b.

Remainder: if a = bq + r, where 0 ≤ r < b for some integer a and natural

number b, then r is the remainder upon division of a by b.

Palindrome: a number which reads the same from left to right and right

to left.

Puzzle: any mind twister.

Alpha numeral: a letter appearing in an equation and taking some nu-

merical value.

Conjecture: a statement which is believed to be true, but without any

substantiating mathematical proof.

Magic square: a square consisting of smaller squares and each smaller

square filled with numbers such that the row sum, the column sum and

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26 Unit 1

the diagonal sum are all equal.

Perfect number: a natural number whose all positive divisors smaller

than the number add up to the number.

Twin-primes: pairs of prime numbers such that the difference of numbers

in each pair is 2.

Mersenne primes: prime numbers of the form 2p − 1. (If 2p − 1 is a prime,

it forces p is also a prime.)

Points to remember

• Any natural number can be written in the generalised form using base

10.

• Given any two integers a and b > 0, there exist unique integers q and

r such that a = bq + r, where 0 ≤ r < b.

• A number is divisible by 4 if and only if the number formed by the

last two digits is divisible by 4.

• A number is divisible by 3 or 9 if and only if the sum of the digits is

divisible by 3 or 9 respectively.

• A number is divisible by 5 if and only if it ends in 0 or 5.

• A number is divisible by 11 if and only if the difference between the

sum of oddly placed digits and the sum of evenly placed digits is

divisible by 11.

We must think and act like a nation of billion people and not like that

of a million people. Dream, dream, dream! —–Abdul Kalam

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CHAPTER 1 UNIT 2

SQUARES, SQUARE ROOTS, CUBES, CUBE ROOTS

After studying this unit, you learn

• perfect squares and square root of perfect squares;

• to recognise the digits in unit’s place of a perfect square;

• to obtain the remainders when a perfect square is divided by 3 and 4;

• different occasions leading to perfect squares;

• some methods of finding perfect squares and square-roots of perfect

squares;

• perfect cubes and cube-roots of perfect cubes;

1.2.1 IntroductionLook at the numbers of the form 1, 4, 9, 16, 25 and so on. What do

you recognise? For example, 4 = 2 × 2, 25 = 5 × 5. You see that each

such number is the product of two equal numbers. Similarly, you see

that 8 = 2 × 2 × 2, 64 = 4 × 4 × 4. Each number is the product of three

equal numbers. These numbers are given special names. In this chapter

we study some properties of such numbers. We also study the reverse

process: whenever a number is the product of two equal numbers or three

equal numbers, can we find these equal numbers?

1.2.2 Perfect squaresLook at the following diagrams.

How many dots do you find in each figure? You recognise them as 1, 4,

9, 16.

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28 Unit 2

B

C

4321 5 6 8 9 107

D

A

2

3

4

6

7

8

9

10

5

1

Suppose you have a square ABCD

of side-length 10 units. Divide

the square in to smaller unit

squares(as in the adjoining figure)

using lines parallel to the sides.

Can you count that there are 100

unit squares?

Activity 1: Repeat this with squares of side-length 8, 12, 15 units and

tabulate your findings.

Observe that 1 = 1× 1, 4 = 2× 2, 9 = 3× 3, 16 = 4× 4, 100 = 10× 10.

If a is an integer and b = a× a, we say b is a perfect square.

Hence 1, 4, 9, 16, 100 are all perfect squares. Since 0 = 0 × 0, we see that

0 is also a perfect square.

If a is an integer, we denote a × a by a2(we read this as Square of a or

simply a-square). Thus 36 = 62, 81 = 92. Thus a perfect square is of the

form m2, where m is an integer.

Here you may be observing some thing more. For example 4 = 2 × 2

and 4 = (−2) × (−2); in the second representation, you again have equal

integers, but negative this time. There is nothing strange about this. The

property is inherent in the number system: if you multiply two negative

integers, you get a positive integer. Thus for any natural number m, we

get m2 = m×m = (−m)× (−m). This also tells something about the nature

of perfect squares. If m is a natural number, m2 = m ×m is also a natural

number and hence m2 is positive. If m = 0, then m2 = 0 × 0 = 0. If m

is a negative integer, then m = −n for some natural number n. Hence

m2 = (−n)× (−n) = n2, which again is positive.

Thus a perfect square is either equal to 0 or must be a positive

integer. It can never be a negative integer.

We have seen that 1 = 12 and 4 = 22 are perfect squares. Can 2 and

3 also be written as a product of two equal integers? Your intuition tells

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Squares and cubes 29

that this cannot happen. Can we see this mathematically? Suppose m

and n are two natural numbers such that m < n. Then it is easy to see

that m2 < n2(why?).

If at all 2 is a perfect square, 2 = n2 for some natural number. Then

1 < 2 < 4 gives 1 = 12 < n2 < 22 = 4. This forces 1 < n < 2(why?). Thus n

is a natural number strictly between 1 and 2. But we know that given a

natural number k, there is no natural number between k and its successor

k + 1. Thus no natural number exists between 1 and 2. We conclude that

2 is not a perfect square.

Similarly, you can conclude that 3 is not a perfect square. This may be

extended to prove that any natural number n, such that m2 < n < (m+1)2,

cannot be a perfect square.

Look at the following table:

a 1 2 3 8 −7 −12 20 −15a2 1 4 9 64 49 144 400 225

Do you see that squares of 2,8,-12, 20 are even numbers, where as the

squares of 1,3, −7, −15 are all odd numbers. What do you infer?

Statement 1. The square of an even integer is even and the square

of an odd integer is odd.

This is not hard to prove. If m is even, then m = 2n for some integer n

and m2 = (2n)× (2n) = 4n2 is an even integer. If m is odd, then m = 2k + 1

for some integer k, so that

m2 = (2k + 1)(2k + 1)

=((2k + 1)× (2k)

)+ (2k + 1)× 1

= (2k)× (2k) + (1× 2k) + (2k × 1) + (1× 1)

= 4k2 + 2k + 2k + 1

= 4k2 + 4k + 1,

which is an odd number.

Consider the first ten perfect squares as in the table.

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30 Unit 2

12 22 32 42 52 62 72 82 92 102

1 4 9 16 25 36 49 64 81 100

If you observe the unit’s place in these squares, you see that they are

1,4,9,6,5,6,9,4,1,0 in that order. Thus the only digits which can occupy

the unit’s place in a perfect square are 0,1,4,5,6,9. If you take any num-

ber, its unit’s digit is one of 0,1,2,3,4,5,6,7,8,9. Hence the square of that

number(multiply the number with it self in your mind to get the digit in

unit’s place) ends with one of the numbers 0,1,4,5,6,9. Can you see that

the digits 2,3,7,8 can never occur as the last digit of a perfect square? We

can make a formal statement:

Statement 2. A perfect square always ends in one of the digits

0,1,4,5,6,9. If the last digit of a number is 2,3,7 or 8, it cannot be a

perfect square.

Think it over ! If a number ends in 0,1,4,5,6 or 9, then it is not

necessary that the number is a perfect square.

In precise mathematical language, we say that a necessary condition

for the given number to be a perfect square is that it should end with one

of the digits 0,1,4,5,6 or 9, but this condition is not a sufficient condition

to ensure that the given number is a perfect square. This helps us to

recognise perfect squares.

Exercise 1.2.2

1. Express the following statements mathematically:

(i) square of 4 is 16; (ii) square of 8 is 64; (iii) square of 15 is 225.

2. Identify the perfect squares among the following numbers:

1,2,3,8,36,49,65,67,71,81,169,625,125,900,100,1000,100000.

3. Make a list of all perfect squares from 1 to 500.

4. Write 3-digit numbers ending with 0,1,4,5,6,9, one for each digit, but

none of them is a perfect square.

5. Find numbers from 100 to 400 that end with 0,1,4,5,6 or 9, which

are perfect squares.

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1.2.3 Some facts related to perfect squares

There are some nice properties about perfect squares. We study them

here.

(a) Look at the following table:

a 4 10 20 25 100 300 1000

a2 16 100 400 625 10000 90000 1000000

The number of

zeros at the 0 2 2 0 4 4 6

end of a2

What do you observe? The number of zeros at the end of a square is

always an even number(it may be equal to 0, but still an even number).

Moreover the number of zeros at the end of each square is twice the num-

ber of zeros at the end of the number whose square is considered. Can

you now formulate this as a statement?

Statement 3. If a number has k zeros at the end, then its square

ends in 2k zeros.

Thus, if a number ends in odd number of zeros, it cannot be a perfect

square. This helps us to rule out certain numbers from the list of perfect

squares.

(b) Look at the adjoining table:

a a2 The remainder of a2 The remainder of a2

when divided by 3 when divided by 4

1 1 1 1

2 4 1 0

3 9 0 1

5 25 1 1

8 64 1 0

11 121 1 1

−6 36 0 0

Do you see that the remainder of a perfect square when divided by 3 is

either 0 or 1? Similarly, the remainder of a perfect square when divided by

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32 Unit 2

4 is either 0 or 1. When you divide a number by 3, the possible remainders

are 0, 1 and 2. But when you divide a perfect square by 3, the remainder

is either 0 or 1, but never 2. Similarly, when you divide a number by 4,

the possible remainders are 0,1,2 or 3. However, when you divide a perfect

square by 4, the remainder is either 0 or 1, but it can never be 2 and 3.

Statement 4. The remainder of a perfect square, when divided by

3, is either 0 or 1, but never 2. The remainder of a perfect square,

when divided by 4, is either 0 or 1, but never 2 and 3.

Think it over !

The remainder of a perfect square, when divided by 8, is either

0 or 1 or 4. It can never be equal to 2,3,5,6,7.

(c)

Activity 2:

Take any four consecutive natural numbers and form their product. Add

1 to this product. Check whether it is a perfect square. Repeat this with

some more sets of four consecutive natural numbers. Do the same with

some sets of consecutive negative integers. What is your observation if one

of the four consecutive integers is 0 ?

For example:

(1× 2× 3× 4) + 1 = 24 + 1 = 25 = 52;

(8× 9× 10× 11) + 1 = 7920 + 1 = 7921 = 892.

Statement 5. When the product of four consecutive integers is

added to 1, the resulting number is a perfect square.

(d) Read the following table:

1 = 1 = 12,

1 + 3 = 4 = 22,

1 + 3 + 5 = 9 = 32,

1 + 3 + 5 + 7 = 16 = 42,

1 + 3 + 5 + 7 + 9 = 25 = 52.

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Continue this process for some more rounds, adding the next odd number

to the previous sum. You see that you go on getting perfect squares. A

careful observation also reveals some thing more. The sum of the first 4

odd natural numbers is 42; the sum of the first 5 odd natural numbers is

52. Check this with the sum of the first 8 and 12 odd natural numbers.

Can you formulate this as a new statement?

Statement 6. The sum of the first n odd natural numbers is equal

to n2, for every natural number n.

(e)

Activity 3:

Consider the numbers 11, 101, 1001, 10001 and compute their squares.

Do the same inserting some more zeros. Do you see some pattern?

For example:

112 = 121,

1012 = 10201,

10012 = 1002001,

and so on. You see that the middle number of each square is always 2;

on both sides of 2, zeros appear as digits; and the end digits are equal to

1. The number of zeros both sides of 2 are equal and equal to the number

of zeros in the original number. Thus we can formulate a statement as

follows.

Statement 7. Consider the number N = 1000 · · ·01, where zeros

appear k times. (For example, for k = 6, you get N = 10000001; there

are 6 zeros in the middle.) Then N2 = 1000 · · ·02000 · · ·01, where the

number of zeros on both sides of 2 is k.

(f) Look at the following patterns:

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34 Unit 2

To generate such a pattern, put a dot in row 1, put 2 dots in row 2,

put 3 dots in row 3, and so on, as shown in the figure. The dots are now

arranged in the shape of a triangle. Count the number of dots in each of

the triangle.(Single dot is considered as a degenerate triangle.) They are

1,3,6, 10, 15, 21, 28, 36 and so on. These numbers are called triangular

numbers (you know the reason). You can see how the triangular numbers

are formed. For n-th triangular number, you form a triangle of dots with

n rows and each row contains as many points as the index of that row. If

you want to find the 8-th triangular number, the number of points in the

8-th triangle is1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36.

Here are the first few triangular numbers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91.

Take any two consecutive triangular numbers and find their sum. For

example: 10 + 15 = 25 = 52; 28 + 36 = 64 = 82; 55 + 66 = 121 = 112; 36 + 45 =

81 = 92. You see that the sum of any two consecutive triangular numbers

is a perfect square. You can also observe some thing more. Note that 28

is 7-th triangular number and 36 is 8-th one; their sum is 82. Similarly

66 is 11-th triangular number and 78 is 12-th one; their sum is 144 = 122.

Verify this property for some more pairs.

Statement 8. The sum of n-th and (n + 1)-th triangular numbers is

(n+ 1)2.

Exercise 1.2.3

1. Find the sum 1 + 3 + 5 + · · ·+ 51(the sum of all odd numbers from 1 to

51) without actually adding them.

2. Express 144 as a sum of 12 odd numbers.

3. Find the 14-th and 15-th triangular numbers, and find their sum.

Verify the Statement 8 for this sum.

4. What are the remainders of a perfect square when divided by 5?

1.2.4 Methods for squaring a numberMany times, it is easy to find the square of a number without actually

multiplying the number with itself. Consider 422. We may write 42 = 40+2.

Thus

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422 = (40 + 2)(40 + 2)

= 402 + (40× 2) + (2× 40) + 22

= 402 + (2× 40× 2) + 22.

Here we have used the distributive property of integers. Now it is easy to

recognise 402 = 1600; 2×40×2 = 160; and 22 = 4. Hence 422 = 1600+160+4 =

1764. (You can compute 402, 2× 40× 2 and 22 in mind and add.)

Note. The basis for this method is the identity (a+ b)2 = a2+2ab+ b2,

which you will study later.

Activity 4:

Find 892, 682, 962 using the above method.

There is an easy way of computing the square of a number ending with

5. For example, consider 352. Take the digit in unit’s place, namely 5.

Put 25 (= 52) first. Remove the unit’s digit from the given number and

consider the number formed by the remaining digits, which is 3. Consider

the product of 3 and its next number 4; 3 × 4 = 12. Prefix 12 to 25 to get

1225. Check that 352 = 1225.

Take another example, say 1052. Here the number formed by the re-

maining digits after the removal of the digit in the unit’s place is 10 and

its successor is 11. Their product is 10×11 = 110. Now you may check that

1052 = 11025. You may formulate this as a statement.

Statement 9. If n = a1a2 · · · ak5(represented in base 10), then n2 is

equal to 25 prefixed by

(a1a2 · · · ak)× (a1a2 · · · ak + 1)

Exercise 1.2.4

1. Find the squares of:

(i) 31, (ii) 72; (iii) 37 ; (iv) 166.

2. Find the squares of: (i) 85; (ii) 115; (iii) 165.

3. Find the square of 1468 by writing this as 1465 + 3.

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36 Unit 2

1.2.5 Square roots

As you know, if the side-length of a square ABCD is l, then its area is

l2. Can you reverse the process? Given the area of a square, can we find

its side-length?

Suppose the area of a square is 16 cm2. To find its side-length, we write

l2 = 16 = 42 and conclude l = 4 cm. Here the square of a number is given

and we have to find the number. Do you see that we are moving in the

opposite direction?

Again consider the following perfect squares:

1 = 12, 4 = 22, 9 = 32, 16 = 42, 49 = 72, 81 = 92, 196 = 142.

In each case the number is obtained by product of two equal numbers.

Here we say 1 is square root of 1; 2 is square root of 4; 7 is square root of

49 and so on.

Suppose N is a natural number such that N = M2. The number

M is called a square root of N.

We have seen earlier m2 = m ×m = (−m) × (−m) = (−m)2. Thus m2 has

two square roots: m and −m. Which one should be taken? For example

16 = 42 = (−4)2. Thus both 4 and −4 are square roots of 16. It is not clear,

which one of these should be taken. Many times, physical context clarify

the matter. As in the above example, if the area of a square is 16 units,

then its side-length is necessarily 4 units(−4 is not admissible as it cannot

be length). However, mathematically both 4 and −4 are acceptable as a

square root of 16. We make the following convention.

Whenever the word square root is used, it is always meant to

be the positive square root. The square root of N is denoted by√N.

Activity 5: Fill in the following blanks looking at the similarity of the

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statements:

12 = 1 =⇒√1 = 1;

22 = 4 =⇒√4 = 2;

52 = 25 =⇒√25 = −−−−;

112 = 121 =⇒ −−−− = 11;

−−−− = 225 =⇒ −−−−− = 15.

Activity 6: Fill in the blanks with appropriate word or number looking at

the similarity of the statements:

2 4 3 9 4 16

square square

square root square root

6

square

49

square

We have learnt earlier that the square of a non-zero integer is always a

positive integer. Hence square root is meaningful only for positive integers

or possibly 0 (whose square root is 0).

Square root of a perfect square by factorisation

We know that 3 =√9 and 4 =

√16. However 9 = 3×3 and 16 = 2×2×2×2 =

4 × 4. Thus we can factorise the given perfect square in terms of their

prime factors, combine these prime factors appropriately to write the given

perfect square as a product of two equal integers. This will help us to read

off the square root of the given perfect square.

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Example 1. Find the square root of 5929.

Solution: We do this in several steps.

Step 1. We express 5929 as a product of prime numbers:

7 5929

7 847

11 121

11

Thus 5929 = 7× 7× 11× 11.

Step 2. We arrange these prime factors suitably to write 5929 = (7× 11)×(7× 11) = 77× 77.

Step 3. Since 5929 = 77× 77, a product of two equal integers, we conclude√5929 = 77.

Example 2. Find the square root of 6724.

Solution: We observe that 6724 is even so that 2 is its prime factor.

Thus 6724 = 2 × 3362. Again 3362 is even so that 3362 = 2 × 1681. Thus

6724 = 2 × 2 × 1681. Now there is no easy way of finding prime factors of

1681. We must go on checking whether 1681 is divisible by the primes

in increasing order, starting from 3. We see that it is not divisible by

3,5,7,11,13,17,19,23,29,31,37, but it is divisible by 41 and 1681 = 41× 41.

Thus we obtain 6724 = 2 × 2 × 41 × 41 = (2 × 41) × (2 × 41) = 82 × 82. We

conclude:√6724 = 82.

There is no easy way of finding the prime factors of a given

number. There are some algorithms which can be used on com-

puters to find the prime factors of a given large number. How-

ever, these algorithms also use up lot of computer time. The

fact that there is no easy way of factorising a large number is

the basis for modern day security systems used in banks and

other financial institutions.

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Why some numbers are not perfect squares?

You might have noticed, while finding the square root of a perfect

square by factorising it, that each prime factor of the perfect square oc-

curs even number of times. For example, if you take 1296, you factorise

it as 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3. It has two distinct prime factors,

2 and 3. Now 2 occurs four times and 3 also occurs 4 times. This helps

you to write 1296 = (2 × 2 × 3 × 3) × (2 × 2 × 3 × 3) = 36 × 36, as a product

of two equal integers. You conclude that√1296 = 36. The success of this

method depends on the fact that the prime factors can be properly paired

to get the given number as a product of two equal integers. This is possible

because each prime factor occurs even number of times.

Activity 7:

Write all perfect squares between 1000 and 1500. In each case factorise

it as a product of prime numbers. Check that in every case, each of the

prime factor occurs even times in the product.

Now you can see why a number fails to be a perfect square. In its

prime factorisation, some primes may not occur even number of times.

Then there is no way of pairing the factors such that the given number

is equal to a product of two equal integers. However, you can make it a

perfect square on multiplication by a suitable factor or on division by a

suitable factor.

Suppose a number is not a perfect square. Take for example, 48. We

see that 48 = 2× 2× 2× 2× 3. Here 2 occurs four times, where as 3 occurs

only once. Hence we cannot properly pair the prime factors. However, if

we multiply 48 by 3, we see that

48× 3 = 2× 2× 2× 2× 3× 3 = (2× 2× 3)× (2× 2× 3) = 12× 12,

and we get a perfect square. Of course you may as well multiply 48 with

3× 2× 2 and get

48× 12 = (2× 2× 3× 2)× (2× 2× 3× 2) = 24× 24,

leading to a perfect square. In fact you can multiply 48 by 3k2 where k

is any positive integer and get as earlier 48 × 3k2 = (12k) × (12k), which is

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40 Unit 2

a perfect square. However 3 is the least number with which you have to

multiply 48 to get a perfect square.

Example 3. Find the least positive integer whose product with 9408 gives

a perfect square.

Solution: You can start with 2 and go on dividing by 2 till you get an odd

number: 9408 = 2× 2× 2× 2× 2× 2× 147. Now 147 = 3× 49 = 3× 7× 7. Thus

we get

9408 = 2× 2× 2× 2× 2× 2× 3× 7× 7.

Here both 2 and 7 occur even number of times, where as 3 occurs only

once. Hence we have to multiply 9408 by 3 to get a perfect square(you get

9408× 3 = (168)2).

Let us go back to 48 = 2 × 2 × 2 × 2 × 3. Instead of multiplying by 3, we

can as well divide by 3:

48

3=

2× 2× 2× 2× 3

3= 4× 4.

Still we get a perfect square. Here again you may divide by 3 × 2 × 2 and

get48

3× 2× 2=

2× 2× 2× 2× 3

3× 2× 2= 2× 2,

a perfect square. However 3 is the least number with which you have to

divide 48 to get a perfect square.

Example 4. Find the smallest positive integer with which one has to divide

336 to get a perfect square.

Solution: We observe that 336 = 2 × 2 × 2 × 2 × 3 × 7. Here both 3 and 7

occur only once. Hence we have to remove them to get a perfect square.

We divide 336 by 3× 7 = 21 and get

336

21= 16 = 42.

The least number required is 21.

Exercise 1.2.5

1. Find the square root of the following numbers by factorisation:

(i)196; (ii) 256; (iii) 10404; (iv) 1156; (v) 13225.

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2. Simplify:

(i)√100 +

√36; (ii)

√1360 + 9; (iii)

√2704 +

√144 +

√289; (iv)

√225−√

25; (v)√1764−

√1444; (vi)

√169×

√361.

3. A square yard has area 1764 m2. From a corner of this yard, an-

other square part of area 784 m2 is taken out for public utility. The

remaining portion is divided in to 5 equal square parts. What is the

perimeter of each of these equal parts?

4. Find the smallest positive integer with which one has to multiply each

of the following numbers to get a perfect square:

(i) 847; (ii) 450; (iii) 1445; (iv) 1352.

5. Find the largest perfect square factor of each of the following num-

bers:

(i) 48; (ii) 11280; (iii) 729; (iv) 1352.

6. Find a proper positive factor of 48 and a proper positive multiple of

48 which add up to a perfect square. Can you prove that there are

infinitely many such pairs?

1.2.6 Perfect squares near to a given number

Let us start with a non-perfect square, say, 72. Observe that 72 =

2 × 2 × 2 × 3 × 3, so that 2 appears only an odd number of times. We

can multiply 72 by 2 to get 72 × 2 = 144 = 122. Or we may divide 72

by 2 to get72

2= 36 = 62. But there are more perfect squares between

62 and 122, namely 72 = 49, 82 = 64, 92 = 81, 102 = 100 and 112 = 121.

Which one is the nearest to 72 among these perfect squares? You see that

82 = 64 < 72 < 81 = 92 and 72 − 64 = 8 < 9 = 81 − 72. Thus we see that 64 is

nearer to 72 than 81. Hence 64 is the nearest perfect square to 72.

In fact, given a non-perfect square, there is a unique perfect square

nearest to it. Suppose N is the given non-perfect square. You can put

it between two consecutive squares; there is a unique n such that n2 <

N < (n + 1)2.(Can you see why?). Since n and n + 1 are two consecutive

numbers, one of them is even and the other odd. Hence N cannot be

exactly in the middle of n2 and (n + 1)2; if N − n2 = (n + 1)2 − N , then

2N = n2+ (n+1)2 = n2+ n2+2n+1 = 2n2 +2n+1 which is impossible since

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42 Unit 2

2N is even and 2n2 + 2n + 1 is odd. Hence either n2 is the nearest perfect

square to N or (n + 1)2 is the nearest perfect square. If n2 is the nearest

perfect square to N , we say n approximates√N ; if (n + 1)2 happens to be

the nearest square to N , we say n + 1 approximates√N . Thus, even if N

is not a perfect square(so that√N is no more an integer), we can find the

nearest integer to√N and get an integer approximation to

√N .

Example 5. If the area of a square is 90 cm2, what is its side-length,

rounded to the nearest integer?

Solution: Since A = l2, we have l2 = 90. But 81 < 90 < 100 and 81 is nearer

to 90 than 100. Hence the nearest integer to√90 is

√81 = 9.

Example 6. A square piece of land has area 112 m2. What is the closest

integer which approximates the perimeter of the land?

Solution: If l is the side-length of a square, its perimeter is 4l. We know

that l2 = 112. Hence

(4l)2 = 16l2 = (16)× (112) = 1792.

But 422 = 1764 < 1792 < 1849 = 432 and 1764 is nearer to 1792 than 1849.

Hence the integer approximation for√1792 is 42. The approximate value

of the perimeter is 42 cm.

Caution!

If you take√112 to the nearest integer you see that it is 11

and you may be tempted to write the approximate value of the

perimeter as 44(= 4×11) cm. By replacing√112 with 11, you have

already committed an error and when you multiply by 4, the er-

ror increases by 4 times. That is the reason we have multiplied

112 by 16 first and then took the square root to the nearest in-

teger. There is nothing strange in this. The nearest integer to

r = 1/4 is 0. But the nearest integer to 3r = 3/4 is 1, not 3× 0 = 0.

Now you may see the limitations of integers. If you want to find√90,

you have to take it as 9; if you want√94, then you have to take it as 10.

But neither of them gives you a true picture of what is the square root

of a non-perfect square. The limitation is because: there is no integer

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between n and n + 1. However, this property ceases to be true in the

system of rational numbers. You can find a rational number between any

two rational numbers. This will help in moving further close to the square

root of a non-perfect square. You will learn more about this in your higher

classes.

Exercise 1.2.6

1. Find the nearest integer to the square root of the following numbers:

(i) 232; (ii) 600; (iii) 728; (iv) 824; (v) 1729.

2. A piece of land is in the shape of a square and its area is 1000 m2.

This has to be fenced using barbed wire. The barbed wire is available

only in integral lengths. What is the minimum length of the barbed

wire that has to be bought for this purpose?

3. A student was asked to find√961. He read it wrongly and found

√691

to the nearest integer. How much small was his number from the

correct answer?

1.2.7 Perfect cubes

Read the following table:

1 = 1× 1× 1;

8 = 2× 2× 2;

27 = 3× 3× 3;

125 = 5× 5× 5

You observe that each number is written as a product of 3 equal integers..

We say that an integer N is a perfect cube if N can be written

as a product of three equal integers. If N = m×m×m, we say N

is the cube of m and write N = m3(read as cube of m or simply

m-cube).

Consider a few more examples:

(−4)× (−4)× (−4) = −64 = (−4)3,

(−5)× (−5)× (−5) = −125 = (−5)3

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44 Unit 2

(−8)× (−8)× (−8) = −512 = (−8)3Do you see that the negative numbers are also perfect cubes? Contrast

this with perfect squares. A non-zero perfect square is necessarily a posi-

tive integer. However, perfect cubes can as well be negative.

Example 7. Find the cube of 6.

Solution: We have 63 = 6× 6× 6 = 36× 6 = 216.

Example 8. What is the cube of 20?

Solution: Again (20)3 = 20× 20× 20 = (400)× 20 = 8000.

You have studied about a solid called cube. It is a solid having equal

length, breadth and height. If l is the side-length of a cube, then its volume

V = l3 cubic units.

Example 9. If a cube has side-length 10 cm, what is its volume?

Solution: We have V = 10× 10× 10 = 1000 cm3.

Example 10. Find the smallest integer larger than 1 which is a perfect

square as well as a perfect cube.

Solution: Start with any number n and multiply it with itself 6 times to

get a number N . We observe that

N = n× n× n× n× n× n = (n× n)× (n× n)× (n× n)

= (n2)× (n2)× (n2) = (n2)3.

Thus N is the cube of n2. On the other hand you may also observe that

N = n× n× n× n× n× n = (n× n× n)× (n× n× n)

= (n3)× (n3) = (n3)2.

Hence N is also the square of n3. Thus N is both a perfect cube and a

perfect square. Taking n = 2, we get the least number: N = 2× 2 × 2 × 2 ×2× 2 = 64. You may verify that 64 = 43 and 64 = 82.

Example 11. Show that 6 is not a perfect cube.

Solution: We observe that 1 < 6 < 8 so that 13 < 6 < 23. Since no integer

exists between 1 and 2, 6 cannot be written as a product of three equal

integers. Hence 6 is not a perfect cube.

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Exercise 1.2.7

1. Looking at the pattern, fill in the gaps in the following:

2 3 4 −5 — 8 —

23 = 8 33= — —= 64 — = — 63= — — = — — = −729

2. Find the cubes of the first five odd natural numbers and the cubes

of the first five even natural numbers. What can you say about the

parity of the odd cubes and even cubes?

3. How many perfect cubes you can find from 1 to 100? How many from

−100 to 100?

4. How many perfect cubes are there from 1 to 500? How many are

perfect squares among these cubes?

5. Find the cubes of 10, 30, 100, 1000. What can you say about the

zeros at the end?

6. What are the digits in the unit’s place of the cubes of 1,2,3,4,5,6,7,8,9,

10? Is it possible to say that a number is not a perfect cube by looking

at the digit in unit’s place of the given number, just like you did for

squares?

Activity 8: (More on Hardy-Ramanujan numbers) Express 4104 and

13832 as a sum of two perfect cubes in two different ways. Find some

numbers which can be expressed as a sum of two different perfect squares

in two or more ways. Explore more on this topic.

1.2.8 Cube root

You know how to find the volume of a cube, given its side-length. Can

you reverse the process? Given the volume of a cube, is it possible to find

its side-length?

Suppose the volume of a cube is 125 cm3. If l is its side-length, you

write l3 = 125 and conclude l = 5 cm. Here we say 5 is the cube root of 125

and write 5 = 3√125.

If N is number and n is another number such that N = n3, we say

n is the cube root of N and write n = 3√N .

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Remark: The definition of square root and cube root makes sense even for

non-integers(for square root , the number must be non-negative). At this

stage we confine only to integers and do not get involved in generality.

Contrast the definition of cube root with that of square root. Given a

perfect square, there are two possible square roots; positive and negative.

This is because, the square of a non-zero integer is always positive and

(−n)2 = n2 for any integer n. Such a thing cannot happen for cubes.

If n is positive, then n3 is positive; if n is negative, n3 is also negative.

Hence cube root of a perfect cube is negative or positive depending on the

negativity or positivity of the given perfect cube. This shows that we can,

unambiguously, talk of the cube root of a perfect cube. There is no need

to follow a convention as in the case of square roots.

As in the case of square roots, we can find the cube root of a perfect

cube by prime factorisation.

Example 12. Find the cube root of 216 by factorisation.

Solution: Observe

216 = 2× (108) = 2× 2× (54) = 2× 2× 2× 27 = 2× 2× 2× 3× 3× 3

= (2× 3)× (2× 3)× (2× 3)

= 6× 6× 6.

Hence 3√216 = 6.

Example 13. Find the cube root of −17576 using factorisation.

Solution: Let us first find the cube root of 17576. As earlier, we have

17576 = 2× (8788) = 2× 2× (4394) = 2× 2× 2× (2197)

= 2× 2× 2× 13× (169)

= 2× 2× 2× 13× 13× 13

= (2× 13)× (2× 13)× (2× 13)

= 26× 26× 26.

This shows that −17576 = (−26)× (−26)× (−26). Thus 3√−17576 = −26.

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Srinivasa Ramanujan(1887-1920)

is undoubtedly the greatest Indian

mathematician of all times. He was

self-taught and had an uncanny math-

ematical manipulative ability. He was

not able to pass his school examina-

tions in India, and had to be content

with a clerical position in the Port Trust

of Madras. However, he continued to

create his own mathematics, obtained

lot of hitherto unknown results. He

sent these results G. H. Hardy who at

once recognized Ramanujan’s intrinsic

mathematical ability and arranged for

him to travel to Cambridge.

Because of his lack of formal training, Ramanujan sometimes did not differen-

tiate between formal proof and apparent truth based on intuitive or numerical

evidence. His intuition and computational ability allowed him to determine and

state highly original and unconventional results which continued to defy formal

proof until recently.

Ramanujan had an intimate familiarity with numbers, and excelled especially

in number theory. J. Littelewood( a collaborator of G.H.Hardy) exclaimed that

every integer was a personal friend of Ramanujan. His familiarity with num-

bers may be demonstrated by the following incident. During an illness in Eng-

land, Hardy visited Ramanujan in the hospital. When Hardy remarked that

he had taken taxi number 1729, a singularly dull number, Ramanujan imme-

diately responded that this number was actually quite remarkable: it is the

smallest integer that can be represented in two ways by the sum of two cubes:

1729 = 13 + 123 = 93 + 103(Hardy-Ramanujan number).

Unfortunately, Ramanujan’s health deteriorated rapidly in England, due per-

haps to the unfamiliar climate, food, and to the isolation which Ramanujan felt

as the sole Indian in a culture which was largely foreign to him. Ramanujan

was sent home to recuperate in 1919, but tragically died the next year at the

very young age of 32.

Example 14. What is the least positive integer with which you have to

multiply 243 to get a perfect cube?

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48 Unit 2

Solution: Let us factorise 243. We observe that

243 = 3× (81) = 3× 3× 3× 3× 3.

If we multiply 243 by 3, we see that

243× 3 = 3× 3× 3× 3× 3× 3 = 9× 9× 9,

and we get a perfect cube. The answer is therefore 3.

If you factorise a positive perfect cube, you may observe that the num-

ber of times each prime factor occurs is always a multiple of 3(just like it is

a multiple of 2 for perfect squares). Hence to get the least positive integer

whose product with the given integer makes a perfect cube, you have to

see how much each prime factor is deficient in the prime factorisation of

the given number to be away from a perfect cube.

Often, finding the cube root of a given perfect cube may be time con-

suming. We seek to find easier methods. We can use the behaviour of

the unit’s digit of a cube to fix the cube root. We see that the unit’s dig-

its of the cubes of numbers ending with 1,2,3,4,5,6,7,8,9,0 are uniquely

determined. See the following table:

units’ digit of n 1 2 3 4 5 6 7 8 9 0

unit’s digit of n3 1 8 7 4 5 6 3 2 9 0

We also tabulate the cubes of first nine numbers:

n 1 2 3 4 5 6 7 8 9

n3 1 8 27 64 125 216 343 512 729

Let us see how these help us.

Example 15. Find the cube root of 103823.

Solution: Here the units digit of 103823 is 3. If n3 = 103823, then the units

digit of n must be 7. Let us split 103823 as 103 and 823. We observe that

43 = 64 < 103 < 125 = 53. Hence

403 = 64000 < 103823 < 125000 = 503.

Hence n must lie between 40 and 50. Since the units digit of n is 7, the

only such number is 47. You may check that 473 = 103823.

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Note: This method works only if you know that the given number is

a perfect cube.

Nevertheless, this helps us in estimating the cube root of a non-perfect

cube. We can squeeze the given number between two perfect cubes and

see which is the nearest one.

Example 16. Find the nearest integer to the cube root of 12345.

Solution: We observe that

203 = 8000 < 12345 < 27000 = 303.

Hence 3√12345 must lie between 20 and 30. We do not know whether 12345

is a perfect cube or not. However, we may sharpen the bound: 233 = 12167

and 243 = 13824 and hence 3√12345 must be between 23 and 24. Moreover

12167 is nearer to 12345 than 13824. Hence the closest integer to 3√12345

is 23.

Exercise 1.2.8

1. Find the cube root by prime factorisation:

(i) 10648; (ii) 46656; (iii) 15625.

2. Find the cube root of the following by looking at the last digit and

using estimation:

(i) 91125; (ii) 166375; (iii) 704969.

3. Find the nearest integer to the cube root of each of the following:

(i) 331776; (ii) 46656; (iii) 373248.

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50 Unit 2

Additional problems on“Squares, square roots, cubes and cube roots”

1. Match the numbers in the column A with their squares in the column

B:

A B Answers

—– —– ——–

(1) 5 (a) 25 (1) ——

(2) 8 (b) 144 (2) ——

(3) 2 (c) 36 (3)——

(4) −6 (d) 484 (4)——

(5) −22 (e) 64 (5)——

(6) 12 (f) 4 (6)——

(g) 121

2. Choose the correct option.

(a) The number of perfect squares from 1 to 500 is:

A. 1) B. 16 C. 22 D. 25

(b) The last digit of a perfect square can never be

A. 1 B. 3 C. 5 D. 9

(c) If a number ends in 5 zeros, its square ends in:

A. 5 zeros B. 8 zeros C. 10 zeros D. 12 zeros

(d) Which could be the remainder among the following when a perfect

square is divided by 8?

A. 1 B. 3 C. 5 D. 7

(e) The 6-th triangular number is:

A. 6 B. 10 C. 21 D. 28

3. Consider all integers from −10 to 5, and square each of them. How

many distinct numbers do you get?

4. Write the digit in unit’s place when the following numbers are squared:

4,5,9,24,17,76,34,52, 33, 2319, 18, 3458, 3453.

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5. Write all numbers from 400 to 425 which end in 2,3, 7 or 8. Check if

any of these is a perfect square.

6. Find the sum of the digits of (111111111)2.

7. Suppose x2 + y2 = z2.

(i) if x = 4 and y = 3 find z;

(ii) if x = 5 and z = 13, find y;

(iii) if y = 15 and z = 17, find x.

8. A sum of 2304 is equally distributed among several people. Each

gets as many rupees as the number of persons. How much does each

one get?

9. Define a new operation ⋆ on the set of all natural numbers by m ⋆ n =

m2 + n2.

(i) Is N closed under ⋆?

(ii) Is ⋆ commutative on N?

(iii) Is ⋆ associative on N?

(iv) Is there an identity element in N with respect to ⋆?

10. (Exploration) Find all perfect squares from 1 to 500, each of which is

a sum of two perfect squares.

11. Suppose the area of a square field is 7396 m2. Find its perimeter.

12. Can 1010 be written as a difference of two perfect squares? [Hint:

How many times 2 occurs as a factor of 1010?]

13. What are the remainders when a perfect cube is divided by 7?

14. What is the least perfect square which leaves the remainder 1 when

divided by 7 as well as by 11?

15. Find two smallest perfect squares whose product is a perfect cube.

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52 Unit 2

Glossary

Perfect square: an integer which is the product of two equal integers.

Triangular numbers: the sum of the first n natural number is called n-th

triangular number.

Square-root: a number a is a square root of b if b = a2.

Perfect cube: an integer which is the product of three equal integers.

Cube-root: a number c whose cube is d is called a cube-root of c.

Prime factor: a prime number which divides an integer a is a prime factor

of a.

Irrational number: any real number which is not a rational number.

Points to remember

• A perfect square is the product of two equal integers; a perfect cube

is the product of three equal integers.

• A perfect square is always non-negative(0 is also perfect square); a

perfect cube may be negative, equal to 0 or may be positive.

• Given a positive number, there are two square-roots, positive and

negative. But for any number, there is only one cube-root.

• Given any positive number which is not a square, you can always

squeeze it between two consecutive perfect squares.

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CHAPTER 1 UNIT 3

RATIONAL NUMBERS


• the concept of fraction and rational numbers;

• to add and multiply rational numbers;

• the properties of rational numbers with respect to addition and mul-

tiplication: closure,associativity, commutativity, distributive proper-

ties, and existence of identity and inverse;

• representation of rational numbers on the number-line and density

property of rational numbers;

• the gain while moving from integers to rational numbers and the loss.

1.3.1 Introduction

Earlier, you have studied natural numbers and some of their proper-

ties; the numbers {1, 2, 3, . . .}. This is called as the set of all natural num-

bers and is denoted by N. You have seen that the sum or product of any

two natural numbers is again a natural number: for example 5 + 13 = 18;

12 × 15 = 180. We say that the set of all natural numbers is closed un-

der addition and multiplication(or the closure property of addition and

multiplication of the set of all natural numbers). You have also observed

observed that:

8+12 = 12+8; 13+(9+21) = (13+9)+21; 15×7 = 7×15; 3× (5×6) = (3×5)×6.

Here you can take any natural number. Thus you have also seen that for

all natural numbers m,n, p the following hold:

m+ n = n +m, (commutative property of addition);

m+ (n+ p) = (m+ n) + p, (associative property of addition);

m · n = n ·m, (commutative property of multiplication);

m · (n · p) = (m · n) · p, (associative property of multiplication).

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54 Unit 3

You have learnt how to combine addition and multiplication and get yet

another property called the distributive property. For example

5× (7 + 8) = 5× 15 = 75 = 35 + 40 = (5× 7) + (5× 8).

Thus for all natural numbers m,n, p, you may write

m · (n+ p) = m · n +m · p.

From this you also get

(n + p) ·m = n ·m+ p ·m,

using the commutative property of multiplication and addition. You have

also seen that the natural number 1 satisfies 1 × 8 = 8 × 1 = 8; again 8 is

irrelevant and the relation 1 ·m = m · 1 = m holds for all natural numbers

m.

You must have wondered why such a natural number does not exist for

addition; a number u such that m+u = u+m = m for all natural numbers.

This is the reason you have added the number 0 to the set of all natural

numbers and got the set of all whole numbers, W . Thus W = {0, 1, 2, 3, . . .}.The number 0 satisfies, for example, 8 + 0 = 0 + 8 = 8 and 9× 0 = 0× 9 = 0.

Thus the number 0 obeys certain rules:

m+ 0 = 0 +m = m, for all natural numbers m;

m · 0 = 0 ·m = 0, for all natural numbers m;

0 + 0 = 0;

0 · 0 = 0.

If you multiply two non-zero numbers in W , you get a non-zero number:

for example 14 × 6 6= 0. Thus in W , you must have noticed that m · n = 0

is possible if and only if either m = 0 or n = 0(or may be both). Another

important property you have studied about the natural number is that it

is possible to compare any two natural numbers; for example, if you are

asked to compare 12 and 81, you immediately say that 12 is smaller than

81; or 81 is larger than 12. You write 12 < 81 or 81 > 12. Thus given any

two natural numbers m, n, you know that either m < n or m = n or m > n;

and only one of these properties hold. This is called the ordering on the

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Rational numbers 55

set of all natural numbers; and you write 1 < 2 < 3 < 4 < · · · . Now we place

0 before 1 and get ordering on W ; 0 < 1 < 2 < 3 < 4 < · · · . There is one

fundamental truth about this ordering.

For example, if you consider the set E = {3, 6, 9, . . .}, the set of all mul-

tiples of 3, you see that 3 is the smallest element in it. However this set

does not have the largest element. Suppose you take the set of all marks

scored by students in your class in a test. If you write the marks in as-

cending order, you see that one of the marks is the lowest. Any subset of

N containing at least one element of N is called a non-empty subset of N.

For example, the set of all even natural numbers is a non-empty subset of

N. The set of all numbers which are both even and odd is an empty subset

of N since there is no such number.

Every non-empty subset of natural numbers of N (or W ) has

the smallest element.

This is called the well ordering property of natural numbers.

Activity 1: Write down five finite non-empty subsets of N and find their

smallest elements. Write down two infinite subsets of W and find the least

element of these subsets.

There is one distinct disadvantage inherent in N or W . Consider the

equation x + 5 = 3. You see that there is no natural number m such that

m + 5 = 3. In fact, for any natural number, you know that m + 5 > 5 > 3.

This disadvantage is removed in the set of all integers, Z. You have seen

that you can adjoin to W , another class of numbers called the negative

integers. For each natural number m, you associate another number

−m called the negative of m(or the opposite of m). Thus Z consists of

three parts: the set of all natural numbers; 0; and the set of all negative

numbers. We usually write this as

Z = {· · · − 5,−4,−3,−2,−1, 0, 1, 2, 3, 4, 5, · · ·}.

Here Z is derived from the German word Zahlen(which means number).

You have also seen that you can perform addition and multiplication

on Z. If m and n are two natural numbers, then

(i) (−m) + (−n) = −(m+ n);

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56 Unit 3

(ii) (−m) + 0 = −m = 0 + (−m);

(iii) (−m) + n =

−(m− n), if m > n

n−m, if m < n

0, if m = n

;

(iv) (−m) · n = m · (−n) = −(m · n);

(v) (−m) · (−n) = m · n;

(vi) (−m) · 0 = 0 · (−m) = 0.

Of course, if m and n are two whole numbers, we retain the same old

addition and multiplication for them.

With this extension of definition of addition and multiplication, Z now

enjoys several nice properties:

1. closure property: for all integers a, b, both a + b and a · b are also

integers;

2. commutative property: for all integers a, b

a+ b = b+ a and a · b = b · a;

3. associative property: for all integers a, b, c,

a+ (b+ c) = (a+ b) + c and a · (b · c) = (a · b) · c;4. distributive property: for all integers a, b, c,

a · (b+ c) = a · b+ a · c;5. cancellation law: if a, b, c are integers such that c 6= 0 and ac = bc,

then a = b (effectively you can cancel c on both sides).

Note that the cancellation law holds only if c 6= 0. For example, you may be

tempted to write 3 · 0 = 0 = 5 · 0 and try to cancel 0 on both sides to end up

with an absurd conclusion 3 = 5. Many of the fallacious conclusions are

due to such erroneous cancellations.

What is the advantage in moving from N to Z? You see that 0 has a

special status: a+0 = 0+a = a for all integers. We say that 0 is the additive

identity of Z(or 0 is the identity with respect to addition). Moreover, for

each integer a, we have another integer −a such that a+(−a) = 0 = (−a)+a;

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Rational numbers 57

if a = m is a natural number, −a is the negative integer −m; if a = 0, then

−a = 0; if a is a negative integer, then a = −n for some natural number n

and we take −a = n, so that

a+ (−a) = (−m) +m = 0,

by the way we have defined negative numbers. We say −a is the additive

inverse of a. Every integer has additive inverse. Now you can solve an

equation x + a = b in integers, for any two integers a, b. We can take x =

b+ (−a). Thenx+ a = (b+ (−a)) + a = b+

((−a) + a

)= b+ 0 = b;

we have used associativity of addition, the fact that (−a) is the additive

inverse of a and that 0 is the additive identity.

We can also order the elements of Z. For any natural number n, we

put −n < 0. If m and n are two natural numbers such that m < n, we put

−n < −m. You now see that all the elements of Z can be compared. Every

natural number is an integer and we call the natural numbers as positive

integers.

Here you may be wondering what is the status of subtraction. This is

not introduced as a fundamental operation. You have learnt earlier that

12− 7 = 5. However, you have learnt now that −7 is the additive inverse of

7. Thus you may think 12 − 7 = 12 + (−7) adding 12 and −7. Here you see

that the subtraction is a convenient name we have given to the extension

of our fundamental operation addition to cover negative integers. This

makes sense even when two integers are negative. Suppose you want to

find −8 − 13. You put this as (−8) + (−13) = −(8 + 13) = −21, by the way

we have defined addition of two negative integers. Now what is 15 − 21?.

Your answer is clear: adding 15 and (−21). The definition shows that

15 + (−21) = −(21 − 15) = −6. What is the additive inverse of −m? Since

m+ (−m) = 0, we see that m is the additive inverse of −m; thus you obtain

−(−m) = m. The emphasis here is to the fact that − symbol represents

additive inverse.

It is a universal law that when you gain some thing, you must lose

some thing. In Z, the gain is clear: you are able to solve an equation of the

form x+ a = b, whenever a and b are integers. Such an equation cannot be

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58 Unit 3

solved in N unless b > a, and a, b natural numbers. On the other hand, you

see that in the set of all integers, a non-empty subset need not have the

smallest element: if you take {· · · − 5,−4,−3,−2,−1}, the set of all negative

integers, this does not have the smallest element(why?). But the gain we

have by going to the integers is much more than the loss incurred.

Consider again Z. Here an equation of the form ax = b, where a 6= 0

cannot be solved, in general. (You may solve this provided a divides b.)

Hence Z is also inadequate for our purpose. We look forward for a new

number system in which we can do better things than we are able to do

in Z. We also take care that it has most of the nice properties of Z and

perhaps much more. But remember we have to loose some thing. We shall

see what is the loss and what is the gain?

Exercise 1.3.1

1. Identify the property in the following statements:

(i) 2+ (3+4) = (2+3)+4; (ii) 2 · 8 = 8 · 2; (iii) 8 · (6+5) = (8 · 6)+ (8 · 5).2. Find the additive inverses of the following integers:

6, 9, 123, −76, −85, 1000.

3. Find the integer m in the following:

(i) m+ 6 = 8; m+ 25 = 15; (iii) m− 40 = −26; (iv) m+ 28 = −49.4. Write the following in increasing order:

21, − 8, − 26, 85, 33, − 333, − 210, 0, 2011.

5. Write the following in decreasing order

85, 210, − 58, 2011, − 1024, 528, 364, − 10000, 12.

1.3.2 Rational numbers

In your earlier class, you have learnt about fractions; numbers of the

form p/q, where p and q are natural numbers. For example: 1/2, 1/3, 3/4, 8/3

and such numbers. You have also learnt how to add and multiply such

numbers.

Example 1. Add and multiply1

3and

8

5.

Solution: We have

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1

3+

8

5=

(1× 5) + (8× 3)

3× 5=

5 + 24

15=

29

15.

Their product is 1

3× 8

5=

1× 8

3× 5=

8

15.

If you are given a fraction 10/4, you write this in the form10

4=

5× 2

2× 2=

5

2,

by canceling one 2 in the numerator with another 2 in the denominator.

In other words, if there are common factors between the numerator and

the denominator, you cancel them for convenience. Thus you do not dis-

tinguish between 10/4 and 5/2. You also divide one fraction by another: if

you want to divide 1/3 by 8/5, the result is1/3

8/5=

1× 5

3× 8=

5

24.

You have learnt a lot about working with fractions. Can all these be put in

a formal way? Is it possible to include negative fractions just like we have

included negative integers?

We define a rational number as a number of the formp

q, where p is

an integer and q > 0. Here p is called the numerator and q is called

the denominator of the rational number p/q. Thus the denominator of

a rational number is always a positive integer, where as the numerator

could be positive, negative or possibly 0.

Given two rational numbers a/b and c/d, we say they are equivalent if

a× d = c× b. Thus 10/4 is equivalent to 5/2 since 10× 2 = 20 = 5× 4. We say

a rational number a/b is in its lowest form or irreducible form if a and b

do not have any common factors other than 1. Thus 5/2 is its lowest form,

where as 10/4 is not.

Activity 2: Write ten rational numbers equivalent to 3/4. How many ra-

tional numbers are there which are equivalent to 3/4?

Thus 3/4, 1/5, 6/7, 7/10, −5/8, −6/11 are all rational numbers. You may

observe that each integer is also a rational number: given an integer a,

you may write it as a/1. Thus we do not distinguish between 7 and 7/1.

Suppose you have a fraction 3/4. Just like we have defined negative

integers using natural numbers(or positive integers), we can also define

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60 Unit 3

negative of3

4. This is defined as the rational number

−34. This is denoted

by −34.

We denote the set of all rational numbers by Q. Thus we can describe

Q as:Q =

{p

q

∣∣ p, q are integers and q > 0, HCF(p,q)=1

}.

Here HCF denotes the highest common factor.

At this point, you may wonder why we are not taking negative integers

in the denominator. Suppose you take a number of the formp

−q , where

q > 0 is an integer. You may observe that this is equivalent to the ratio-

nal number−pq, since p × q = (−p) × (−q). Thus you are not losing out

any rational number by restricting denominators to the set of all positive

integers.

We say a rational number is positive if both its numerator and denomi-

nator are positive integers.

The term rational number is derived from the word ratio. A rational

number is a ratio of two integers where the denominator is not equal to

zero.

Exercise 1.3.2

1. Write down ten rational numbers which are equivalent to 5/7 and the

denominator not exceeding 80.

2. Write down 15 rational numbers which are equivalent to 11/5 and the

numerator not exceeding 180.

3. Write down ten positive rational numbers such that the sum of the

numerator and the denominator of each is 11. Write them in decreas-

ing order.

4. Write down ten positive rational numbers such that numerator − de-

nominator for each of them is −2. Write them in increasing order.

5. Is3

−2 a rational number? If so, how do you write it in a form conform-

ing to the definition of a rational number(that is, the denominator as

a positive integer)?

6. Earlier you have studied decimals 0.9, 0.8 . Can you write these as

rational numbers?

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1.3.3 Properties of rational numbers

Closure property

You have learnt that the set of all natural numbers and the set of all inte-

gers have addition and multiplication satisfying the closure, commutative,

associative and distributive properties. Can we similarly define addition

and multiplication on the set of all rational numbers having these prop-

erties? Our starting point is the addition and multiplication of fractions,

which you have learnt in your lower class.

Example 1. Let us find the sum of5

6and

11

13. It is

5

6+

11

13=

(5× 13) + (11× 6)

(6× 13)

=65 + 66

78

=131

78.

Similarly, the sum of4

7and

−35

is:

4

7+−35

=(4× 5) + ((−3)× 7)

7× 5

=20 + (−21)

35

=−135

.

The sum of−74

and−37

is:

−74

+−37

=(−7)× 7 + (−3)× 4

4× 7

=(−49) + (−12)

28

=−6128

.

Example 2. The product of2

11and

8

7is:

2

11× 8

7=

2× 8

11× 7=

16

77.

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Similarly, the product of−35

and−72

is:

−35· −72

=(−3)× (−7)

5× 2=

21

10.

Based on these ideas, we can introduce addition and multiplication of

two rational numbers. Given two rational numbersa

band

c

d, we define

a

b+

c

d=

ad+ cb

bd;

a

b· cd=

ac

bd.

Since b > 0 and d > 0, you see that bd is a natural number. Moreover

ad + cb and ac are integers. Thus you may conclude thatad+ cb

bdand

ac

bdare rational numbers. Thus the sum of two rational numbers is again a

rational number. We say that the set of all rationals numbers is closed

under addition.

The set of all rational numbers is closed under addition and

multiplication.

Activity 3: Take ten pairs of rational numbers. Find the sum of the

numbers in each pair. Check that you will always end up with rational

numbers. Thus satisfy your self that the closure with respect to addition

holds. Similarly, multiply the two numbers in each pair and satisfy your

self that the closure property with respect to multiplication also holds.

Associative property You have seen earlier that integers have the prop-

erties: p + (q + r) = (p + q) + r and p · (q · r) = (p · q) · r. For example:

3 + (5 + 8) = (3 + 5) + 8 and 3× (5× 8) = (3× 5)× 8. Will such things hold for

rational numbers as well?

Example 3. Consider three rational numbers 1/2, 4/5, −6/7. Observe that

1

2+

(4

5+−67

)=

1

2+

(7× 4 + (−6)× 5

5× 7

)=

1

2+

(28− 30

35

)

=1

2+−235

=35× 1 + (−2)× 2

70

=31

70.

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Rational numbers 63

On the other hand, you also have(1

2+

4

5

)+−67

=

(5 + 8

10

)+−67

=13

10+−67

=91− 60

70

=31

70.

Can you conclude1

2+

(4

5+−67

)=

(1

2+

4

5

)+−67?

This is true for any three rational numbers a/b, c/d and e/f . You compute

both a

b+

(c

d+

e

f

)and

(ab+

c

d

)+

e

f.

You get

a

b+

(c

d+

e

f

)=

a

b+

cf + ed

df=

adf + (cf + de)b

bdf

=adf + cfb+ deb

bdf.

Similarly, (ab+

c

d

)+

e

f=

ad+ cb

bd+

e

f=

adf + cbf + ebd

bdf.

Now can you see that adf + cfb + deb = adf + cbf + ebd? What properties of

integers have we used here? Thus both the sums are the same.

Similar result is true for multiplication. Consider2

3,7

8,11

13. We have

2

3×(78× 11

13

)=

2

3× 77

104=

154

312;

(23× 7

8

)× 11

13=

14

24× 11

13=

154

312;

We conclude that2

3×(78× 11

13

)=(23× 7

8

)× 11

13.

For any three rational numbersa

b,c

dand

e

f, we obtain

a

b·(c

d· ef

)=

a

b· cedf

=ace

bdf,

(ab· cd

)· ef

=ac

bd· ef=

ace

bdf,

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64 Unit 3

so that a

b·(c

d· ef

)=(ab· cd

)· ef.

Addition and multiplication are associative on the set of all ra-

tional numbers.

Commutative property

You have studied earlier that addition and multiplication satisfy commu-

tative property on the set of all integers. Given any two integers m and n,

you have m + n = n + m and m · n = n · m; for example 3 + 5 = 5 + 3 and

3× 5 = 5× 3. Do we have similar things for rational numbers?

Example 4. Let us take two rational numbers, say 8/11 and −16/9. Ob-

serve that

8

11+−169

=8× 9 + (−16)× 11

11× 9

=72− 176

99=−10499

.

Similarly,

−169

+8

11=

(−16)× 11 + 8× 9

9× 11

=−176 + 72

99=−10499

.

Thus you will get 8

11+−169

=−169

+8

11.

Example 5. Similarly, you may verify that

8

11· −16

9=−169· 811

, .

Can we put this in a more general setting? Take any two rational numbers

a/b and c/d. Observe that

a

b+

c

d=

ad+ cb

bd, and

c

d+

a

b=

cb+ ad

db.

But you know that ad+ cb = cb+ ad and bd = db(What properties of integers

are used here?). Hence you may conclude thata

b+

c

d=

c

d+

a

b.

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Rational numbers 65

This gives the commutative property of addition. A similar observation can

be made for multiplication:a

b· cd=

ac

bd=

ca

db=

c

d· ab.

We obtain the commutative property of multiplication.

Addition and multiplication are commutative on the set of all

rational numbers.

Distributive property

Consider the rational numbers 2/3, 1/2 and 1/9. Observe that2

3·(1

2+

1

9

)=

2

3· 1118

=22

54=

11

27.

Similarly, we have2

3· 12+

2

3· 19=

2

6+

2

27=

66

162=

11

27.

Note that we have used the property of equivalent fractions. We may con-

clude that 2

3·(1

2+

1

9

)=

2

3· 12+

2

3· 19.

Activity 4:

Take several triples of rational numbers and verify distributive property.

Also prove this as a general statement: if p/q, r/s and u/v are rational

numbers, then p

q·(rs+

u

v

)=

p

q· rs+

p

q· uv,

using the definition of addition and multiplication of rational numbers and

properties of integers.

In the set of all rational numbers, multiplication is distributive

over addition.

Think it over!

Given rational numbers a/b, c/d, e/f , can you have

a

b+

(c

d· ef

)=(ab+

c

d

)·(a

b+

e

f

)?

In other words can addition be distributive over multiplica-

tion?(That is, can we interchange addition and multiplication

in the above distributive property?)

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Additive identity

Consider the rational number0

1. Observe

7

8+

0

1=

7× 1 + 0× 8

8× 1=

7

8.

Similarly, you may verify that

0

1+

7

8=

7

8.

For any rational number a/b, you see that

a

b+

0

1=

a× 1 + 0× b

b× 1=

a

b.

Similarly, you may easily verify that

0

1+

a

b=

a

b.

Thus the rational number0

1acts as additive identity. We denote this sim-

ply by 0.

The set of all rational numbers has 0 as additive identity; that

is r + 0 = 0 + r = 0, for all rational numbers r.

Multiplicative identity

Again consider the rational number1

1. We have, for example,

11

12× 1

1=

11

12.

Thus any rational number a/b, we observe that

a

b· 11=

a

b=

1

1· ab.

Hence the rational number1

1(which again is denoted by 1) is identity with

respect to multiplication.

The set of all rational numbers has 1 as multiplicative identity;

that is r · 1 = r = 1 · r, for all rational numbers r.

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Rational numbers 67

Additive inverse

Take8

13and

−813

. If we add these two, we get

8

13+−813

=8× 13 + (−8 × 13)

169=

0

169= 0.

This is true for any rational number. For each rational numbera

b, consider

the rational number−ab. Let us find their sum:

a

b+−ab

=ab+ (−a)b

b2=

0

b2.

But the rational number0

b2is the same as

0

1as they are equivalent frac-

tions. Thus−ab

is the additive inverse ofa

b.

For each rational number r, there exists a rational number, de-

noted by −r, such that r + (−r) = 0 = (−r) + r.

Multiplicative inverse

You have seen that an integer may not have multiplicative inverse. For

example, 8 has no multiplicative inverse; 8 × a = 1 is not possible for any

integer a. On the other hand, consider7

5. We see that

7

5× 5

7=

35

35= 1.

This is true for any nonzero rational number.

Take any nonzero rational number a/b. Then a 6= 0 and hence b/a is also

a rational number, Observe thata

b· ba=

ab

ba=

1

1,

the multiplicative identity. We have used the fact that 1/1 and ab/ba are

equivalent fractions. You observe here that every nonzero rational number

has multiplicative inverse.

For each rational number r 6= 0, there exists a rational number,

denoted by r−1 (or 1/r), such that r · r−1 = 1 = r−1 · r.

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In integers, you have observed another important property, namely

cancellation law. For example, if 8× a = 48, you write

8× a = 8× 6.

You cancel 8 both sides and get a = 6. Thus if a, b, c are integers such that

a 6= 0 and ab = ac, then b = c. Effectively, you can cancel equal nonzero

integers on both sides of an equality. This also holds in Q.

For example, we have 4/4 = 2/2, as they are equivalent fractions. But

4

4= 2× 2

4, and

2

2= 2× 1

2.

You thus obtain2× 2

4= 2× 1

2.

Canceling 2 on both sides, you get 2/4 = 1/2, which is true as they are

equivalent fractions.

Suppose you have three rational numbers a/b, c/d, e/f such thata

b6= 0

and you have a

b· cd=

a

b· ef.

Sincea

b6= 0, it has multiplicative inverse b/a. Multiply both sides by b/a:

b

a·(ab· cd

)=

b

a·(a

b· ef

).

Use the associative property to write this as(b

a· ab

)· cd=

(b

a· ab

)· ef.

This gives c

d=

e

f,

and we have canceled a/b both sides.

Now we define two more operations on Q: subtraction and division.

Consider the rational numbers4

13and

12

7. We want to give meaning for

4

13− 12

7. Recall, you have the additive inverse of

12

7, which is simply

−127

.

We define 4

13− 12

7=

4

13+−127

.

You can simplify this using the definition of addition:

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Rational numbers 69

4

13+−127

=4× 7 + (−12)× 13

13× 7=−12891

.

In other words subtraction amounts to adding the additive inverse. Ifa

band

c

dare two rational numbers, then

a

b− c

d=

a

b+−cd

=ad− bc

bd.

Similarly, ifa

band

c

dare two rational numbers and if

c

dis not equal to

zero,we define the division ofa

bby

c

das follows:

a

b÷ c

d=

a

b× d

c=

ad

bc.

Note that we are multiplyinga

bwith the multiplicative inverse of

c

d, which

exists as this is a non-zero rational number. If you want to divide8

15by

−711

, it is simply8

15÷ −7

11=

8

15× −11

7=−88105

.

The only fundamental operations are addition and multiplica-

tion. The subtraction and division are defined in terms of addi-

tion and multiplication.

Let us look at what we have gained from enlarging our number system

from Z to Q. If a 6= 0 is an integer, then there is no integer b such that

a · b = b · a = 1, unless a = 1 or a = −1. Thus, apart from 1 and −1, no

other integer has multiplicative inverse. On the other hand, every non-

zero rational number has its multiplicative inverse in Q.

This helps us to solve an equation of the form rx = s, where r 6= 0 and s

are rational numbers. Suppose we have to solve the equation3

8x =

5

9. You

solve this for x by multiplying both sides by 8/3. Thus

8

3× 3

8x =

8

3× 5

9=

40

27.

You get x = 40/27. This you can do for any general equation.

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Suppose r = a/b and s = u/v, where a, u are integers and b, v are natural

numbers. Since r 6= 0 implies that a 6= 0, r has its multiplicative inverse

b/a. Multiplying both sides by b/a, you get

b

a·(ab· x)=

b

a· uv.

This gives x =bu

av.

Exercise 1.3.3

1. Name the property indicated in the following:

(i) 315 + 115 = 430; (ii)3

4· 95=

27

20; (iii) 5 + 0 = 0 + 5 = 5;

(iv)8

9× 1 =

8

9; (v)

8

17+−817

= 0; (vi)22

23· 2322

= 1.

2. Check the commutative property of addition for the following pairs:

(i)102

201,3

4; (ii)

−813

,23

27; (iii)

−79,−1819

.

3. Check the commutative property of multiplication for the following

pairs:

(i)22

45,3

4; (ii)

−713

,25

27; (iii)

−89,−1719

.

4. Check the distributive property for the following triples of rational

numbers:

(i)1

8,1

9,1

10; (ii)

−49,6

5,11

10; (iii)

3

8, 0,

13

7.

5. Find the additive inverse of each of the following numbers:8

5,6

10,−38,−163

,−41.

6. Find the multiplicative inverse of each of the following numbers:

2,6

11,−815

,19

18,

1

1000.

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1.3.4 Representation of rational numbers on the Numberline

Earlier, you have seen how to represent integers on a line. We choose

an infinite line and fix some point on the line. This is denoted by 0. Fix

a unit of length and on both sides of 0, go on marking points at equal

unit distance. On the right side, at unit distance you get 1. If you move a

further unit distance, you get 2 and so on. If you move to the left by unit

distance, you get −1. If you further move unit distance to the left, you get

−2 and so on. Thus all the integers are represented on the line.

1 2 3 4 50−1−2−3−4−5

We can also use the same number line to represent rational numbers.

For example, we can represent 1/2 as the mid-point between 0 and 1.

0−1 1

A

−1/2 1/2

We can obtain A by bisecting the line-segment from 0 to 1. Similarly,

we can represent −12

as the mid-point of the line-segment from −1 to 0.

How can we get7

3?

0 21 3 54 6 87 9

P Q

RP S Q

Again consider the line-segment PQ from 0 to 7. We divide PQ in to 3

equal parts: PR = RS = SQ.(Here you need some geometrical construc-

tions, which you will learn later.) Then PR = 7/3. Hence R represents the

rational number 7/3. We can get −73by locating a point R′ to the left of P

such that R′P = PR.

RP S QR’

−7/3 0 7/3

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Activity 5:

Draw a number line and locate the points1

8,1

6,1

3,4

5,−38

on it.

In this way, you can represent each rational number by a unique point

on the number line. You may observe that 2/4 and 1/2 are represented

by the same point on the number line. (Can you see why?) All rational

numbers which are equivalent to a given rational number get the same

representation on the number line.

0 11/2 3/41/8 1/4 3/8

Can you now observe what is happening when you represent the ratio-

nal numbers on the number line? You have seen that you can locate 1/2

between 0 and 1 as the mid-point of the line-segment from 0 to 1. If you

take the mid-point of 0 and 1/2, you get 1/4; and the mid-point of 1/2 and

1 is 3/4. The mid-point of 1/4 and 1/2 is 3/8. Do you see that the mid-point

of the line-segment joining two points representing two rational numbers

on the number-line is again a rational number? We may observe here that

it is possible to order the rational numbers using the ordering on Z. Sup-

pose a/b and c/d are two rational numbers. We saya

b<

c

d, if ad < bc. Thus

6

7<

7

8, since 48 < 49. On the other hand

−78

<−67, since −49 < −48.

Suppose you consider2

7and

5

8. Then

2

7<

5

8. Then the average of these

two is2

7+

5

82

=51

112.

Now51

112lies between

2

7and

5

8. In fact

2

7<

51

112<

5

8,

which may be easily verified.

This is true for any two rational numbers. If we start with two ratio-

nal numbersa

band

c

dsuch that

a

b<

c

d, then the mid-point of the points

representinga

band

c

drepresents the number

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Rational numbers 73

a

b+

c

d2

=ad+ bc

2bd.

This is also a rational number. You may observe that

a

b<

ad+ bc

2bd<

c

d.

In fact

a

b<

ad+ bc

2bd⇐⇒ a× (2bd) < b(ad + bc)

⇐⇒ 2ad < ad+ bc (cancellation of b)

⇐⇒ ad < bc⇐⇒ a

b<

c

d,

which is given. Similarly, you can prove the other inequality. Thus the

rational numberad+ bc

2bdis strictly between

a

band

c

d.

Between any two distinct rational numbers, there is another

rational number.

Compare this with the property of integers. Given any integer m, there

is no integer between m and its successor m+1. This is no longer true for

rational numbers. We can talk of the next integer, but there is nothing

like the next rational number.

This is precisely the loss we have while moving from integers to rational

numbers. Thus the gain is: we can divide one rational number by another

nonzero rational number which helps us in solving an equation of the form

rx = s, where r 6= 0 and s are rational numbers. The loss is: there is no

more the next number in rationals, which we had in integers.

Exercise 1.3.4

1. Represent the following rational numbers on the number line:−85;3

8;2

7;12

5;45

13.

2. Write the following rational numbers in ascending order:3

4,7

12,15

11,22

19,101

100,−45,−10281

,−137

.

3. Write 5 rational number between2

5and

3

5, having the same denomi-

nators.

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74 Unit 3

4. How many positive rational numbers less than 1 are there such that

the sum of the numerator and denominator does not exceed 10?

5. Suppose m/n and p/q are two positive rational numbers. Where doesm+ p

n+ qlie, with respect to m/n and p/q?

6. How many rational numbers are there strictly between 0 and 1 such

that the denominator of the rational number is 80?

7. How many rational numbers are there strictly between 0 and 1 with

the property that the sum of the numerator and denominator is 70?

1.3.5 Introducing irrational numbers

You have seen that there is no integer whose square is 2. The argument

used was: 12 = 1 < 2 < 4 = 22 and there is no integer between 1 and 2.

However, you have seen now that there are plenty of rational numbers

between 1 and 2. In fact there are infinitely many rational numbers which

lie between 1 and 2. Hence, it is natural to wonder whether one could get

a rational number between 1 and 2 such that its square is 2. And the

astounding answer is no! There is no rational number r such that r2 = 2.

The argument is also simple. Suppose, if possible, there is a rational r

such that r2 = 2. Write r = p/q, such that HCF(p,q)=1 You get the relation

p2 = 2q2.

This shows that p2 is even and hence p itself is even; for, if p is odd, p2

must be odd. Hence you may write p = 2a, for some integer a.

Substitution gives

4a2 = 2q2 which implies q2 = 2a2.

But then, q is also even. Thus p and q are both even and must have a

common factor 2. This contradicts what we know: p/q is in its lowest

form. We conclude that no rational number, whose square is 2, exists.

Do you now see that the set of all rational numbers is also inadequate?

You cannot solve an equation of the form x2 = 2 in Q. This is the reason

that the mathematicians started enlarging the rational number system to

a better number system. One can show that for any natural number n,

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Rational numbers 75

which is not a perfect square, there is no rational number r such that

r2 = n. Thus, we need to give a meaning to√n, where n is a non-perfect

square. This is taken care in the real number system.

The number√2 or such numbers which are not rational numbers are

called irrational numbers. One can also prove that there is no ratio-

nal number whose cube is equal to 2. We denote this by 3√2 called the

cube-root of 2. This again is an irrational number. There is a systematic

construction of the real number system starting from Q, but it involves a

better understanding of the structure of Q.

The set of all real numbers consists of two parts: rational numbers

and irrational numbers. Both parts are infinite. However, in some sense,

the set of all irrational numbers is much larger than the set of all rational

numbers. In this way there are different infinities and a hierarchy among

infinities. You will enter a fascinating world of infinities.

Additional problems on “Rational numbers”

1. Fill in the blanks:

(a) The number 0 is not in the set of —————————-.

(b) The least number in the set of all whole numbers is —————.

(c) The least number in the set of all even natural numbers is———

——.

(d) The successor of 8 in the set of all natural numbers is ————–.

(e) The sum of two odd integers is ——————-.

(f) The product of two odd integers is —————-.

2. State whether the following statements are true or false:

(a) The set of all even natural numbers is a finite set.

(b) Every non-empty subset of Z has the smallest element.

(c) Every integer can be identified with a rational number.

(d) For each rational number, one can find the next rational number.

(e) There is the largest rational number.

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76 Unit 3

(f) Every integer is either even or odd.

(g) Between any two rational numbers, there is an integer.

3. Simplify:

(i) 100(100−3)− (100×100−3); (ii) (20− (2011−201))+(2011− (201−20))

4. Suppose m is an integer such that m 6= −1 and m 6= −2. Which is

largerm

m+ 1or

m+ 1

m+ 2? State your reasons.

5. Define an operation ⋆ on the set of all rational numbers Q as follows:

r ⋆ s = r + s− (r × s),

for any two rational numbers r, s. Answer the following with justifica-

tion:

(i) Is Q closed under the operation ⋆?

(ii) Is ⋆ an associative operation on Q?

(iii) Is ⋆ a commutative operation on Q?

(v) What is a ∗ 1 for any a in Q?

(vi) Find two integers a 6= 0 and b 6= 0 such that a ⋆ b = 0.

6. Find the multiplicative inverses of the following rational numbers:8

13,12

17,26

23,−1311

,−101100

.

7. Write the following in increasing order:10

13,20

23,5

6,40

43,25

28,10

11.

8. Write the following in decreasing order:21

17,31

27,13

11,41

37,51

47,9

8.

9. (a) What is the additive inverse of 0?

(b) What is the multiplicative inverse of 1?

(c) Which integers have multiplicative inverses?

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Rational numbers 77

10. In the set of all rational numbers, give 2 examples each illustrating

the following properties:

(i) associativity; (ii) commutativity; (iii) distributivity of multiplica-

tion over addition.

11. Simplify the following using distributive property:

(i)2

5×(1

9+

2

5

); (ii)

5

12×(25

9+

32

5

); (iii)

8

9×(11

2+

2

9

).

12. Simplify the following:

(i)

(25

9+

12

3

)+

3

5; (ii)

(22

7+

36

5

)× 6

7;

(iii)

(51

2+

7

6

)÷ 3

5; (iv)

(16

7+

21

8

)×(15

3− 2

9

).

13. Which is the property that is there in the set of all rationals but which

is not in the set of all integers?

14. What is the value of

1 +1

1 +1

1 + 1

?

15. Find the value of (1

3− 1

4

)/(1

2− 1

3

).

16. Find all rational numbers each of which is equal to its reciprocal.

17. A bus shuttles between two neigbouring towns every two hours. It

starts from 8 AM in the morning and the last trip is at 6 PM. On one

day the driver observed that the first trip had 30 passengers and each

subsequent trip had one passenger less than the previous trip. How

many passengers travelled on that day?

18. How many rational numbers p/q are there between 0 and 1 for which

q < p?

19. Find all integers such that3n+ 4

n + 2is also an integer.

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78 Unit 3

20. By inserting parenthesis(that is brackets), you can get several values

for 2 × 3 + 4 × 5. (For example((2 × 3) + 4

)× 5 is one way of inserting

parenthesis.) How many such values are there?

21. Suppose p/q is a positive rational in its lowest form. Prove that1

q+

1

p+ qis also in its lowest form.

22. Show that for each natural number n, the fraction14n+ 3

21n+ 4is in its

lowest form.

23. Find all integers n for which the number (n+3)(n−1) is also an integer.

Glossary

Ordering; the comparability of numbers.

Well ordering property; every non-empty subset has the least element.

Positive integers; the integers which can be identified with natural num-

bers.

Negative integers; those integers which are additive inverses of positive

integers.

Cancellation law; the law which enables us to cancel two non-zero equal

quantities on both sides of an equality.

Rational number; the numbers which are of the formp/q, where p is an

integer and q is a natural number.

Lowest form or irreducible form of a rational number; that form p/q of

a rational number such that p and q have no common factor.

Additive identity; that number which added to a given number does not

change the given number.

Additive inverse; given a number, that number which added to the given

number gives the additive identity.

Multiplicative identity; that number which multiplied with the given

number does not change the given number.

Multiplicative inverse; given a number, that number which when multi-

plied with the given number gives the multiplicative identity.

Density property; the inseparable property of numbers; for example ra-

tional numbers are dense in number line.

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Rational numbers 79

Successor; the next number; integers have this property, where as ratio-

nals do not have this property.

Points to remember

• A rational number is a ratio of the form p/q, where p is an integer and

q is a natural number.

• The set of all rational numbers is closed under addition and multipli-

cation.

• The set of all rational numbers has associative property and com-

mutative property with respect to both addition and multiplication.

Moreover, multiplication is distributive over addition.

• The rational number 0 is the additive identity; 1 is the multiplicative

identity.

• Every rational number has its additive inverse; every non-zero ratio-

nal number has its multiplicative inverse.

• Between any two distinct rational numbers, there are infinitely many

rational numbers.

• Every integer has the next integer, but there is no next rational num-

ber for any given rational number.

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CHAPTER 1 UNIT 4

COMMERCIAL ARITHMETIC

After studying this unit, you learn to:

• identify the processes of mathematics in commercial transactions;

• define the term percentage;

• solve the numerical problems involving percentage;

• identify profit or loss in commercial transactions;

• workout problems on profit, loss, selling price and marked price;

• calculate discount, discount percentage and selling price after dis-

count;

• solve problems on commission and commission percentage;

• define simple interest and other terminologies associated with simple

interest;

• calculate simple interest, principal, time, rate of interest and amount

in numerical problems.

1.4.1 Introduction

In this unit, we study several aspects of commercial mathematics which

are useful in daily life. You go to a market and buy some essential item.

You pay money for it. But behind this transaction a separate world of

arithmetic exists. The vendor brings the goods from somewhere and he

has to decide how much he should charge for the goods before selling.

Depending on the market strategy, he has to fix his selling prices. Some

times, he may have to sell it for a price lower than the price for which he

has procured the goods. Some other times, he has to announce attractive

incentives to get a good number of customers. When you do some transac-

tions in buying and selling land, houses and cattle, you may have to give

some money for the service you get from the agent. All these things are

generally dealt with money. This is the reason why money is an important

part of human life. Commercial mathematics is an insight into this kind

of transactions in daily life.

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Commercial arithmetic 81

1.4.2 Percentage

You are familiar with the meaning of percent. Per cent means for ev-

ery hundred. Thus, 4 percent (or 4%) means 4 for every hundred or4

100.

Percentage is also a fraction whose denominator is 100. The numerator

of the fraction is called rate percent. Thus 12% means 12 out of hun-

dred or12

100. The concept of percentage is used in business transactions,

calculating interest, comparison of quantities and the like.

Suppose a basket has, say, 6 pineapples and 14 oranges. Then the

number of pineapples and oranges can be compared using the fraction6

14=

3

7. The number of pineapples is

3

7times the number of oranges. In

the same way, the number of oranges is7

3times the number of pineapples.

Comparison can also be done using percentages.

There are 6 pineapples out of

20 (=6+14) fruits. Therefore

percentage of pineapples

6

20=

30

100= 30%

(here denominator is made

100)

Out of 20 fruits, number of

pineapples is 6. So out of 100

fruits, number of pineapples

is6

20× 100 = 30%.

This is called unitary method.

The basket contains only pineapples and oranges.

Therefore percentage of pineapples + percentage of oranges = 100.

Or 30% +% of oranges =100 or % of oranges = 100 -30= 70.

Thus the basket has 30% pineapples and 70% oranges.

Percentage is a convenient way of comparing quanti-

ties.

Example 1. A man spends 78% of his monthly income and saves 1,100.

What is his monthly income?

Solution: Let his monthly income be 100. Then his expenditure is 78.

Therefore, his savings is 100-78 = 22.

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Let us put this in reverse way. If the savings is 22, then the income

is 100. Hence if the savings is 1, then the income is100

22.

If the savings is 1,100, then the income is100

22× 1100 = 5, 000 rupees.

Hence his monthly income is 5,000.

Aliter: Given expenditure is 78%. Therefore savings is (100− 78) = 22%.

Let the monthly income of the man be x. Then 22% of x is 1,100.

This means22

100x = 1100.

Solving for x, we obtain

x = 1100× 100

22= 5000.

Hence his monthly income is 5,000.

Verification: 78% of 5,000 =78

100× 5000 = 3900. Hence, his savings is

5,000 -3,900 = 1,100.

Example 2. An athlete won 8 events out of a number of events. If the win

percentage was 40, how many events were there in total?

Solution: We are given that win percentage is 40. So,

40 events were won out of 100 events.

⇒ 8 events were won out of100

40× 8 = 20 events.

Hence, there were 20 events in total.

Activity 1: Try this in another method.

Example 3. Ravi’s income is 25% more than that of Raghu. What percent

is Raghu’s income less than that of Ravi?

Solution: Let Raghu’s income be 100. Then, Ravi’s income is 125. Put

this in the reverse way. If Ravi’s income is Rs 125, Raghu’s income is

100. Hence if Ravi’s income is Rs 1, Raghu’s income is100

125. Changing the

scale to 100, if Ravi’s income is 100, then Raghu’s income is100

125× 100

= 80. Therefore Raghu’s income is 100 - 80 = 20% less than that of Ravi.

Example 4. The salary of an employee is increased by 15%. If his new

salary is 12,650, what was his salary before enhancement?

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Solution: Let the salary before enhancement be 100. Since his incre-

ment is 15%, his salary after enhancement is 100 + 15 = 115.

Now we reverse the role. If the new salary is 115, salary before enhance-

ment is 100.

If the new salary is 12,650, salary before enhancement is100

115× 12650 =

11,000.

Therefore, his salary before enhancement is 11,000.

Exercise 1.4.2

1. In a school, 30% of students play chess, 60% play carrom and the rest

play other games. If the total number of students in the school is 900,

find the exact number of students who play each game.

2. In a school function 360 remained after spending 82% of the money.

How much money was there in the beginning? Verify your answer.

3. Akshay’s income is 20% less than that of Ajay. What percent is Ajay’s

income more than that of Akshay?

4. A daily wage employee spends 84% of his weekly earning. If he saves

384, find his weekly earning.

5. A factory announces a bonus of 10% to its employees. If an employee

gets 10,780, find his actual salary.

1.4.3 Profit and Loss

We purchase goods from shops. The shopkeeper purchases goods ei-

ther directly from a manufacturer or through a wholesale dealer. The

money paid to buy goods is called Cost Price and abbreviated as C.P.

The price at which goods are sold in shops is called Selling Price abbre-

viated as S.P. When an article is bought, some additional expenses such

as freight charges, labour charges, transportation charges, maintenance

charges etc., are made before selling. These expenses are known as Over-

head Charges. These expenses have to be included in the cost price.

Hence, you may conclude that

the real cost price = total investment

= price for buying goods + overhead charges.

If S.P > C.P, there is a gain or profit. If S.P < C.P, there is a loss.

Therefore,

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84 Unit 4

profit = S.P - C.P. and loss = C.P. - S.P.

Gain (or profit) or loss on 100 is called as gain percent or loss percent.

Note: Profit or loss is always calculated on cost

price.

Remember the following formulae:

1. Gain or profit % =gain

C.P.× 100;

2. Loss % =loss

C.P.× 100;

3. Selling Price =(100+gain%)

100×

C.P.;

4. Selling Price =(100-loss%)

100×

C.P.;

5. Cost Price =100

(100+gain%)×

S.P.;

6. Cost Price =100

(100-loss%)× S.P..

Example 5. The cost price of a computer is 19,500. An additional 450

was spent on installing a software. If it is sold at 12% profit, find the selling

price of the computer.

Solution: Cost price of the com-

puter is 19,500 + 450 (over-

head expenses) = 19,950. The

computer is sold at a profit of 12%.

Therefore,

selling price =(100+gain%)

100×C.P.

=(100 + 12)

100× 19950

Aliter: (Unitary method)

Profit = 12%. Hence selling

price is 100 + 12 = 112.

If the cost price is 100, sell-

ing price is 112.

If the cost price is 19,950,

selling price =112

100× 19950

= 22344.

=112

10× 1995

= 22344.

Thus the selling price is 22,344.

Hence, the selling price of

computer is 22,344.

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Example 6. On selling a bicycle for 4,300, a dealer loses 14%. For how

much should he sell it to gain 14%?

Solution: Selling price of the bicy-

cle is 4,300, and the loss is 14%.

Therefore,

cost price =100

(100-loss%)× S.P.

=100

(100− 14)× 4300

=100

86× 4300

= 5000.

Hence the cost price of the bicycle

is 5,000.

Aliter: (Unitary method)

We know that the loss is 14%.

Therefore selling price =

100 - 14 = 86.

If the selling price is 86,

cost price is 100. If the sell-

ing price is 4300,

cost price =100

86×4300 = 5000.

Now, the cost price of the bicycle is 5,000. Let us find out what should

be the selling price to get 14% profit.

Expected gain = 14%.

Therefore,

selling price =(100+gain%)

100×C.P.

=(100 + 14)

100× 5000

=114

100× 5000

= 5700.

Aliter:(Unitary method)

Expected gain = 14%.

Therefore, selling price

= 100 + 14 = 114.

If the cost price is 100, sell-

ing price = 114.

If the cost price is 5,000,

selling price =114

100×5000 = 5700.

Hence, the selling price of the bicycle to gain 14% is 5,700.

Example 7. Two cows were sold for 12,000 each, one at a gain of 20%

and the other at a loss of 20%. Find the loss or gain in the entire transac-

tion.

Solution: We first find out the cost price of each cow and then add them

to find the total amount spent. We know the total money received. Com-

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86 Unit 4

paring them, we will know whether there is a loss or gain in the whole

transaction.

First cow:

S.P. = 12,000.

Gain = 20%. Therefore,

Cost price =100

(100+gain%)× S.P.

=100

(100 + 20)× 12000

=100

120× 12000

= 10000.

Second Cow:

S.P = 12,000.

Loss = 20%. Therefore,

Cost price =100

(100-loss%)× S.P.

=100

(100− 20)× 12000

=100

80× 12000

= 15000.

Activity 2:

Find the cost price of the cow in both the cases using unitary method.

Let us compute the total cost price and selling price in the combined trans-

action.

Total C.P. of both cows = (10,000 + 15,000) = 25,000. Total S.P. of both

cows = 12,000 × 2 = 24,000. Here, we observe that S.P. < C.P.

Therefore, loss = (25,000 - 24,000 ) = 1,000. Hence,

loss percentage =loss

C.P.× 100 =

1000

25000× 100 = 4.

Therefore, there is a loss of 4% in the whole transaction.

Example 8. If the cost price of 21 cell phones is equal to selling price of

18 cell phones, find the profit percent.

Solution: Let the C.P. of each cell phone be Re. 1. Then the C.P. of 21 cell

phones is 21.

By the given data, S.P. of 18 cell phones = C.P. of 21 cell phones = 21.

Therefore, S.P. of 1 cell phone is21

18. This gives

Profit = S.P. -C.P. =21

18− 1 =

3

18=

1

6.

Thus, there is a profit of 1/6 on each cell phone. Now we can calculate

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profit percentage:

Profit percentage =profit

C.P.× 100 =

1/6

1× 100 =

50

3= 16

2

3.

Therefore, the profit percentage is 1623%.

To find the overall profit or loss, we need to find the combined

C.P. and combined S.P.

Example 9. A dealer sells a radio at a profit of 8%. Had he sold it for 85

less, he would have lost 2%. Find the cost price of the radio.

Solution: Let the cost price of the radio be x. Let us calculate the selling

price with 8% profit and 2% loss, separately.

With 8% profit:

We have here

S.P. =(100+gain%)

100× C.P.

=(100 + 8)

100× C.P.

=54

50x.

With 2% loss:

We obtain here

S.P. =(100-loss %)

100× C.P.

=(100− 2)

100×C.P.

=49

50x.

Activity 3:

Calculate S.P. in both the cases using unitary method.

Thus the difference in the selling price with 8% profit and 2% loss is

54

50x− 49

50x =

5

50x =

x

10.

But, this difference is given to be equal to 85, so thatx

10= 85. This

implies that, x = 850. Hence, the cost price of radio is 850.

Exercise 1.4.3

1. Sonu bought a bicycle for 3,750 and spent 250 on its repairs. He

sold it for 4,400. Find his loss or profit percentage.

2. A shop keeper purchases an article for 3,500 and pays transport

charge of 100. He incurred a loss of 12% in selling this. Find the

selling price of the article.

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3. By selling a watch for 720, Ravi loses 10%. At what price should he

sell it, in order to gain 15%?

4. Hari bought two fans for 2,400 each. He sold one at a loss of 10%

and the other at a profit of 15%. Find the selling price of each fan and

find also the total profit or loss.

5. A store keeper sells a book at 15% gain. Had he sold it for 18 more,

he would have gained 18%. Find the cost price of the book.

6. The cost price of 12 pens is equal to selling price of 10 pens. Find the

profit percentage.

1.4.4 Discount

A reduction on the marked price of articles is called discount. Gen-

erally discount is given to attract customers to buy goods or to promote

the sale of goods. The following completely describe the facts related to

discount:

• discount is always given on the marked price of the article;

• discount = marked price - selling price;

• marked price is sometimes called list price;

• discount = rate of discount times the marked price;

• net price = marked price - discount.

Example 10. A computer marked at 18,000 was sold at 15,840. Find

the percentage of discount.

Solution: Marked price is 18,000 and selling price is 15,480. There-

fore, discount = 18,000 - 15,840 = 2,160. Thus, for a marked price of

18,000, discount is 2,160.

For a marked price of 100, discount =2160

18000× 100 = 12%. Therefore the

percentage of discount is 12.

Example 11. A tape recorder is sold at 5,225 after being given a discount

of 5%. What is its marked price?

Solution: We are given that the discount is 5%. This means that for

100, the discount is 5.

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Therefore, selling price = 100 - 5= 95.

Thus on a selling price of 95, the marked price is 100.

On a selling price of 5,225, the marked price =100

95× 5225 = 5500.

Therefore, marked price of the tape recorder is 5,500.

Example 12. A shop keeper buys an article for 500. He marks it at 20%

above the cost price. If he sells it at 12% discount, find the selling price.

Solution: Cost price = 500;

profit = 20% of 500 =20

100× 500 = 100;

marked price = cost price + profit = 500 + 100 = 600.

Now the discount given is 12%.

For 600, the discount =12

100× 600 = 72 rupees.

Therefore, selling price = cost price - discount= 600 - 72 = 528.

Hence the selling price of the article is 528.

Example 13. A cloth seller marks a dress at 45% above the cost price and

allows a discount of 20%. What profit does he make in selling the dress?

Solution: Suppose the cost price of the dress material is 100. Since the

seller marks it 45% above the C.P., the marked price would be

100 + 45 = 145.

Discount of 20% on this marked price is = 20% of 145 =20

100× 145= 29.

Therefore, selling price = 145 - 29 = 116. We get

profit = S.P. - C.P. = 116 - 100 = 16.

Hence, profit percent =profit

C.P.× 100=

16

100× 100 =16.

Thus the merchant makes a profit of 16% on the marked price. (Observe

it is not 45-20=25 percent.)

Exercise 1.4.4

1. An article marked 800 is sold for 704. Find the discount and

discount percent.

2. A dress is sold at 550 after allowing a discount of 12%. Find its

marked price.

3. A shopkeeper buys a suit piece for 1,400 and marks it 60% above

the cost price. He allows a discount of 15% on it. Find the marked

price of the suit piece and also the discount given.

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4. A dealer marks his goods 40% above the cost price and allows a dis-

count of 10%. Find the profit percent.

5. A dealer is selling an article at a discount of 15%. Find:

(i) the selling price if the marked price is 500;(ii) the cost price if he makes 25% profit.

1.4.5 Commission

You might have observed some advertisements in news papers regard-

ing the availability of houses, sites, vehicles etc., for sale. Many a times,

these transactions are mediated by a person other than the owner and

the buyer. This mediator who helps in buying and selling is called com-

mission agent or broker. The money that the broker or agent receives in

the deal is called brokerage or commission. Commission is calculated on

the transaction amount in percentage. Commission per hundred rupees

is called commission rate.

Example 14. A real estate agent receives a commission of 1.5% in selling

a land for 1,60,000. What is the commission amount?

Solution: Selling price of the land is 1,60,000 and the commission rate

is 1.5%.

If the selling price is 100, commission = 1.5.

If the selling price is 1,60,000, commission =1.5

100× 160000= 2,400.

Therefore commission amount is 2,400.

The commission amount can also be calculated directly.

Commission = commission rate × selling price

In the example above,

commission = 1.5% of 1,60,000 =1.5

100× 160000 = 2,400.

Example 15. The price of a long note book is 18. A shop keeper sells

410 note books in a month and receives 1,033.20 as commission. Find

the rate of commission.

Solution: Price of one note book is 18.

=⇒ Price of 410 note books = 410× 18 = 7,380.

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Now the commission received for this amount is 1,033.20. Hence for

100, the commission is

1033.20

7380× 100 = 14.

Therefore, the rate of commission is 14%.

Example 16. Abdul sold his house through a broker by paying 6,125

as brokerage. If the rate of brokerage is 2.5%, find the selling price of the

house.

Solution: Brokerage given is 6,125. The brokerage rate is 2.5%.

If the brokerage is 2.5, selling price would be 100.

If the commission is 6,125, selling price is

100

2.5× 6125 = 245000.

Thus the selling price of the house is 2,45,000.

The selling price can also be calculated directly;

selling price =100

commission rate× commission.

In this example, S.P. =100

2.5× 6125 = 2,45,000.

Exercise 1.4.5

1. Sindhu sells her scooty for 28,000 through a broker. The rate of

brokerage is 212%. Find the commission that the agent gets and the

net amount Sindhu gets.

2. A share agent sells 2000 shares at 45 each and gets the commission

at the rate of 1.5%. Find the amount the agent gets.

3. A person insures 26,000 through an insurance agent. If the agent

gets 650 as the commission, find the rate of commission.

4. A selling agent gets 10,200 in a month. This includes his monthly

salary of 6000 and 6% commission for the sales. Find the value of

goods he sold.

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1.4.6 Simple interest

People borrow money from banks or financial institutions or money

lenders for various purposes. While returning the borrowed money after a

period of time, they need to pay some extra amount. This extra amount

paid on the borrowed money after a period of time is called interest. In

this context, we define the following terms.

1. Principal: the money borrowed is called principal or sum.

2. Interest: the extra money paid on the principal after a period of time

is called interest.

3. Amount: the total money paid is called amount. Thus Amount =

Principal + Interest.

4. Rate: interest for every 100 for one year is known as rate percent

per annum.

5. Time: time is the duration for which the borrowed money is utilised.

Time is expressed in years or months or days.

6. Simple interest: the interest calculated uniformly on the principal

alone throughout the loan period is called simple interest. In other

words, it is the interest paid on the principal alone.

In the world of finance (Bankers rule), time is often

expressed in days also.

Formula to find the simple interest:

Let P = principal; R = rate of interest per annum; T = time in years; I =

simple interest. These are related by the formula

I =P × T × R

100

From the above formula, we obtain different formulae:

P =100× I

T ×R, T =

100× I

P ×R, R =

100× I

P × T.

Amount = principal + interest.

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Example 17. Calculate the interest on 800 at 612% per annum, for 312years.

Solution: Given: P = 800; T = 312 years =7

2years; R = 612% =

13

2%. We

use the formula for I;

I =PTR

100=

800× (7/2)× (13/2)

100= 2× 7× 13 = 182.

Thus the interest is 182.

Example 18. Find the simple interest on 3,000 at 16% per annum for

the period from 4th February 2010 to 16th June 2010.

Solution: Here the principal is P = 3,000, and the rate of interest R =

16% p.a. However, we are not given time in years; we have only the period,

4th February 2010 to 16th June 2010. We have to convert this to years.

Observe that

February 5th to 28th = 24 days(2010 is not a leap year)

March = 31 days

April = 30 days

May = 31 days

June = 16 days.

Adding, we get the total time is equal to 132 days. Converting this to years,

T =132

365years. Now we have all the required data to apply formula:

I =PTR

100=

(3000× 132× 16)

365× 100= 173.58.

Thus the interest is approximately 174.

Note: 1. For calculating interest, the day on which money de-

posited is not counted, while the day on which money is withdrawn

is counted.

2. When the time is given in days or months, it is to be expressed in

years.

Example 19. A sum at a simple interest of 1212% amounts to 2,502.50

after 3 years. Find the sum, and the interest.

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Solution: Given: Amount = 2,502.50, and R = 1212% =25

2%. Let the sum

be x. Using the formula, we have

I =PTR

100=

x× 3× 25

2× 100=

75x

200=

3x

8.

But we know that

Amount = Principal + Simple interest = x+3x

8=

(8x+ 3x)

8=

11x

8.

This amount is given to be 2,502.50. Therefore, we obtain 2502.50 =11x

8.

Solving for x, we get

x =2502.5× 8

11= 1820.

Hence, the sum is 1,820. The interest is

2,502.50 - 1,820= 682.50.

Example 20. 800 amounts to 920 in 3 years at a certain rate of

interest. If the rate of interest is increased by 3%, what would the amount

will become?

Solution: Recall, interest(I) = amount(A) - principal(P ). Hence,

I = 920 - 800 = 120.

This interest is accrued in T= 3 years, for the principal 800. Using,

I =PTR

100we get,

R =100× I

P × T=

100× 120

800× 3= 5.

Thus the original rate of interest is 5%. After the increase of interest by

3%, the new rate of interest, which we again denote by R = 5% + 3% =

8%. The principal P = 800 and the period T = 3 years remain the same.

Therefore,

I =PTR

100=

800× 3× 8

100= 192.

Therefore, the new amount = 800 + 192 = 992.

Exercise 1.4.6

1. Find the simple interest on 2,500 for 4 years at 614% per annum.

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2. Find the simple interest on 3,500 at the rate of 212% per annum for

165 days.

3. In what period will 5,200 amounts to 7,384 at 12% per annum

simple interest?

4. Ramya borrowed a loan from a bank for buying a computer. After

4 years she paid 26,640 and settled the accounts. If the rate of

interest is 12% per annum, what was the sum she borrowed?

5. A sum of money triples itself in 8 years. Find the rate of interest.

1.4.7 Tax

The Government requires money for its functioning. Money required

for a Government is collected from the public in the form of taxes. One

such method of collecting money is Sales Tax.

Sales Tax is the tax we pay when we buy goods/articles from a shop.

Sales Tax is charged by the Government on the sale of every good/article.

Sales tax is called indirect tax as it is collected from the manufacturer,

wholesaler and retailer (shopkeeper) who in turn collects it from the cus-

tomer.

Value added tax (VAT)

Value Added Tax(VAT) is a revised version of sales tax. Normally an

article, before it reaches consumer, passes through various stages as given

here:

Manufacturer→ Wholesaler → Retailer → Consumer.

The person/company who/which manufactures an article is Manufac-

turer. Depending on the cost of producing an article, manufacturer marks

the price higher than the cost price. On this marked price manufacturer

has to charge sales tax, which he pays to the Government.

The person who purchases large quantities of articles from the manufac-

turer is the Wholesaler. He marks a price higher than the price he pur-

chased from the manufacturer(as he has to get his profit). On this marked

price, he also charges sales tax.

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The person who buys articles in smaller quantities from the wholesaler

is the Retailer. He marks the price higher than his cost price(he again

takes his profit). On this marked price, he charges sales tax.

The common people who purchase the articles from the shop are Con-

sumers. For the consumer, the cost price is the marked price of the re-

tailer plus the sales tax on the marked price. Thus, VAT is a tax on the

value added at each stage for a product that has to pass through various

stages in the channel of distribution.

Remember:

No shopkeeper sells any article at loss. Even when a high dis-

count is given, he makes a profit. The discount is given to

attract customers and to expedite sales so that he can reinvest

his principal.

Example 21. Abdul purchases a pair of clothes with a marked price

1,350. If the rate of sales tax is 4%, calculate the amount to be paid by

him.

Solution: Marked price of the item is 1,350, and sales tax is 4%, on the

marked price. Hence the total sales tax on the item is4

100× 1350 = 54.

Amount to be paid = Marked price + tax = 1,350 + 54= 1,404. Hence,

the amount to be paid by him is 1,404.

Example 22. A blazer marked at 1,600 is billed at 1,696. Find the

rate of sales tax.

Solution: Selling price is 1,696 and marked price is 1,600. Hence

sales tax paid is 96. On 1,600, sales tax is 96. Hence the percentage

of sales tax is96

1600× 100 = 6.

Hence, the rate of sales tax is 6%.

Exercise 1.4.7

1. A person purchases the following items from a mall for which the

sales tax is mentioned against:

(a) Stationary materials for 250 and sales tax of 4% there on;

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(b) Electronic goods worth 2,580 and sales tax of 10% there on;(c) Groceries worth 1,200 on which sales tax of 3% is levied;(d) Medicines worth 200 with sales tax of 6%.

Find the bill amount for each item.

2. A person buys electronic goods worth 10,000 for which the sales

tax is 4% and other material worth 15,000 for which the sales tax

is 6%. He manufactures a gadget using all these and sells it at 15%

profit. What is his selling price?

3. A trader purchases 70 kg of tea at the rate of 200 /kg and another

30 kg at the rate of 250/kg. He pays a sales tax of 4% on the

transaction. He mixes both of them and sells the product at the rate

of 240/kg. What is the percentage gain or loss(find approximate

value)?

Additional problems on “Commercial Arithmetic”

1. Four alternative options are given for each of the following statements.

Select the correct option.

(a) Nine percent of 700 is:

A. 63 B. 630 C. 6.3 D. 0.63

(b) What percent of 50 metres is 12 metres?

A. 20% B. 60% C. 24% D. 32%

(c) The number whose 8% is 12 is:

A. 120 B. 150 C. 130 D. 140

(d) An article costing 600 is sold for 750. The gain percentage is:

A. 20 B. 25 C. 30 D. 35

(e) By selling note book for 22 a shopkeeper gains 10%. The cost

price of the book is:

A. 18 B. 30 C. 20 D. 22

(f) The percentage of loss, when an article worth 10,000 was sold

for 9,000 is:

A. 10 B. 20 C. 15 D. 25

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98 Unit 4

(g) A radio marked 1000 is given away for 850. The discount is:

A. 50 B. 100 C. 150 D. 200

(h) A book marked 250 was sold for 200 after discount. The

percentage of discount is:

A. 10 B. 30 C. 20 D. 25

(i) The marked price of an article is 200. If 15% of discount is

allowed on it, its selling price is:

A. 185 B. 170 C. 215 D. 175

(j) One sells his bike through a broker by paying 200 brokerage.

The rate of brokerage is 2%. The selling price of the bike is:

A. 12,000 B. 10,000 C. 14,000 D. 12,500

(k) The brokerage amount for a deal of 25,000 at 2% rate of com-

mission is:

A. 500 B. 250 C. 5,000 D. 2,500

(l) If 1,600 is the commission at 8% for goods sold through a broker

, the selling price of the goods is:

A. 18,000 B. 20,000 C. 22,000 D. 24,000

(m) The simple interest on 5,000 at 2% per month for 3 months is:

A. 100 B. 200 C. 300 D. 400

(n) The time in which simple interest on a certain sum be 0.15 times

the principal at 10% per annum is:

A. 1.5 years B. 1 year C. 2 years D. 2.5 years

(o) The principal that yields a simple interest of 1,280 at 16% per

annum for 8 months is:

A. 10,000 B. 12,000 C. 12,800 D. 14,000

2. A time interval of 3 minutes and 20 seconds is wrongly measured as

3 minutes and 25 seconds. What is the percentage error?

3. Hari reads 22% of the pages of a book on the first day, 53% on the sec-

ond day and 15% on the third day. If the number of pages remaining

to be read is 30, find the total number of pages in the book.

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4. If 55% of students in a school are girls and the number of boys is 270,

find the number of girls in the school.

5. By selling an article for 920, a shop keeper gains 15%. Find the cost

price of the article.

6. Amit sells a watch at 20% gain. Had he sold it for 36 more, he would

have gained 23%. Find the cost price of the watch.

7. On selling apples at 40 per kg, a vendor incurs 10% loss. If he incurs

a total loss of 120, calculate the quantity(in kg) of apples he sold.

8. A dealer allows a discount of 20% and still gains 20%. Find the

marked price of an article which costs the dealer 720.

9. A shop keeper buys an article for 600 and marks 25% above the cost

price. Find (i) the selling price if he sells the article at 10% discount;

(ii) the percentage of discount if it is sold for 690.

10. A retailer purchases goods worth 33, 600 and gets a discount of 14%

from a whole seller. For paying in cash, the whole seller gives an

additional discount of 1.5% on the amount to be paid after the first

discount. What is the net amount the retailer has to pay?

11. An old car was disposed through a broker for 42, 000. If the brokerage

is 212%, find the amount the owner gets.

12. A milk-man sells 20 litres of milk everyday at 22. He receives a com-

mission of 4% for every litre. Find the total commission he receives in

a month of 30 days.

13. A bike was sold for 48, 000 and a commission of 8, 640 was received

by the dealer. Find the rate of commission.

14. In how many years will a sum of money becomes three times at the

rate of interest 10% per annum?

15. In what time will the simple interest on a certain sum be 0.24 times

the principal at 12% per annum?

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100 Unit 4

16. Find the amount of 30, 000 from 15-th January, 2010 to 10-th Au-

gust, 2010 at 12% per annum.

17. A person purchases electronic items worth 2,50,000. The shop

keeper charges him a sales tax of 21% instead of 12%. The purchaser

does not realise that he has over paid. But after some time he finds

that he has paid excess and asks the shop-keeper to return the ex-

cess money. The shop-keeper refuses and the purchaser moves the

consumer court. The court with due hearing orders the shop-keeper

to pay the purchaser the excess money paid by the way of sales tax,

with an interest of 12% per annum. If the whole deliberation takes 8

months, what is the money that the purchaser gets back?

Glossary

Percentage: it means for every hundred; thus 6% means 6 for hundred.

Rate percent: when the percent is expressed as a fraction with denomi-

nator equal to 100, the numerator is called rate percent.

Cost price: the money paid to procure an item is called its cost price and

abbreviated as C.P.

Overhead charges: it is the additional charges on the item before it is

ready for sale, like, labour, transportation, etc.

Selling price: it is the price at which goods are sold; it is abbreviated as

S.P.

Profit: it is the gain in a transaction; it is equal to S.P- C.P. whenever S.P.

> C.P.

Loss: if S.P. is less than C.P., then C.P.-S.P. is the loss.

Discount: it is the reduction on the marked price of an item given by the

seller to attract customers.

Brokerage or commission: it is the money charged by a mediator for the

service provided for a smooth transaction of sales between a seller and a

buyer.

Simple interest: it is the money charged by a lender on the receiver for

the amount lent for a period of time.

Principal: it is the money lent for a period of time by a lender to a receiver.

Rate: it is the money charged as interest for every 100 for one year.

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Time: it is the period for which money is lent.

Amount: principal plus interest is called amount.

Sales tax: on every sales done, the Government levies a certain amount of

money; it is called sales tax; this is a certain percent on the marked price

of an item.

VAT: Value Added Tax; the sales tax goes on adding when there is multiple

sales of goods.

Manufacturer: a company or a factory at which goods originate for sales.

Wholesaler: the purchaser who buys goods in bulk, and in turn sells it to

small vendors.

Retailer: the vendor who buys goods in small quantities and sells it.

Consumer: the ultimate user of the goods.

Points to remember

• Percentage is a method of comparing quantities of the same kind.

• If SP > CP, then SP − CP is the profit; if SP < CP, then CP − SP is the

loss.

• Profit or loss is always calculated on the cost price.

• Discount is the reduction given in the marked price.

• The money that an agent (or broker) receives in a deal is the commis-

sion(or brokerage).

• While calculating the simple interest, the day on which money is de-

posited is not counted where as the day on which it is withdrawn is

taken in to account.

• VAT means value added tax.

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CHAPTER 1 UNIT 5

STATISTICS

After learning this chapter, you learn to:

• explain the terms data, observation, range, frequency, class interval,

exclusive and inclusive class intervals size of class interval, mid-point

of class interval;

• construct frequency distribution table for exclusive and inclusive class

intervals;

• draw histogram for the given frequency distribution;

• define mean, median and mode;

• calculate mean for grouped and un-grouped data;

• calculate median for grouped and un-grouped data;

• identify mode of un-grouped and grouped data.

1.5.1 Introduction

Statistics is considered to be a mathematical science pertaining to the

collection, analysis, interpretation and presentation of data.

Statistics is useful in drawing conclusions from numerical data. It is also

useful to predict weather, to obtain information concerning business, im-

port, export, education etc.. All research and investigation require statis-

tical interpretation.

A collection of numerical facts with particular information is called data.

Consider, the marks obtained by 20 students in mathematics in 8th

standard mid-term examination:

56, 31, 44, 78, 67, 74, 38, 60, 56, 59, 87, 73, 38, 77, 84, 80, 49, 60, 60, 71.

The above data is a collection numerical entries. It is called observation.

Such a collection of data is called raw data.

The data can be arranged in ascending or descending order. Arranged in

descending order, we get,

87, 84, 80, 78, 77, 74, 73, 71, 67, 60, 60, 60, 59, 56, 56, 49, 44, 38, 38, 31.

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Statistics 103

From this, we can infer that the highest score is 87 and lowest score is 31.

The difference between the highest and the lowest score is called range.

The range of above data is (87 -31) = 56.

We can observe that the scores 38 and 60 are repeated. The number

38 is repeated twice and 60 is repeated thrice. We say that the frequency

of 38 is 2 and the frequency of 60 is 3. The frequency of rest of the scores

is 1. The number of times a particular observation (score) occurs in a data

is called its frequency. The above data may be represented in a tabular

form, showing the frequency of each distribution. This representation in

tabular form is called Frequency Distribution Table. Tallies are used to

mark the counts; III represents three(3) counts, where as I I I I represents 5

counts.

Example 1. The marks scored by 20 students in a unit test out of 25

marks are given below.

12, 10, 08, 12, 04, 15, 18, 23, 18, 16, 16, 12, 23, 18, 12, 05, 16, 16, 12, 20.

Prepare a frequency distribution table.

Solution: The table looks like:

Marks Tally No. of students

marks (frequency)

23 II 2

20 I 1

18 III 3

16 IIII 4

15 I 1

12 I I I I 5

10 I 1

08 I 1

05 I 1

04 I 1

Total 20 20

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104 Unit 5

1.5.2 Grouping Data

Organising the data in the form of frequency distribution table is called

grouped frequency distribution of raw data. Sometimes, we have to deal

with a large data.

Example 2. Consider the following marks (out of 50) scored in mathemat-

ics by 50 students of 8th class:

41, 31, 33, 32, 28, 31, 21, 10, 30, 22, 33, 37, 12, 05, 08, 15, 39, 26, 41,

46, 34, 22, 09,

11, 16, 22, 25, 29, 31, 39, 23, 31, 21, 45, 47, 30, 22, 17, 36, 18, 20, 22,

44, 16, 24, 10,

27, 39, 28, 17.

Prepare a frequency distribution table.

Solution: If we prepare a frequency distribution table for each observa-

tion, then the table would be too long. So for convenience, we make groups

of observations like 0 - 9, 10 - 19 and so on. We obtain a frequency of dis-

tribution of the number of observations coming under each group. In this

way, we prepare a frequency distribution table for the above data as below:

Groups Tally marks Frequency

0 - 9 III 03

10 - 19 I I I I , I I I I 10

20 - 29 I I I I , I I I I , I I I I , I 16

30 - 39 I I I I , I I I I , I I I I 15

40 - 49 I I I I , I 06

50 - 59 0

Total 50 50

The data presented in this manner is said to be grouped and the dis-

tribution obtained is called grouped frequency distribution. Grouped

frequency distribution table helps us to draw meaningful inferences like:

1. most of the students have scored between 20 and 29;

2. only 3 students have scored less than 10;

3. no student has scored 50 or more than 50.

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Statistics 105

In the above table, marks are grouped into 0 - 9, 10 - 19 and the like. No

score overlaps in any group. Each of these groups is called a class interval

or a class. This method of grouping data is called Inclusive Method.

Class Limit: In the class interval, say (10 - 19), 9.5 is called the lower class

limit and 19.5 is called the upper class limit.

Note: To find the class limit, in inclusive method, subtract 0.5 from

lower score to get lower class limit and add 0.5 to the upper score to

get upper class limit.

Class Size: The number of scores in the class interval say, (10-19), in-

cluding 10 and 19, is called the class size or width of the class. In this

example, the class size is 10.

Class Mark: The midpoint of a class is called its class mark (or midpoint

of class interval). It is obtained by adding the two limits and dividing by 2.

For example, the class mark of (10-19) is (10+19)/2 = 14.5. The class mark

of (10-20) is (10+20)/2 = 15.

The data in Example 2 can also be grouped in class intervals like 0 -

10, 10 - 20, 20 - 30 and so on. The frequency distribution table will then

be as follows:

Groups Tally marks Frequency

0 - 10 III 03

10 - 20 I I I I , I I I I 10

20 - 30 I I I I , I I I I , I I I I , III 18

30 - 40 I I I I , I I I I , III 13

40 - 50 I I I I , I 06

Total 50 50

Here observe that 10 occurs in both the classes (0 - 10) as well as (10

-20). But it is not possible that an observation (say 10) can belong to two

classes (0 - 10) and (10 - 20) simultaneously. In order to avoid this, we

follow a convention that the common observation (here 10) will belong to

the higher class, that is 10 belongs to (10 - 20) and not to (0 - 10). Similarly

30 belongs to (30 - 40) and not (20 -30). This method of grouping the data

is called Exclusive method.

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106 Unit 5

Class Limit: In the class interval (10 - 20), 10 is called the lower limit and

20 is called the upper limit.

Class Size: The difference between the upper limit and the lower limit is

called the class size or width. The width of class in (10 - 20) is 20 - 10 =

10.

Example 3. Forty candidates from 10th class of a school appear for a

test. The number of questions (out of 60) attempted by them in forty five

minutes is given here.

52, 42, 40, 36, 12, 28, 15, 37, 35, 22, 39, 50, 54, 39, 21, 34, 46, 31, 10,

09,

13, 24, 29, 31, 49, 58, 40, 44, 37, 28, 13, 16, 29, 36, 39, 41, 47, 55, 52,

09.

Prepare a frequency distribution table with the class size 10 and answer

the following:

(i) Which class has the highest frequency? (ii) Which class has the lowest

frequency? (iii) Write the upper and lower limits of the class (20- 29).

(iv) Which two classes have the same frequency?

Solution: Let us first prepare the frequency distribution table pertaining

to this data.

Class-Interval Tally marks Frequency

0 - 9 II 2

10 - 19 I I I I , I 6

20 - 29 I I I I , II 7

30 - 39 I I I I , I I I I , I 11

40 - 49 I I I I , III 8

50 - 59 I I I I , I 6

Total 40 40

Using this table, we can observe:

(i) (30-39) has the highest frequency;

(ii) (0-9) has the lowest frequency;

(iii) upper limit is 29.5 and lower limit is 19.5;

(iv) (10-19) and (50-59) have the same frequency.

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Statistics 107

Example 4. The heights of 25 children in centimetre are given below:

174, 168, 110, 142, 156, 119, 110, 101, 190, 102, 190, 111, 172, 140,

136, 174, 128, 124, 136, 147, 168, 192, 101, 129, 114.

Prepare a frequency distribution table, taking the size of the class interval

as 20, and answer the following:

(i) Mention the class intervals of highest and lowest frequency.

(ii) What does the frequency 6 corresponding to class interval (160-180)

indicate?

(iii) Find out the class mark (or midpoint) of (140-160).

(iv) What is the range of heights?

Solution: The frequency distribution for the given data is as follows:

Class-Interval Tally marks Frequency

100 - 120 I I I I , III 8

120 - 140 I I I I 5

140 - 160 III 3

160 - 180 I I I I , I 6

180 - 200 III 3

Total 25 25

Answers:

(i) Highest frequency: (100-120); lowest frequency: (140-160). and (180-

200).

(ii) There are 6 children whose heights are in the range 160 cm to 180

cm.

(iii) Mid-point = (140+160)/2 = 150

(iv) Range = highest score- lowest score= 192 - 101 = 91.

Exercise 1.5.2

1. The marks scored by 40 candidates in an examination (out of 100) is

given below:

75, 65, 57, 50, 32, 54, 75, 67, 75, 88, 80, 42, 40, 41, 34, 78, 43, 61,

42, 46, 68, 52, 43, 49, 59, 49, 67, 34, 33, 87, 97, 47, 46, 54, 48, 45,

51, 47, 41, 43.

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108 Unit 5

Prepare a frequency distribution table with the class size 10. Take

the class intervals as (30 - 39), (40 - 49), . . . and answer the following

questions:

(i) Which class intervals have highest and lowest frequency?

(ii) Write the upper and lower limits of the class interval 30-39.

(iii) What is the range of the given distribution?

2. Prepare the frequency distribution table for the given set of scores:

39, 16, 30, 37, 53, 15, 16, 60, 58, 26, 28, 19, 20, 12, 14, 24, 59,

21, 57, 38, 25, 36, 34, 15, 25, 41, 52, 45, 60, 63, 18, 26, 43, 36, 18,

27, 59, 63, 46, 48, 25, 33, 46, 27, 46, 42, 48, 35, 64, 24. Take class

intervals as (10 -20), (20 - 30), . . . and answer the following:

(i) What does the frequency corresponding to the third class interval

mean?

ii) What is the size of each class interval? Find the midpoint of the

class interval 30 - 40.

iii) What is the range of the given set of scores?

1.5.3 Histogram

A histogram is a representation of a frequency distribution by means

of rectangles whose widths represent class intervals and whose areas are

proportional to the corresponding frequencies. In a histogram, frequency

is plotted against class interval. Thus, a histogram is a two-dimensional

graphical representation of data. However, if the length of all the class

intervals are the same, then the frequency is proportional to the height of

the rectangle.

Construction of a histogram

We will show how to construct histograms taking some examples.

Example 5. Draw the histogram of the following frequency distribution:

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Statistics 109

Class - Interval Frequency

0 - 9 5

10 - 19 8

20 - 29 12

30 - 39 18

40 - 49 22

50 - 59 10

Solution: The given distribution is in inclusive form. It should be con-

verted into exclusive form. This can be done by applying a correction

factord

2, where

d = (lower limit of a class) - (upper limit of a class before it)

Here, we have

actual upper limit = stated limit +d

2;

actual lower limit = stated limit -d

2.

For example, consider the class limit 10 - 19. You get

d = lower limit of the class interval -upper limit of class before it =10-9 =1.

Hence, d = 1 ord

2= 0.5. Now,

actual upper limit = (stated upper limit) +d

2= 19 + 0.5 =19.5.

actual lower limit = (stated lower limit) -d

2= 10 - 0.5= 9.5

Converting into exclusive form, we get the table as below:

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110 Unit 5

Stated Actual Frequency

Class Class

interval interval

0 - 9 -0.5 - 9.5 5

10 - 19 9.5 - 19.5 8

20 - 29 19.5 - 29.5 12

30 - 39 29.5 - 39.5 18

40 - 49 39.5 - 49.5 22

50 - 59 49.5 - 59.5 10

9.5 19.5 29.5 39.5 49.5 59.5−0.5

15

10

5

20

22

X

Y

frequ

ency

class intervals

Construction of histogram:

1. Draw x-axis and y-axis. Choose a proper scale for x and y axes, say

on x-axis: 1cm = 10 and on y-axis: 1cm = 5.

2. Mark the class intervals on x-axis; (0.5 - 9.5), (9.5 - 19.5) and the like.

3. Draw a rectangle of height 5cm on the first class interval (0.5 - 9.5).

4. Draw a second rectangle of height 8 cm on the second class interval

and follow the same procedure for the rest of the class intervals and

the corresponding frequencies.

Then, the histogram takes the form shown above.

In the above histogram, we observe that,

• there are no gaps between rectangles, showing that the distribution

is continuous;

• the heights of rectangles represent the frequencies and the base rep-

resents class intervals.

Remember:

• In a bar graph, the height of the bar represents the data. The

bars may be separated or closed.

• In a histogram, the area of each rectangle represents correspond-

ing data (frequency). There should be no gaps between rectangles.

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Statistics 111

Example 6. Draw the histogram for the following frequency distribution.


0 - 5 5

5 - 10 8

10 - 15 15

15 - 20 4

20 - 25 104

8

12

16

5 10 15 20 25

Y

X

frequ

ency

class intervals

Solution: The given distribution is in exclusive form. So, we can take the

class-intervals as( 0 - 5), (5 - 10) etc., along the x-axis and frequency along

y-axis.Choosing a proper scale, we can construct a histogram as explained

in the previous example.

Note: Histogram is drawn for class intervals which are exclusive

(continuous). If the class intervals are given in inclusive (discrete), it

should be converted into exclusive form by applying the correction

factor.

Exercise 1.5.3

1. Draw a histogram to represent

the following frequency distri-

bution.


20 -25 5

25 - 30 10

30 - 35 18

35 - 40 14

40 - 45 12

2. Draw a histogram to represent

the following frequency distri-

bution.


10 - 19 7

20 - 29 10

30 - 39 20

40 - 49 5

50 - 59 15

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112 Unit 5

1.5.4 Mean, Median and Mode

Now we study three important quantities associated with a statistical

data. They give a clear picture of the behaviour of an experiment. They

are generally called measures of central tendencies.

Mean

Mean is a commonly used as a measure, in a given statistical experi-

ment, to get an idea how the experiment is behaving. This is simply the

average of the numerical data collected during an experiment.

Mean for an un-grouped data:

It is the sum of the numerical values of all the observations divided by

the total number of observations. If x1, x2, x3, . . . , xN are the values of N

observations, then

Mean =(sum of all values of observations)

(the number of observations)=

x1 + x2 + x3 + · · ·+ xNN

.

The sum of N values of x is represented by∑

x. Here∑

stands for sum-

mation notation. Therefore,

X =

∑x

N

Note: Sum is denoted by∑

and read as

sigma.

Mean is denoted by X. This is read as X-

bar.

Example 7. Find the mean of first six even natural numbers.

Solution: The first six even natural numbers are 2, 4, 6, 8, 10, 12. There

are six scores. Therefore, N = 6. The observations are x1 = 2, x2 = 4,

x3 = 6, x4 = 8, x5 = 10, x6 = 12. Hence

∑x = 2 + 4 + 6 + 8 + 10 + 12 = 42

Hence the mean is given by

X =

∑x

N=

42

6= 7.

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Statistics 113

Example 8. Marks scored by Hari in 5 tests (out of 25 marks) are given:

24, 22, 23, 23, 25. Find his average score.

Solution: We observe that there are 5 scores, so that∑

x = 24 + 22 + 23 +

23 + 25 = 117. Hence the mean is given by

X =

∑x

N=

117

5= 23.4.

In the above examples, the number of values is very less. So we could

find the mean easily. If large number of values is given and we need to find

the mean, it is difficult. In these cases, where the given data is more, we

group the data and prepare a frequency distribution table. From frequency

distribution table, we can find the mean.

Mean of a grouped data:Example 9: The number of goals scored by a hockey team in 20 matches

is given here:

4, 6, 3, 2, 2, 4, 1, 5, 3, 0, 4, 5, 4, 5, 4, 0, 4, 3, 6, 4.

Find the mean.

Solution: To find the mean, let us prepare a frequency distribution table

first. We observe that some scores are repeated. So to find the sum of all

scores, we have to multiply each score with its frequency and then find

the sum.

Scores Tally marks Frequency

0 II 2

1 I 1

2 II 2

3 III 3

4 I I I I , II 7

5 III 3

6 II 2

N = 20

Scores Frequency fx

(x) (f )

0 2 0

1 1 1

2 2 4

3 3 9

4 7 28

5 3 15

6 2 12

N = 20∑

fx = 69

To do this, let us denote the scores by x and frequency by f , then multiply

f and x and add the product fx. Here∑

fx denotes the sum of all the

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114 Unit 5

products f × x. Now the mean is

X =sum of the scores

number of scores=

∑fx

N=

69

20.

Thus we get X = 3.45.

Example 10. Find the mean for the given frequency distribution table.


0 - 4 3

5 - 9 5

10 - 14 7

15 - 19 4

20 - 24 6

N = 25

Solution: To find the mean, first we

have to find the mid-point of each class

interval. Mid-point of 0 - 4 = (0+4)/2 =

2; mid-point of 5 - 9 = (5+9)/2= 7 and

the like. Denote the midpoint of the

class interval by x. Write down the fre-

quencies f corresponding to each class

interval.

Class interval Mid-point of CI (x) Frequency(f ) fx

0 - 4 2 3 6

5 - 9 7 5 35

10 - 14 12 7 84

15 - 19 17 4 68

20 - 24 22 6 132

N = 25∑

fx = 325

Multiply f and x to get fx. Add all fx and find out∑

fx. Now the mean is

calculated using the formula,

X =sum of all scores

total number of scores=

∑fx

N.

Thus we get

X =

∑fx

N=

325

25= 13.

Therefore mean is 13.

Activity 1:

Mark the height in centimetres on a wall in your school. (Take the help of

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Statistics 115

your teacher). Measure and record the height of 10 of your friends. Find

the mean height.

Median

Median is the mid-point of the data(raw scores), after being arranged in

ascending or descending order. Median divides the given set of scores into

two equal halves, that is there are as many scores below the median as

above the median.

The mean depends on the nature of scores. If some scores are

very high(or low), that will influence the mean. For example;

consider the data 5,8, 6, 9, 12, 110,130. If you compute the

mean, it is 40(the sum is 280 and there are 7 scores). But there

are 5 scores below 40 and only two scores above 40. Hence it is

not central. On the other hand the median is 9 and you see that

it is in the centre. Thus mean is influenced by unusually high

scores in the given data and may not be a true representative of

an experiment. In such cases median is preferred.

Median for an un-grouped data:

Arrange the given set of scores in ascending or descending order of magni-

tudes (values). If the total number of scores is odd, then the middle most

score is the median. If the total number of scores is even, then the average

of the two middle most scores is the median.

Example 11. Find the median of the data: 26, 31, 33, 37, 43, 8, 26, 33.

Solution: Arranging the scores in ascending order, we have 26, 31, 33,

37, 38, 42, 43.

Here the number of terms is 7. The middle term is the 4-th one and it is

37. Therefore, median is 37.

Think it over:

Do you get the same median if the scores are arranged in de-

scending order?

Example 12. Find the median of the data: 32, 30, 28, 31, 22, 26, 27, 21.

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116 Unit 5

Solution: Arranging in descending order, we obtain 32, 31, 30, 28, 27,

26, 22, 21.

There are 8 terms. Therefore, median is the average of the two middle

terms, which are 27 and 28. Thus the median is(27 + 28)

2= 27.5.

Think it over:

Do you get the same median if the scores are arranged in as-

cending order?

Note: When N scores are given, we can use the following method to find

the median:

First arrange the scores in ascending or descending order; (i) If N is odd,

then median is the score at(N + 1)

2-th place; (ii) If N is even, median is

1

2(the score at

N

2-th place + the score at

(N

2+ 1

)-th place

).

Median for a grouped data

In the case of an un-grouped data, you could compute its median as the

middle score if the number of scores is odd(or the mean of its two middle

scores in the case of even number of scores). We have to adopt a different

way of computing the median of a grouped data. We describe it through

examples.

Example 13. Find the median for the following grouped data.


1 - 5 4

6 - 10 3

11 - 15 6

16 - 20 5

21 - 25 2

N = 20

Solution: Median is the middle

score. Here actual scores are not

given. So we have to use a differ-

ent procedure to find the median.

We have N = 20, an even num-

ber. Therefore, there are two mid-

dle scores: one corresponding to

N/2 = 20/2 = 10-th score and the

other is 11-th score.

We have to locate the median between 10-th and 11-th scores. Since we

are not given individual scores, we have to get some idea about 10-th and

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Statistics 117

11-th scores. For this it is required to find the cumulative frequency.

Observe the following table, to know how to find cumulative frequency.

Class Frequency(f ) Cumulative

Interval frequency(fc)

1 - 5 4 4 4

6 - 10 3 7 4+3=7

11 - 15 6 13 7+6=13

16 - 20 5 18 13+5=18

21 - 25 2 20 18+2=20

N = 20

Observe that the cumulative frequency corresponding to the last class

interval is equal to N. Counting frequencies from first class interval

downwards, we find that the 10-th score lies in the class interval (11 - 15).

This class interval (11 - 15) is called the median class. The frequency cor-

responding to this is 6. Its lower real limit (LRL) is 10.5. The cumulative

frequency above this class is 7. Now, knowing:

(a) lower real limit(LRL)=10.5;

(b) frequency of the median class (fm) = 6;

(c) cumulative frequency above the median class (fc) = 7; and

(d) size of the class interval(i) = 5;

we can find the median, using the formula

median = LRL+((N/2)− fc)

fm× i.

Thus we get

median = 10.5 +(20/2)− 7

6× 5 = 10.5 +

(10− 7)

6× 5

= 10.5 +3

6× 5 = 10.5 + 2.5 = 13.

Note: Median = LRL+(N/2− fc)

fm× i. This formula can be derived from the

basic principles.

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118 Unit 5

Example 14. Calculate the median for an exclusive(continuous) distribu-

tion given below.

Class Interval Frequency

10 - 20 11

20 - 30 13

30 - 40 13

40 - 50 9

50 - 60 4

Class Frequency Cumulative

Interval (f ) frequency (fc)

10 - 20 11 11

20 - 30 13 24

30 - 40 13 37

40 - 50 9 46

50 - 60 4 50

Solution: We first prepare the cumulative frequency table(see above). Here

you see that the total number of observation is N = 50. Therefore (30 - 40)

is the median class. We also observe that LRL = 30, fc = 24, fm = 13 and

i = 20− 10 = 10. Now we are in a position to use the formula for median:

median = LRL+(N/2)− fc

fm× i = 30 +

(25− 24)

13× 10 = 30 +

10

13= 30.77,

approximately.

Mode

There is another measure of central tendency which is used occasion-

ally, called mode. Mode is the score that occurs frequently in a given set

of scores. Mode is the value around which the other scores cluster around

densely.

Mode for an un-grouped data:

Example 15. Find the mode for the data: 15, 20, 22, 25, 30, 20, 15, 20,

12, 20.

Solution: Here 20 appear maximum times (4 times). Therefore, mode is

20.

Example 16. Find the mode of the data: 5, 3, 3, 5, 7, 6, 3, 4, 3, 5, 8, 5.

Solution: Here 3 and 5 appear 4 times. Therefore modes are both 3 and

5.

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Statistics 119

Note: A collection of data can have more than one mode. If the

data has only one mode, we say it has uni-mode, if it has 2

modes, we say it has bi-mode and if it has more than 2 modes,

we say it has multi-mode.

Mode for a grouped data

For a grouped data, the same score having maximum frequency is

the mode.

Example 17. Find the mode for the following data.

Number 12 13 14 15 16 17

Frequency 7 9 6 22 20 19

Solution: Here the maximum frequency is 22. Therefore, the number 15

corresponding to maximum frequency is the mode.

Hence, the mode is 15.

Exercise 1.5.4

1. Runs scored by 10 batsmen in a one day cricket match are given.

Find the average run scored.

23, 54, 08, 94, 60, 18, 29, 44, 05, 86

2. Find the mean weight from the following table.

Weight(kg) 29 30 31 32 33

No. of children 02 01 04 03 05

3. Calculate the mean for the following frequency distribution:

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency 3 7 10 6 8 2 4

4. Calculate the mean for the following frequency distribution:

Marks 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44

Frequency 6 5 9 12 6 2

5. Find the median of the data: 15, 22, 9, 20, 6, 18, 11, 25, 14.

6. Find the median of the data: 22, 28, 34, 49, 44, 57, 18, 10, 33, 41,

66, 59.

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120 Unit 5

7. Find the median for the following frequency distribution table:

Class 110-119 120-129 130-139 140-149 150-159 160-169

Intervals

Frequency 6 8 15 10 6 5

8. Find the median for the following frequency distribution table:

Class-interval 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30

Frequency 5 3 9 10 8 5

9. Find the mode for the following data:

(i) 4, 3, 1, 5, 3, 7, 9, 6

(ii) 22, 36, 18, 22, 20, 34, 22, 42, 46, 42

10. Find the mode for the following data:

x 5 10 12 15 20 30 40

f 4 8 11 13 16 12 9

Additional problems on “Statistics”

1. Four alternative options are given for each of the following statements.

Select the correct option.

(a) The size or width of the Class interval (0 - 4) is :

A. 4 B. 5 C. 3 D. 0

(b) The midpoint of class interval (10 - 19) is:

A. 10 B. 14 C. 15 D. 14.5

(c) The difference between highest and lowest score of a distribution

gives:

A. class interval B. class width C. range D. class limit

(d) The number of times a particular observation (score) occurs in a

data is called its:

A. frequency B. range C. class interval D. class limit

(e) In inclusive form, the actual upper limit and lower limit of class

interval (0 - 4) are:

A. -0.5 & 3.5 B. 0.5 & 4.5 C. −1 & 5 D. 1 & 5

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Statistics 121

(f) The height of a rectangle in a histogram represents:

A. class interval B. midpoint C. frequency density D. fre-

quency

(g) In a histogram, the width of the rectangle indicates:

A. class interval B. midpoint C. frequency density D. fre-

quency

(h) The mean of scores 10, 15, 12, 15, 15 is:

A. 15 B. 13 C. 13.4 D. 14.3

(i) Class interval grouping of data is done when:

A. the range of data is small B. the range of data is large

C. the class intervals are small D. class intervals are large

(j) The mean of 6, 4, 7, x and 10 is 8. The value of x is:

A. 10 B. 12 C. 14 D. 13

(k) If n = 10 and Mean = 12, then∑

fx is:

A. 120 B. 1200 C. 12 D. 13

(l) The mean of first three multiples of 5 is :

A. 5 B. 10 C. 15 D. 30

(m) The median of 37, 83, 70, 29, 32, 42, 40 is:

A. 29 B. 30 C. 40 D. 42

(n) In an inclusive class interval (10 - 14), the lower real limit is:

A. 9.5 B. 10.5 C. 13.5 D. 14.5

(o) In an exclusive class interval (10 - 20), the lower real limit is:

A. 20 B. 10 C. 10.5 D. 20.5

(p) The mode of 2, 3, 3, 5, 3, 5, 7, 3, 5 is:

A. 3 B. 5 C. 3 and 5 D. 3,5,7

(q) For given two values of x, 16, 18 the frequencies are respectively

12 and 20. Then the mode is:

A. 16 B. 18 C. 12 D. 20

(r) A collection of data having more than 3 modes is said to be:

A. uni-mode B. bi-mode C. tri-mode D. multi-mode

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122 Unit 5

2. Prepare a frequency distribution table for the scores given:

42,22,55,18,50,10,33,29,17,29,29,27,34,15,40,42,40,41,35, 27,

44,31,38,19,54,55,38,19,20,30,42,59,15,19,27,23,40,32,28,51.

Take the class intervals as 10-20, 20-30, 30-40, 40-50, 50-60. From

the frequency distribution table answer the following questions:

(i) What does the frequency corresponding to the class interval 20-

30 indicate?

(ii) In which class intervals are the scores 10,20 and 30 included?

(iii) Find the range of the scores.

3. The following are the marks scored in a unit test(out of 25). Prepare

a frequency distribution table, taking the class intervals as 0-4, 5-9,

10-14, 15-19, 20-24:

21,14,3,7,23,18,24,16,18,17,20,10,17,18,21,23,19,12,14,9,16,18,12, 14,11.

From the table (i) find the mid-points of each class interval; (ii) find

the class interval having maximum frequency; (iii) find the range of

the scores.

4. Draw histogram for the fol-

lowing frequency distribution:


5-15 2

15-25 8

25-35 14

35-45 14

45-55 12

5. Draw histogram for the fol-

lowing frequency distribution:


0-10 4

11-20 18

21-30 12

31-40 6

41-50 20

51-60 10

6. The marks obtained by 12 students in a mathematics examination

are given below.

48,78,93,90,66,54,83,58,60,75,89,84.

Find (i) the mean of the marks; (ii) the mean mark of the students if

each student is given 4 grace marks.

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Statistics 123

7. If the mean of 8,12,21,42,x is 20, find the value of x.

8. Find the mean for the following distribution:

12,14,10,12,15,12,18,10,15,11,19,20,12,15,19,10,18,16,20,17.

Glossary

Data: collection of numerical facts with a particular information during

an experiment.

Observation: numerical display of data.

Score: the numerical entries in an observation.

Range: the difference between the highest and the lowest scores in an

observation.

Frequency: the number of times a particular score appears in an obser-

vation.

GFD: Grouped frequency distribution; the data is collected in to several

groups and the frequency of scores in each group is recorded.

Class interval: in a grouped frequency distribution, each group is called

a class interval.

Cumulative frequency: the sum of the frequencies up to the current class

interval.

FDT: Frequency Distribution Table; it displays the frequencies of scores

corresponding to various class intervals.

Inclusive method: while grouping, the end points of the groups do not

overlap.

Exclusive method: the end points of consecutive groups overlap.

Class limit: the end points of a class in exclusive method; the end points

of a class with correction factors in inclusive method.

Class size(or width): the difference between the upper class limit and the

lower class limit.

Class mark: the mid point of the class interval; it is equal to the average

between the upper class limit and the lower class limit.

Histogram: a graphical way of representing grouped data using rectangu-

lar bars in which the frequency is proportional to the area of the rectangle.

Mean: average of the scores; It is equal to the sum of the scores divided

by the number of scores.

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124 Answers

Median: the middle score.

Mode: the score which appears maximum number of times in an obser-

vation.

Median class: in a grouped data, it is difficult to say which is the median

as scores are not given explicitly; but one can say in which class interval,

the middle score lies. This is called the median class.

Points to remember

• Statistics is a branch of science which helps us to analyse the col-

lected data in a systematic way.

• Mean, median and mode are three different measures which, in some

sense, represent the given data. They are called measures of central

tendency.

Answers to Exercises in Chapter 1.

Exercise 1.1.2

1. (i) (3× 10) + (9× 1); (ii) (5× 10) + (2× 1); (iii) (1× 100) + (6× 1);

(iv) (3 × 100) + (5 × 10) + (9 × 1); (v) (6 × 100) + (2 × 10) + (8 × 1); (vi)

(3× 1000) + (4 × 100) + (5 × 10) + (8× 1); (vii) (9× 1000) + (5 × 100) + (2 × 1);

(viii) (7 × 1000). 2. (i) 56; (ii) 758; (iii) 6058; (iv) 7006; (v) 1010.

3. 555 = (4210)5 = (4 × 125) + (2 × 25) + (1 × 5) + (0 × 1). 4. 1024 =

(10000000000)2 = (1× 210).

Exercise 1.1.3

1. (i) B = 4; (ii) A = 5, B = 4; (iii) A = 5; (iv) A = 0; (v) two solutions:

A = 0, B = 0 and A = 1, B = 2; (vi) A = 6, B = 1. 2. A = 3, B = 4, C = 5.

3. A = 1, B = 7, C = 9. 4. Not possible. (Hint: AA = 11× A.)

Exercise 1.1.4

1. If s −→ (q, r) denotes the quotient q and remainder r, when s is divided

by 13, then:

8 −→ (0, 8); 31 −→ (2, 5); 44 −→ (3, 5); 85 −→ (6, 7); 1220 −→ (93, 11).

2. If s −→ (q, r) denotes the quotient q and remainder r, when s is divided

by 304, then

128 −→ (0, 128); 636 −→ (2, 28); 785 −→ (2, 177); 1038 −→ (3, 126); 2236 −→

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Answers 125

(7, 108); 8858 −→ (29, 42); 13765 −→ (45, 85); 58876 −→ (304, 204).

3. 107. 4. 62. 5. 46.

Exercise 1.1.5

1.

12 5 10

7 9 11

8 13 6

Magic sum is

27. Central

number is 9.

We have 27 =

3× 9.

2.

16 9 14

11 13 15

12 17 10

Magic sum is

39. Central

number is 13.

We have 39 =

3× 13.

3.

2 9 4

7 5 3

6 1 8

This is only one magic

square. You can construct

different magic squares for

different positions of 1, as

you have seen earlier.4.

15 1 11

5 9 13

7 17 3

5.

34 48 2 16 30

46 10 14 28 32

8 12 26 40 44

20 24 38 42 6

22 36 50 4 18

Exercise 1.1.6

1. 250 numbers. 3. The only numbers in the required form and divisible

by 11 are 4939, 4037, 4136, 4235, 4334, 4433, 4532, 4631, 4730. Hence

a + b = 18(in the case of 4939), or a + b = 7. 5. There are 4 numbers:

58476, 48576, 57684, 67584. (Use divisibility by 4 and 11 together.)

Additional problems on “Playing with numbers”

1. (a) A; (b) B; (c) C; (d) C; (e) A; (f) D; (g) D. 2. 24365. 3. 12.

4. 2, 5, 7, 8. 5. 1 = 3+3−5, 2 = 5−3, 3 = 5+5+5−3−3−3−3−3+3,

4 = 3+3+3−5, 5 = 5+5+5+5−3−3−3−3−3, 6 = 5+5+5+3+3−3−3−3−3−3,7 = 5 + 5 − 3, 8 = 5 + 3, 9 = 5 + 5 + 5 + 3 + 3 + 3 − 3 − 3 − 3 − 3 − 3,

10 = 5 + 5 + 5 + 5 + 5− 3− 3− 3− 3− 3. 6. 10, 20, 30, 40, 50, 60, 70, 80,

90, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72, 81, 84 7. 108

pages. 8. 70.

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126 Answers

9.

8 9 4

3 7 11

10 5 6

10. Only such number is 108. 11. x = 4,

y = 2 or x = 0, y = 6 or x = 9, y = 6. 12. Take

one group as {1, 2, 3, 5, 8, 7} and the other group as

{4, 6, 10}. Then (1×2×3×5×8×7)/(4×6×10) = 7 and

this is the minimum quotient. 13. 27314925

and 27364975. 15. 8066. 16. 12. 17. 62.Exercise 1.2.2

1. (i) 42 = 16; (ii) 82 = 64; (iii) 152 = 225. 2. 1, 36 = 62, 49 = 72, 81 = 92,

169 = 132, 625 = 252, 900 = 302, 100 = 102.

3. 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324, 361,

400,441,484. 4. You can take 200,201,204,205,206,209. None of these

is a square as they lie between 196 = 142 and 225 = 152.

5. 100,121,144,169,196,225,256,289,324,361,400.

Exercise 1.2.3

1. 1 + 3 + 5 + · · ·+ 51 = 262 = 676. 2. 144 = 122 = 1 + 3 + 5 + · · ·+ 23.

3. 105 and 120. Their sum is 225 = 152. 4. 0, 1 or 4.

Exercise 1.2.4

1. (i) 961; (ii) 5184; (iii) 1369; (iv) 27556. 2. (i) 7225; (ii) 13225;

(iii) 27225. 3. 2155024.

Exercise 1.2.5

1. (i) 14; (ii) 16; (iii) 102; (iv) 34; (v) 115. 2. (i) 16; (ii) 37;

(iii) 81; (iv) 10; (v) 4; (vi) 247. 3. 56 m. 4. (i) 7; (ii) 2; (iii)

5; (iv) 2. 5. (i) 16; (ii) 16; (iii) 729; (iv) 676. 6. You can take

16 as factor of 48 and 240 as a multiple of 48; 16 + 240 = 256 = 162. You

can generate infinitely many by taking l = m(3m+2), m = 1, 2, 3, . . .; 16 as a

factor of 48 and 48l as a multiple of 48. Observe

16 + 48l = 16 + 48m(3m+ 2) = 16 + 96m+ 144m2 = (4 + 12m)2.

Exercise 1.2.6

1. (i) 15; (ii) 24; (iii) 27; (iv) 29; (v) 42. 2. 127 m. 3. 5.

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Answers 127

Exercise 1.2.7

1.

2 3 4 −5 6 8 −923 = 8 33 = 27 43 = 64 (−5)3 = −125 63 = 216 83 = 512 (−9)3 = −729

2.

13 33 53 73 93 23 43 63 83 103

1 27 125 343 729 8 64 216 512 1000

The cube of an odd number is odd and the cube of an even number is

even.

3. There are 4 perfect cubes from 1 to 100; there are 9 perfect cubes from

−100 to 100(recall 0 is also a perfect cube). 4. There are 7 perfect cubes

from 1 to 500. Of these 64 is the only perfect square; 64 = 43 = 82.

5. The number of zeros at the end is always a multiple of 3. 6. Each

digit occurs at the end of some cube. Hence one cannot conclude that

some number is not a cube by looking at the last digit(compare this with

perfect squares).

Exercise 1.2.8

1. (i) 22; (ii) 36; (iii) 25. 2. (i) 45; (ii) 55; (iii) 89. 3. (i) 69; (ii)

36; (iii) 72.

Additional problems on “Squares square roots,

cubes and cube roots”

2. (a) C; (b) B; (c) C; (d) A; (e) C. 3. 11. 4. 6,5,1,6,9,6,6,4,9,1,4,4,9.

5. None of them is a perfect square. 6. 81. 7. (i) z = ±5; (ii) y = ±12;(iii) x = ±8. 8. 48. 9. (i) closed; (ii) commutative; (iii) associative;

(iv) no, because m2 + k2 = m2 implies k = 0 and N does not contain 0.

11. 344 m. 12. If 1010 = a2 − b2 for some integers a and b, then either

both a and b are odd or both even. Hence a2 − b2 is divisible by 4. But

1010 is not divisible by 4. Hence 1010 is not the difference of two perfect

squares. 13. 0,1,6. 14. 1156 = 342. 15. 4 and 16.

Exercise 1.3.1

1. (i) associative property of addition in Z; (ii) commutative property of

multiplication in Z; (iii) distributivity of multiplication over addition in

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128 Answers

Z.

2. −6, −9, −123, 76, 85, −1000. 3. (i) m = 2; (ii) m = −10; (iii)

m = 14; m = −77. 4. −333, −210, −26, −8, 0, 21, 33, 85, 2011. 5. 2011,

528, 364, 210, 85, 12, −58, −1024, −10000.Exercise 1.3.2

1. You can take10

14,15

21,20

28,25

35,30

42,35

49,40

56,45

63,50

70,55

77.

2.Take22

10,33

15,44

20,55

25,66

30,77

35,88

40,99

45,110

50,121

55,132

60,143

65,154

70,165

75,176

80.

3.10

1,9

2,8

3,7

4,6

5,5

6,4

7,3

8,2

9,

1

10.

4.1

3,2

4,3

5,4

6,5

7,6

8,7

9,

8

10,

9

11,10

12.

5. The number3

−2 is the same as−32, as these are equivalent fractions;

recall that the denominator is always positive as a convention.

6. 0.9 =9

10; 0.8 =

8

10=

4

5.

Exercise 1.3.3

1. (i) closure property of addition; closure property of multiplication; (iii)

0 is the additive identity; (iv) 1 is the multiplicative identity.

5.−85,−610

(=−35

),3

8,16

3,4

1. 6.

1

2,11

6,−158

,18

19, 1000.

Exercise 1.3.4

2.−137

<−10281

<−45

<7

12<

3

4<

101

100<

22

19<

15

11.

3. We can take13

30,14

30,15

30,16

30,17

30. If you increase both the numerator and

denominator and use the property of equivalent fractions, you can get any

number of such collections.

4. There are 15 such rational numbers:1

2,1

3,1

4,1

5,1

6,1

7,1

8,1

9,2

3,2

5,2

7,3

4,3

5,3

7,4

5.

5. Ifm

nand

p

qare distinct, then

m+ p

n+ qlies between

m

nand

p

q. If

m

n=

p

q,

thenm+ p

n+ q=

m

n=

p

q. 6. 79. 7. 69.

Additional problems on “Rational numbers”

1. (a) natural numbers; (b) 0; (c) 2; (d) 9; (e) even; (f) odd. 2. (a) false;

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Answers 129

(b) false; (c) true; (d) false; (e) false; (f) true; (g) false. 3. (i) 297; (ii)

39. 4. m/(m + 1) < (m + 1)/(m + 2); do not forget two cases m < −2 and

m > −1. 5. (i) yes; (ii) yes; (iii) yes; (iv) a ∗ 1 = 1; (e) a = 2, b = 2.

6.13

8,17

12,23

26,−1113

,−100101

. 7.10

13<

5

6<

20

23<

25

28<

10

11<

40

43.

8.21

17>

13

11>

31

27>

9

8>

41

37>

51

47. 9. (a) 0; (b) 1; (c) 1, −1.

11 (i)46

225; (ii)

413

108; (iii)

225

6; (iv)

11825

504. 13. Every non-zero rational

number is invertible, but only ±1 are invertible integers. 14.5

3.

15.1

2. 16. ±1. 17. 140. 18. No rational p/q between 0 and 1 for

which q < p. 19. n = 0, −1, −3, −4. 20. 4 values: 26, 46, 50, 70.

23. n = 2, 3, 5, 0, −1, −3.Exercise 1.4.2

1. chess: 270, carom: 540, other games: 90 2. 2,000. 3. 25%.

4. 2,400. 5. 9,800.

Exercise 1.4.3

1. 10% profit. 2. 3,168. 3. 920. 4. Profit 120 5. 600 6. 20%.

Exercise 1.4.4

1. 12%. 2. 625. 3. 1904(marked price) and 336(discount).

4. 26%. 5. (i) 425. (ii) 340.

Exercise 1.4.5

1. 700(commission); Sindhu gets 27,300. 2. 1,350. 3. 2.5%.

4. 70,000.

Exercise 1.4.6

1. 625. 2. 39.55 3. 312 years 4. 18,000. 5. 1212%.

Exercise 1.4.7

1.(i) 260; (ii) 2,838; (iii) 1,236. (iv) 212. 2. 30,245

3. 7.33% gain.

Additional problems on “Commercial Arithmetic”

1. (a) A.; (b) C.; (c) B.; (d) B. ; (e) C.; (f) A.; (g) C.; (h) C.;

(i) B.; (j) B.; (k) A.; (l) B.; (m) C.; (n) A.; (o) B..

2. 212. 3. 325. 4. 330. 5. 800. 6. 1,200. 7. 27 kg.

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130 Answers

8. 1,080. 9. (i) 675; (ii) 8% . 10 28,462.56. 11. 41,050.

12. 528. 13. 18%. 14. 20 years. 15. 2 years. 16. 32,041.65.

17. 24,300.

Exercise 1.5.2

1.

Class Tally Frequency

Interval marks

30-39 IIII 4

40-49 I I I I , I I I I , 16

I I I I , I

50-59 I I I I , II 7

60-69 I I I I 5

70-79 IIII 4

80-89 III 3

90-99 I 1

(i) highest: 40 - 49, lowest: 90 - 99

(ii) 30.5 and 39.5 (iii) 65.

2.

Class Tally Frequency

Interval marks

10-20 I I I I , IIII 9

20-30 I I I I , I I I I , II 12

30-40 I I I I , I I I I 10

40-50 I I I I , IIII 9

50-60 I I I I , I 6

60-70 IIII 4

(i) 10 scores between 30 and 40.

ii) 10, 35. (iii) 52.

Exercise 1.5.31.

25

30

35

40

20

15

10

5

20 30 35 40 4525

2.

25

30

35

40

20

15

10

5

19.5 29.5 39.5 49.5 59.59.5

Exercise 1.5.4

1. 42.1. 2. 29.73. 3. 42.75. 4. 22.63. 5. 15. 6. 37.5.

7. 136.8. 8. 16.7. 9. (i) 3; (ii) 22 and 42(bi-mode distribution).

10. 20.

Additional problems on “Statistics”

1. (a) B. (b) D. (c) B. (d) A. (e) A. (f) D. (g) A. (h) C. (i) B. (j) D.

(k) A. (l) B. (m) C. (n) A. (o) B. (p) A. (q) D. (r) B.

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Answers 131

2.

Class-Interval Tally Frequency

marks

10 - 20 I I I I , III 8

20 - 30 I I I I , I I I I 10

30 - 40 I I I I , III 8

40 - 50 I I I I , III 8

50 - 60 I I I I , I 6

Total 40 40

(i) The largest number of

scores lie in the class inter-

val 20-30.

(ii) 10, 20, 30 are included re-

spectively in the class inter-

vals 10-20, 20-30, 30-40.

(iii) The range of the scores is

59-10=49.

3.

Class-Interval Tally Frequency

marks

0 - 4 I 1

5 - 9 II 2

10 - 14 I I I I , II 7

15 - 19 I I I I , IIII 9

20 - 24 I I I I , I 6

Total 25 25

(i) Mid-points of 0-4, 5-9, 10-

14, 15-19, 20-24 are respec-

tively 2,7,12,17,22.

(ii) The class interval having

maximum frequency is 15-

19.

(iii) The range of the scores is

24-3=21.

4.

5 15 25 35 45 55

20

15

10

5

2

8

14 14

12

5.

20

5

10

15

0 10 20 30 40 50 60

4

10

18

12

20

6

6. (i) 73.17(approximately); (i) 77.17(approximately). 7. 17. 8. 14.75.

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CHAPTER 2 UNIT 1

ALGEBRAIC EXPRESSIONS


• the meaning and types of polynomials;

• addition and subtraction of polynomials;

• multiplication of polynomials: monomials by monomials, binomial by

monomials, binomial by binomial (x + a)(x + b), (a + b)2, (a − b)2 and

(a+ b)(a− b) etc.

2.1.1 Introduction

Let us first review some of the things you have learnt earlier.

A symbol which has a fixed value is called a constant.

Examples: 5, -7, 235,√5, 2 +

√3 etc.

A symbol which does not have any fixed value, but may be assigned value

(values) according to the requirement is called a variable or a literal.

Examples: p, q, x, y, z etc.

Note:

1. Combination of a constant and a variable is a variable.

Examples: 3x, (4 + p), 6/x, x/7, x− 4, 9x etc.

2. Combination of two or more variables is either a variable or a constant.

Examples: xy, x/y, (x − y), (y − x), −x, (x + y), xyz, xy/z , 13 + x − y,

14x − y, 10 − xy, 7x/y, 8/xy etc. Observe (4 + x) + (4 − x) = 8 which is a

constant.

A term is a number (constant), a variable, or a combination of (product or

quotient of) numbers and variables.

Examples: 9, x, 3x, 4xy, 7x/15y, 21/xy, yz/x etc.

A single term or a combination of two or more terms connected by additive

(both addition and subtraction) and multiplicative(both multiplication and

division) symbols form an algebraic expression.

Examples: 7− y, 3x2 − 4y, 6xy, 6 + x2 − 3x, (7x/9) + 4y − 6z etc.

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Algebraic Expressions 133

Note: The signs of multiplication and division do not separate terms. For

example: 9x24y or 4x2/7y are single terms.

An algebraic expression which contains only one term is called a mono-

mial.

Examples: 4, 5/ 11, x, 6x, 8xy, 7x2/y, yz/x, 5x2y/7z etc.

An algebraic expression which contains two terms is called a binomial.

Examples: 7 + x, xy − 7, (5x/y)− 3x, 3x2 − 6xy, yz2 + 2z.

An algebraic expression which contains three terms is called a trinomial.

Examples: 4 + x+ y, 6x+ 15− y2, ax2 + bx+ c, ax+ by + 2.

An algebraic expression in which each term contains only the variable(s)

with non negative integral exponent(s) is called a polynomial.

Examples: x2 − 4x, x− 4xy + y2, 6− 5y + xy + x2y, 4.

Note:x

y+ 2 is not a polynomial, it is only an algebraic expression.

Exercise 2.1.11. Separate the constants and variables from the following:

12 + z, 15, −x/5, −3/7, √x,√3, (2/3)xy, (5xy)/2, 7, 7 − x, 6x+ 4y,

−7z, (8yz)/3x, y + 4, y/4 and (2x)/(8yz).

2. Separate the monomials, binomials and trinomials from the following:

7xyz, 9−4y, 4y2−xz, x−2y+3z, 7x+ z2 , 8x/y, (8/5)x2y2, 4+5y−6z.

2.1.2 Algebraic expressions

Consider 9x. It contains two factors: 9 and x. We say 9 is the numerical

coefficient(or constant or arithmetical coefficient);x is called as vari-

able(or literal factor or literal coefficient). Or consider 9xy. Here there

are two variables x and y. We say 9x is the coefficient of y and 9y is the

coefficient of x. Observe the following tabular columns:

Product Coefficient Numerical Literal

Coefficient Coefficient

(constant) (variable)

−8xy −8x is the coefficient of y −8 x

−8y is the coefficient of x −8 y

xy is the coefficient of −8 – xy

−8 is the coefficient of xy −8 –

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134 Unit 1

Note: 1. If the literal factor (variable) has no sign, it is taken as

positive.

2. If the variable has no power, it is taken as 1.

3. If the variable has no numerical coefficient, it is taken as 1.

For example, x means +1x.

4. The highest power of the variable in a polynomial with single

variable is called the degree of the polynomial.

Like and unlike terms

Terms having the same variable with same exponents are called like terms.

Examples:

5x, 2x, 7x, −9x, (1/3)x etc.;

x2, 2x2, 6x2, 9x2,1

7x2 etc.;

x3, 3x3, 7x3, −9x3, 1

9x3 etc.

Terms having the same variable with different exponents or different

variables with same/different exponents are called unlike terms.

Examples: x, x2, x3, x4, x5 etc.; x, m, n, p etc.; −x, xy, xy2 etc.

2.1.3 Addition and Subtraction of polynomials

Let us first review properties of addition and multiplication in the set of

all integers:

1. Sum of two positive integers is a positive integer; (+7)+(+5) = +7+5 =

+12.

2. Sum of two negative integers is a negative integer; (−7) + (−5) = −7 −5 = −12.

3. Sum of a positive integer and a negative integer is positive if the abso-

lute value of the negative integer is smaller than the positive integer;

(+7) + (−5) = +7− 5 = +2.

4. Sum of a positive integer and a negative integer is negative if the ab-

solute value of the negative integer is larger than the positive integer;

(−7) + (+5) = −7 + 5 = −25. Product of two positive integers is also a positive integer; (+7)× (+5) =

+35.

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6. Product of two negative integers is a positive integer; (−7)×(−5) = +35

7. Product of a positive integer and a negative integer is a negative inte-

ger; (+7)× (−5) = −35.8. Product of a negative integer and a positive integer is a negative inte-

ger; (−7)× (+5) = −35.

Make the following rules for addition and subtraction of two polynomials:

1. like terms can be added or subtracted;

2. unlike terms cannot be added or subtracted;

3. while adding or subtracting like terms, their numerical coefficients

are added or subtracted;

Example 1. Add 5x2y, −7x2y and 9x2y.

Solution: Recall rule 1. We have

(5x2y) + (−7x2y) + (9x2y) =(5 + (−7) + 9

)x2y

=(5− 7 + 9

)x2y

= 7x2y.

This is horizontal addition. We can also have vertical addition:

+5 x2y

−7 x2y

+9 x2y

7 x2y

You may observe that we are adding the coefficients and retaining the

variable term as it is.

Example 2. Add: 7x2 − 4x+ 5 and 9x− 10.

Solution: Here there are unlike

terms. We can add only like terms.

We write like terms one below other

to facilitate easy addition.

7x2 −4x +5

+9x −107x2 +5x −5

Example 3. Add 8xy + 4yz − 7zx, 6yz + 11zx− 6y and −5xz + 6x− 2yx.

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136 Unit 1

Solution: Here again there are

many unlike terms. We write

like terms one below other to

facilitate easy addition. We

are also using the commuta-

tive property: xy = yx and xz =

zx.

+8xy +4yz −7zx+6yz +11zx −6y

−2xy −5xz +6x

6xy +10yz −xz +6x −6y

Example 4. Subtract 2x3 − x2 + 4x− 6 from x3 + 5x2 − 4x+ 6.

Solution: We write like terms

one below other to facilitate

easy subtraction. Note that we

are subtracting relevant coeffi-

cients. Subtracting a negative

+1x3 +5x2 −4x +6 → Minuend

+2x3 −x2 +4x −6 → Subtrahend

(−2) (+1) (−4) (+6)

−1x3 +6x2 −8x +12

number is equivalent to adding the negative of that number. Hence we

have changed the signs of coefficients in the subtrahend and added the

coefficients. You can also do this in a quick way, once you understand the

meaning of subtraction, as follows:

(x3 + 5x2 − 4x+ 6)−

(2x3 − x2 + 4x− 6

)

= x3 + 5x2 − 4x+ 6− 2x3 + x2 − 4x+ 6

= (1− 2)x3 + (5 + 1)x2 + (−4− 4)x+ (6 + 6)

= −1x3 + 6x2 − 8x+ 12

= −x3 + 6x2 − 8x+ 12.

Exercise 2.1.3

1. Classify into like terms:

4x2,1

3x, −8x3, xy, 6x3, 4y, −74x3, 8xy, 7xyz, 3x2.

2. Simplify:

(i) 7x− 9y + 3− 3x− 5y + 8; (ii) 3x2 + 5xy − 4y2 + x2 − 8xy − 5y2.

3. Add:

(i) 5a+ 3b, a− 2b and 3a+ 5b; (ii) x3 − x2y+ 5xy2+ y3, −x3 − 9xy2+ y3,

and 3x2y + 9xy2.

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4. Subtract:

(i) −2x2y + 3xy2 from 8x2y; (ii) a− b− 2c from 4a+ 6b− 2c.

2.1.4 Multiplication of Polynomials

Observe the following products: (i) 5x× 6x2 = (5× 6)× (x× x2) = 30x3;

(ii) 2x× 6y × 8z = (2x× 6y)× (8z)

=((2× 6)× (x× y)

)× (8z) =

(12xy

)× (8z)

=(12× 8

)× (xy × z) = 96xyz.

We can also write this in one step: 2x×6y×8z = (2×6×8)×(x×y×z) = 96xyz.

Note: coefficient of the product= the product of the coefficients of

expressions;

algebraic factor of the product= the product of all the algebraic fac-

tors.

Example 5. Find the product of 6x and −7x2y.Solution: We have (6x)× (−7x2y) =

(6× (−7)

)× (x× x2y) = (−42)x3y.

Note: We are using x × x2y = (x × x2)y = x3y. In other words, we

multiply similar variables and use the law of indices for simplifying

the expression: xm × xn = xm+n for all integers m,n, which you will

study later.

Multiplying a monomial by a monomial

Example 6. Find the product 4x× 5y × 7z.

Solution: We have 4x× 5y × 7z = (4× 5× 7)× (x× y × z) = 140xyz.

Example 7. What is the product 2l2m× 3lm2?

Solution: We have 2l2m× 3lm2 = (2× 3)× (l2× l)× (m×m2) = 6l3m3. We are

using the law of indices: ak × al = ak+l.

Multiplying a monomial by a binomial

Consider the product 9× 103 = 927. We may also write this in the form

9× 103 = 9× (100 + 3) = (9× 100) + (9× 3) = 900 + 27 = 927.

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138 Unit 1

Do you see that we have used the distributive property of multiplication

over addition? We adopt the same strategy when a binomial is involved.

Thus

2x(3x+ 5xy) =((2x)× (3x)

)+((2x)× (5xy)

)= 6x2 + 10x2y.

Example 8. Determine the product (8y + 3)× 4x.

Solution: We have

(8y + 3)× (4x) = (4x)× (8y + 3)

= (4x× 8y) + (4x)× 3

= 32xy + 12x.

Here we have used the commutativity of the product and right distributiv-

ity. We can also use left distributive law and get

(8y + 3)× (4x) =((8y)× (4x)

)+(3× (4x)

)

= 32yx+ 12x

= 32xy + 12x,

because xy = yx, the commutative property of the product.

Important points:

We are using all the properties of numbers to the algebraic vari-

ables; associativity, commutativity, law of indices, distributive

property. This is because, when we substitute numbers for these

variables, all these properties are true. As an extrapolation, we

expect them to hold for variables as well.

Multiplying a binomial by a binomial

Consider the product of two binomials (4a+ 6b) and (5a+ 7b). We have

(4a+ 6b)(5a + 7b) = 4a(5a+ 7b) + 6b(5a + 7b)

=((4a)(5a)

)+((4a)(7b)

)+((6b)(5a)

)+ (6b)(7b)

= 20a2 + 28ab+ 30ab+ 42b2

= 20a2 + 42b2 + 58ab.

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Exercise 2.1.4

1. Complete the following table of products of two monomials:

First → 3x −6y 4x2 −8xy 9x2y −11x3y2Second ↓

3x

−6y4x2

−8xy9x2y

−11x3y2

2. Find the products:

(i) (5x+8)3; (ii) (−3pq)(−15p3q2−q3); (iii)6x

5(a3−b3); (iv) −x(x−15).

3. Simplify the following:

(i) (2x2y−xy)(3xy−5); (ii) (3x2y2+1)(4xy−6xy2); (iii) (3x2+2x)(2x2+3);

(iv) (2m3 + 3m)(5m− 1).

2.1.5 Special product

Now we study a special product, a product of two binomials. Consider the

product:

(x+ a)(x+ b) = x(x+ b) + a(x+ b) = x2 + xb+ ax+ ab

= x2 + ax+ bx+ ab

= x2 + (a+ b)x+ ab.

We have used commutative property and the distributive property: xb = bx,

(ax+ bx) = (a + b)x. We say (x+ a)(x+ b) = x2 + (a + b)x+ ab is an identity.

Thus

(x+ 2)(x+ 1) = x2 + (2 + 1)x+ (2× 1) = x2 + 3x+ 2.

We can also represent the the identity (x+2)(x+1) = x2+3x+2, pictorially.

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140 Unit 1

A E F B

G

CH

J

ID

(x+2)

(x+1)

x

x

x x x

1 1

1

1 1

11 1 1

1x

x

x

x x

1

x 2

1

K L

You observe that the

area of the rectangle is

(x+2)(x+1). The rectan-

gle is split in to smaller

squares and rectangles:

squares AEKJ, KLHI

and LGCH having ar-

eas x2, 1 and 1 respec-

tively; rectangles EFLK,

FBGL, JKID having ar-

eas x, x, x respectively.

Thus the area of ABCD

is x2 + 1 + 1 + x+ x+ x =

x2 + 3x + 2. We thus ob-

tain

(x+2)(x+1) = x2+3x+2.

What is the value of (x + a)(x + b)? If we replace x by y what

happens to the identity: (x+ a)(x+ b) = x2 + (a+ b)x+ ab? Can you

see that you again get an identity?

Example 9. Find the product (x+ 6)(x+ 7).

Solution: We observe that (x+6)(x+7) = x2+(6+7)x+(6×7) = x2+13x+42.

Example 10. Determine the product (x+ 8)(x− 4)

Solution: Using the identity (x+ a)(x+ b) = x2 + (a+ b)x+ ab, we get

(x+ 8)(x− 4) = x2 + (8− 4)x+ 8× (−4) = x2 + 4x− 32.

Example 11. Compute (2x+ 5)(2x+ 3).

Solution: We know that (x+ a)(x+ b) = x2 + (a+ b)x+ ab. Thus

(2x+ 5)(2x+ 3) = (2x)2 + (5 + 3)(2x) + (5× 3) = 4x2 + 16x+ 15,

Example 12. Find the product 103× 98 using the above identity.


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103× 98 = (100 + 3)(100− 2)

= (100)2 +((3) + (−2)

)100 +

(3× (−2)

)

= 10000 + (1× 100) + (−6) = 10094,

where we have taken x = 100, a = 3 and b = −2 in the expansion: (x+a)(x+

b) = x2 + (a + b)x+ ab.

Example 13. Find the product (p2 − 5)(p2 − 3).

Solution: We have

(p2 − 5)(p2 − 3) = (p2)2 +((−5) + (−3)

)(p2) + (−5)× (−3) = p4 − 8p2 + 15.

Identities

An identity is an equality which is true for every value of the variable in

it.

For example (x+ 3)(x+ 2) = x2 + 5x+ 6 is an identity. If you give any value

for the variable x, you see that the left side and right side coincide.

We have some special identities, which are helpful in solving problems.

Consider

(a+ b)2 = (a + b)(a+ b) = a(a+ b) + b(a + b) = a2 + ab+ ba + b2

= a2 + ab+ ab+ b2

= a2 + b2 + 2ab.

Observe that we have used the commutative law: ab = ba. We can have a

pictorial proof of this identity using geometrical squares.

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142 Unit 1

KH F

CGD

A E B

(a+b

)

(a+b)

a b

b b

a

a b

b²ab

a² aba

Observe that the square

ABCD has area equal to

(a + b)2, since its side-length

is a + b. Now we divide the

square in to two smaller

squares and two rectangles;

squares HKGD having area

a2 and the square EBFK

having area b2; the rectangle

KFCG having area ab and the

rectangle AEKH having area

ab. Thus the area of ABCD is

a2 + b2 + ab+ ab = a2 + b2 + 2ab.Similarly we obtain

(a− b)2 = a2 + b2 − 2ab.

And its pictorial proof is as follows.

KH F

CGD

A E Bb

b b

b

b²

a

a

b(a−b)

(a−b)

(a−b

)

(a−b

)(a−b)² b(a−b)

(a−b)

Consider a square ABCD with

side-length a so that its area

is equal to a2. Now we divide

the square in to two smaller

squares and two rectangles;

squares HKGD having area

(a − b)2 and the square EBFK

having area b2; the rectan-

gle KFCG having area b(a − b)

and the rectangle AEKH hav-

ing area b(a−b). (We assume a is

greater than b.) Now the area of

HKGD is obtained by subtract-

ing the areas of EBKF , KFCG

and AEKH from that of ABCD.Hence we get

(a− b)2 = a2 − b2 − b(a− b)− b(a− b) = a2 − b2 − ba + b2 − ba + b2

= a2 − 2ab+ b2.

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We also have(a+ b)(a− b) = a2 − ab+ ba− b2 = a2 − b2.

These are called standard identities.

Example 14. Find (2x+ 3y)2.

Solution: We use the identity: (a + b)2 = a2 + 2ab + b2. Taking a = 2x and

b = 3y, we get

(2x+ 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2 = 4x2 + 12xy + 9y2.

Example 15. What is the expansion of (4p− 3q)2?

Solution: Here we use (a− b)2 = a2 − 2ab+ b2. Taking a = 4p and b = 3q, we

obtain(4p− 3q)2 = (4p)2 − 2(4p)(3q) + (3q)2 = 16p2 − 24pq + 9q2.

Example 16. Compute (4.9)2.

Solution: We can use identities for this problem. Observe

(4.9)2 = (5− 0.1)2 = 52 − 2(5)(0.1) + (0.1)2 = 25− 1 + 0.01 = 24.01.

You may verify this by directly computing (4.9)2.

Example 17. Compute 54× 46.

Solution: Here again, identities are useful. We make use of (a+ b)(a− b) =

a2 − b2. Taking a = 50 and b = 4, we see that

54× 46 = (50 + 4)(50− 4) = (50)2 − (4)2 = 2500− 16 = 2484.

Activity 1:

Represent the identity (a+ b)(a− b) = a2− b2 pictorially as has been done in

the case of other identities.

Exercise 2.1.5

1. Find the product:

(i) (a+3)(a+5), (ii) (3t+1)(3t+4); (iii) (a−8)(a+2); (iv) (a−6)(a−2).

2. Evaluate:

(i) 53× 55; (ii)102× 106; (iii) 34× 36; (iv) 103× 96.

3. Can you imitate the case of the identity (x+a)(x+ b) = x2+(a+ b)x+ab,

and get a similar expression for the product (x+ a)(x+ b)(x+ c)?

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144 Unit 1

4. Using the identity (a + b)2 = a2 + 2ab+ b2, simplify the following:

(i) (a + 6)2, (ii) (3x+ 2y)2; (iii) (2p+ 3q)2; (iv) (x2 + 5)2.

5. Evaluate using the identity (a+ b)2 = a2 + 2ab+ b2:

(i) (34)2; (ii) (10.2)2; (iii) (53)2; (iv) (41)2.

6. Use the identity (a− b)2 = a2 − 2ab+ b2 to compute:

(i) (x− 6)2; (ii) (3x− 5y)2; (iii) (5a− 4b)2;

(iv) (p2 − q2)2.

7. Evaluate using the identity (a− b)2 = a2 − 2ab+ b2:

(i) (49)2; (ii) (9.8)2; (iii) (59)2; (iv) (198)2.

8. Use the identity (a+ b)(a− b) = a2 − b2 to find the products:

(i) (x + 6)(x − 6); (ii) (3x + 5)(3x − 5); (iii) (2a + 4b)(2a − 4b); (iv)(2x

3+ 1

)(2x

3− 1

).

9. Evaluate these using identity:

(i) 55× 45; (ii) 33× 27; (iii) 8.5× 9.5; (iv) 102× 98.

10. Find the product:

(i) (x−3)(x+3)(x2+9); (ii) (2a+3)(2a−3)(4a2+9); (iii) (p+2)(p−2)(p2+4);

(iv)

(1

2m− 1

3

)(1

2m+

1

3

)(1

4m2 +

1

9

); (v) (2x−y)(2x+y)(4x2+y2); (vi)

(2x− 3y)(2x+ 3y)(4x2 + 9y2).

Additional problems on “Algebraic expressions”

1. Choose the correct answer.

(a) Terms having the same literal factors with same exponents are

called

A. exponents B. like terms C. factors D. unlike terms

(b) The coefficient of ab in 2ab is:

A. ab B. 2 C. 2a D. 2b

(c) The exponential form of a× a× a is:

A. 3a B. 3 + a C. a3 D. 3− a

(d) Sum of two negative integers is:

A. negative B. positive C. zero D. infinite

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(e) What should be added to a2 + 2ab to make it a complete square?

A. b2 B. 2ab C. ab D. 2a

(f) What is the product (x+ 2)(x− 3)?

A. 2x− 6 B. 3x− 2 C. x2 − x− 6 D. x2 − 6x

(g) The value of (7.2)2 is( use an identity to expand):

A. 49.4 B. 14.4 C. 51.84 D. 49.04

(h) The expansion of (2x− 3y)2 is:

A. 2x2 + 3y2 + 6xy B. 4x2 + 9y2 − 12xy

C. 2x2 + 3y2 − 6xy D. 4x2 + 9y2 + 12xy

(i) The product 58× 62 is(use an identity):

A. 4596 B. 2596 C. 3596 D. 6596

2. Take away 8x− 7y − 8p+ 10q from 10x+ 10y − 7p+ 9q.

3. Expand:

(i) (4x+ 3)2; (ii) (x+ 2y)2; (iii) (x+ 1/x)2; (iv) (x− 1/x)2.

4. Expand:

(i) (2t+ 5)(2t− 5); (ii) (xy + 8)(xy − 8); (iii) (2x+ 3y)(2x− 3y).

5. Expand:

(i) (n− 1)(n+ 1)(n2 + 1); (ii) (n− 1/n)(n+ 1/n)(n2 + 1/n2);

(iii) (x− 1)(x+ 1)(x2 + 1)(x4 + 1); (iv) (2x− y)(2x+ y)(4x2 + y2).

6. Use appropriate formulae and compute:

(i) (103)2; (ii) (96)2; (iii) 107× 93; (iv) 1008× 992; (v) 1852 − 1152.

7. If x+ y = 7 and xy = 12, find x2 + y2.

8. If x+ y = 12 and xy = 32, find x2 + y2.

9. If 4x2 + y2 = 40 and xy = 6, find 2x+ y.

10. If x− y = 3 and xy = 10, find x2 + y2.

11. If x+ (1/x) = 3, find x2 + (1/x2) and x3 + (1/x3).

12. If x+ (1/x) = 6, find x2 + (1/x2) and x4 + (1/x4).

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146 Unit 1

13. Simplify: (i) (x+ y)2 + (x− y)2; (ii) (x+ y)2 × (x− y)2.

14. Express the following as difference of two squares:

(i) (x + 2z)(2x + z); (ii) 4(x + 2y)(2x + y); (iii) (x + 98)(x + 102); (iv)

505× 495. (Hint. ab =

(a + b

2

)2

−(a− b

2

)2

.)

15. If a = 3x− 5y, b = 6x+ 3y and c = 2y − 4x, find

(i) a + b− c; (ii) 2a− 3b+ 4c.

16. The perimeter of a triangle is 15x2 − 23x + 9 and two of its sides are

5x2 + 8x− 1 and 6x2 − 9x+ 4. Find the third side.

17. The two adjacent sides of a rectangle are 2x2−5xy+3z2 and 4xy−x2−z2.Find its perimeter.

18. The base and the altitude of a triangle are (3x − 4y) and (6x + 5y)

respectively. Find its area.

19. The sides of a rectangle are 2x+ 3y and 3x+ 2y. From this a square of

side length x+y is removed. What is the area of the remaining region?

20. If a, b, c are rational numbers such that a2 + b2 + c2 − ab − bc − ca = 0,

prove that a = b = c.

Glossary

Constant: any symbol which has a fixed value.

Variable: a symbol which can be given any value as desired.

Algebraic expression: a combination of constants and variables con-

nected by algebraic operations.

Polynomial: an algebraic expression in which the variables have non-

negative integral powers.

Term: a part of an algebraic expression which does not involve addition

and subtraction, but may be connected by multiplication and division.

Coefficient: the companion term of a variable.

Monomial: an algebraic expression containing only one term.

Binomial: an algebraic expression containing two terms.

Trinomial: an algebraic expression containing three terms.

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Degree: the largest power of a variable in a polynomial(if it has more than

one variable, then one has to take the sum of the powers of variables in

each term and take the maximum of all these sums).

Identity: equality of two algebraic expressions which is valid for all the

values of the variables in it.

Points to remember

• An expression involving variables and constants combined using the

algebraic operations, namely addition, multiplication, subtraction and

division is an algebraic expression.

• While adding two algebraic expressions, we add only like terms.

• While multiplying two expressions, we multiply term-by-term and use

the laws of exponents to simplify it.

• In a polynomial, the powers of variable(s) in it are non-negative inte-

gers.

• An identity is an equality of two algebraic expressions which is valid

for all values of the variable(s) in it.

All the religions of the world, while they may differ in other respects,

unitedly proclaim that nothing lives in this world but Truth.

——Gandhiji

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CHAPTER 2 UNIT 2

FACTORISATION


• factorise an algebraic expression by taking out common factors;

• factorise an expression by grouping appropriate terms;

• factorise an expression which is a difference of two square expres-

sions;

• factorise a trinomial expression;

• factorise a square trinomial expression by using known identities.

2.2.1 Introduction

The process of writing a given algebraic expression as a product of

two or more expressions is called factorisation. Each of the expres-

sion(constant or variable) which form the product is called a factor of

the given algebraic expression.

For example: (i) Consider 7xy. It has factors: 7, x, y, 7x, 7y, xy and 7xy;

(ii) (x+ 3) and (x+ 2) are factors of x2 + 5x+ 6.

Recall the process of factorisation of a number. Given an integer, we

write this as a product of other integers. This ultimately leads to prime

factorisation of a number. In some sense, algebraic expressions also follow

similar way. Given an expression, you will be able to write the expression

as a product of its factors.

Note: (1) Factorisation and multiplication are reverse processes. In

multiplication, we multiply different expressions and get a new ex-

pression. In factorisation, we split the given expression in to simpler

expressions whose product turns out to be the given expression.

(2) You can always use 1 as a factor: (x + 5) = 1 × (x + 5). But this

does not give any thing new. This is called the trivial factorisation.

Many times one has to consider such factorisations as well.

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Factorisation 149

2.2.2 Different methods of factorisation

There are many ways of factorising a given expression. We study them

here.

1. Factorisation taking common factors

Consider the following examples:

Example 1: 5x2 − 10x

Here you see that 5x is the HCF of 5x2 and 10x. We can remove them from

both the terms. Thus we get

5x2 − 10x = (5x)(x)− (5x)2 = (5x)(x− 2).

Example 2: 4a+ 12b = 4(a+ 3b).

Example 3: 3x2y − 6xy2 + 9xy = 3xy(x− 2y + 3).

Example 4: a3 − a2 + a = a(a2 − a+ 1).

You see that we look at the HCF of all the terms in the expression and

take out the HCF to get a factorisation.

2. Factorisation by grouping

We follow several steps in this process:

Step I: Arrange the terms of the given expression in suitable groups

such that each group has a common factor;

Step II: Factorise each group;

Step III: Take out the factor which is common to each group.

Example 5: Factorise ax− bx+ ay − by.

Solution: We group them as

(ax− bx) + (ay − by) = (a− b)x+ (a− b)y = (a− b)(x+ y).

Do you see that the distributive law is used in a reverse way? We can also

do this as follows:

ax− bx+ ay − by = (ax+ ay)− (bx+ by)

= a(x+ y)− b(x+ y)

= (a− b)(x+ y).

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150 Unit 2

Example 6: Get the factors of y3 − 3y2 + 2y − 6− xy + 3x.

Solution: Here again we group the terms in the expression and factorise:

y3 − 3y2 + 2y − 6− xy + 3x = (y3 − 3y2) + (2y − 6)− (xy − 3x)

= y2(y − 3) + 2(y − 3)− x(y − 3) = (y2 + 2− x)(y − 3).

3. Factorisation of difference of two squares

We know from earlier unit that (a + b)(a − b) = a2 − b2 for all a, b. This

leads to a nice factorisation when the given expression can be written as

difference of two squares.

Example 7: Factorise 36a2 − 49b2.

Solution: Observe that 36a2 = (6a)2 and 49b2 = (7b)2. Thus we get

36a2 − 49b2 = (6a)2 − (7b)2 = (6a+ 7b)(6a− 7b).

Example 8: Factorisex2

y2− 9

16.

Solution: Here again, we write

x2

y2− 9

16=

(x

y

)2

−(3

4

)2

=

(x

y+

3

4

)(x

y− 3

4

).

Example 9: Compute (4.5)2 − (1.5)2.

Solution: We have

(4.5)2 − (1.5)2 = (4.5 + 1.5)(4.5− 1.5) = 6× 3 = 18.

Exercise 2.2.2

1. Resolve in to factors:

(i) x2 + xy; (ii) 3x2 − 6x; (iii) (1.6)a2 − (0.8)a; (iv) 5− 10m− 20n.

2. Factorise:

(i) a2+ ax+ ab+ bx; (ii) 3ac+7bc− 3ad− 7bd; (iii) 3xy− 6zy− 3xt+6zt;

(iv) y3 − 3y2 + 2y − 6− xy + 3x.

3. Factorise:

(i) 4a2 − 25; (ii) x2 − 9

16; (iii) x4 − y4; (iv)

(7 310

)2−(2 110

)2;

(v) (0.7)2 − (0.3)2; (vi) (5a− 2b)2 − (2a− b)2.

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Factorisation 151

2.2.3 Factorisation of trinomials

We have seen earlier how to multiply two binomials of the from (x + a)

and (x+ b):

(x+ a)(x+ b) = x2 + (a+ b)x+ ab.

We can also proceed in the reverse direction. Given the trinomial of the

form x2+(a+b)x+ab, we can factorise this to get x2+(a+b)x+ab = (x+a)(x+b).

But generally, the trinomial is not given in this form. You may be given in

the form x2 +mx + n, where m,n are some numbers. You must be able to

write m = (a+b) and n = a×b to bring the given trinomial in to a factorisable

form. This needs some properties of numbers. We study them here.

The sum and product of two numbers are positive if and

only if both the numbers are positive.

This says that if a + b and ab are positive then so are a, b. The converse

is also true. Thus 6 and 5 are positive; 5 = 3 + 2 and 6 = 3 × 2; both 3 and

2 are positive.

The sum of two numbers is negative and their product pos-

itive if and only if both the numbers are negative.

Thus a+b negative and ab positive if and only if both a and b are negative.

If we are given numbers 21 and −10, we see that −10 = (−7) + (−3) and

21 = (−7)(−3).We say 7 is the absolute value of both 7 and −7. Thus given an integer

a, we define its absolute value by |a| = a if a > 0; |a| = −a if a < 0; and

|a| = 0 if a = 0. Observe −8 < −6 but | − 8| = 8 > 6 = | − 6|.

The sum of two numbers is positive and their product neg-

ative if and only if one of the numbers is positive and the

other negative, and the positive number has larger absolute

value than the negative number.

This means a + b is positive and ab negative only if one of a, b is positive

and the other negative; and if a is positive and b is negative, then |a| > |b|;if a is negative and b positive, then |a| < |b|. For example, we see that if

a+ b = 7 and ab = −18, then a = 9 and b = −2 or a = −2 and b = 9.

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152 Unit 2

The sum of two numbers is negative and their product neg-

ative if and only if one of the numbers is positive and the

other negative, and the positive number has smaller abso-

lute value than the negative number.

Thus a+ b is negative and ab negative if and only if one of a, b is positive

and the other negative; if a is positive and b is negative, then |a| < |b|; if ais negative and b positive, then |a| > |b|. For example, if a + b = −12 and

ab = −28, we can write a = 2 and b = −14 or a = −14 and b = 2.

Here we have not mentioned any thing about the nature of numbers.

They can be integers, rational numbers or even real numbers, which you

will study in your next class.

Example 10. Factorise 6x2 + 11x+ 3.

Solution: Here you can adopt the standard method of splitting and group-

ing:

6x2 + 11x+ 3 = 6x2 + 9x+ 2x+ 3 = (6x2 + 9x) + (2x+ 3)

= 3x(2x+ 3) + 1(2x+ 3) = (3x+ 1)(2x+ 3).

The genuine problem is how to arrive at this splitting. Suppose we are

looking for a factorisation of the form 6x2 + 11x + 3 = (ax + b)(cx + d). This

gives, after expansion,

6x2 + 11x+ 3 = acx2 + (ad+ bc)x+ bd.

After comparing terms of different degrees,

ac = 6, ad+ bc = 11, bd = 3.

Thus acbd = 18 or (ad)(bc) = 18; and ad + bc = 11. You have two numbers

whose product is 18 and their sum is 11. You will immediately conclude

that ad = 9, bc = 2 or ad = 2, bc = 9. Thus you may write

6x2 + 11x+ 3 = acx2 + (ad+ bc)x+ bd = 6x2 + 9x+ 2x+ 3,

which is the splitting we have used. Observe we can also use the other

pair of values for (ad, bc); ad = 2, bc = 9. We get

6x2 + 11x+ 3 = (6x2 + 2x) + (9x+ 3) = 2x(3x+ 1) + 3(3x+ 1) = (2x+ 3)(3x+ 1),

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Factorisation 153

which is the same as the original factorisation.

Rule: If you want to factorise a trinomial of the form x2 + px + q, you

must be able to find numbers a and b such that a · b = q and a + b = p.

Then x2 + px+ q = (x+ a)(x+ b).

You may notice here that a polynomial of degree 2 is written as

a product of two polynomials, each of degree 1, in the process

of factorisation. This helps us in understanding the structure

of a polynomial of degree 2. You will see in later classes that

this is an easy way of solving a polynomial equation, if you are

able to factorise the polynomial.

Example 11. Factorise x2 − 9x+ 20.

Solution: Again you must be able to find two numbers a and b such that

ab = 1×20(the product of the coefficients of x2 term and the constant term)

and a+b = −9. Here the product is positive and the sum is negative. Hence

both the numbers must be negative. This can be achieved by taking a = −5and b = −4. Thusx2 − 9x+ 20 = (x2 − 5x) + (−4x+ 20) = x(x− 5)− 4(x− 5) = (x− 4)(x− 5).

Factorising a square trinomial

Any algebraic expression, which can be written either in the form a2 +

2ab + b2 or in the form a2 − 2ab + b2, is called a square trinomial. For

example, x2 + 2x + 1 is a square trinomial. We can have an immediate

factorisation for such a trinomial using a2 + 2ab + b2 = (a + b)(a + b) or

a2 − 2ab+ b2 = (a− b)(a− b).

Example 12. Factorise 4x2 + 12xy + 9y2.


4x2 + 12xy + 9y2 = (2x)2 + 2(2x)(3y) + (3y)2 = (2x+ 3y)2.

Thus its factors are two equal expressions: 2x+ 3y and 2x+ 3y.

Example 13. Is x2 − 6xy + 36y2 a square trinomial?

We observe that

x2 − 6xy + 36y2 = (x)2 − (x)(6y) + (6y)2,

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154 Unit 2

which is not in the standard form,. Thus x2 − 6xy + 36y2 is not a square

trinomial.

Think it over!

Is it possible to factorise x2+1? Alternatively, is it possible to find

two numbers whose sum is zero and whose product is one?

Exercise 2.2.3

1. In the following, you are given the product pq and the sum p + q.

Determine p and q:

(i) pq = 18 and p + q = 11; (ii) pq = 32 and p + q = −12; (iii) pq = −24and p+q = 2; (iv) pq = −12 and p+q = 11; (v) pq = −6 and p+q = −5;(vi) pq = −44 and p+ q = −7.

2. Factorise:

(i) x2 + 6x+ 8; (ii) x2 + 4x+ 3; (iii) a2 + 5a+ 6; (iv) a2 − 5a+ 6; (v)

a2 − 3a− 40; (vi) x2 − x− 72.

3. Factorise:

(i) x2+14x+49; (ii) 4x2+4x+1; (iii) a2−10a+25; (iv) 2x2−24x+72;

(v) p2 − 24p+ 144; (vi) x3 − 12x2 + 36x.

Additional problems on “Factorisation”

1. Choose the correct answer:

(a) 4a+ 12b is equal to

A. 4a B. 12b C. 4(a+ 3b) D. 3a

(b) The product of two numbers is positive and their sum negative

only when

A. both are positive

B. both are negative

C. one positive the other negative

D. one of them equal to zero

(c) Factorising x2 + 6x+ 8, we get

A. (x+1)(x+8) B. (x+6)(x+2) C. (x+10)(x−2) D. (x+4)(x+2)

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Factorisation 155

(d) The denominator of an algebraic fraction should not be

A. 1 B. 0 C. 4 D. 7

(e) If the sum of two integers is −2 and their product is −24, the

numbers are

A. 6 and 4 B. −6 and 4 C. −6 and −4 D. 6 and −4(f) The difference (0.7)2 − (0.3)2 simplifies to

A. 0.4 B. 0.04 C. 0.49 D. 0.56

2. Factorise the following:

(i) x2 + 6x+ 9; (ii) 1− 8x+ 16x2; (iii) 4x2 − 81y2; (iv) 4a2 + 4ab+ b2; (v)

a2b2 + c2d2 − a2c2 − b2d2.


(i) x2 + 7x+ 12; (ii) x2 + x− 12; (iii) x2 − 3x− 18; (iv) x2 + 4x− 21; (v)

x2 − 4x− 192; (vi) x4 − 5x2 + 4; (vii) x4 − 13x2y2 + 36y4.


(i) 2x2+7x+6; (ii) 3x2−17x+20; (iii) 6x2−5x−14; (iv) 4x2+12xy+5y2;

(v) 4x4 − 5x2 + 1.


(i) x8 − y8; (ii) a12x4 − a4x12; (iii) x4 + x2 + 1; (iv) x4 + 5x2 + 9.

6. Factorise x4 + 4y4. Use this to prove that 20114 + 64 is a composite

number.

Glossary

Common factor: given two or more expressions, the factor of each ex-

pression which is common to all the expressions.

Factorisation: the process of writing an algebraic expression as a product

of more than one algebraic expressions.

Points to remember:

• Factorisation is the reverse process to the formation of the products;

• One can factorise some expressions using proper grouping and split-

ting of its terms.

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CHAPTER 2 UNIT 3

LINEAR EQUATIONS IN ONE VARIABLE

After studying this unit, you learn:

• the meaning of linear equation in one variable;

• to solve a linear equation in one variable;

• to formulate a linear equation from a verbal problem;

• to verify the solution.

2.3.1 Introduction

In this chapter, we study linear equations in one variable with rational

co-efficients and solve them in rational number system.

Equality of two different algebraic expression is called an equation. The

standard form of an equation is a statement that an algebraic expression

is equal to 0. The statement need not be true for any value of the variables

in it; or may be true only for certain values of the variables in it. For

example: consider 3x − 5 = 0. If we are looking for integers x for which

this statement is true, you see that no integer satisfies it. On the other

hand, if you look for a rational number, then x = 5/3 makes the statement

3x− 5 = 0 true. The value of the variable which makes the statement true

is called a solution of the equation.

Some times you may come across statements like 2x− 5 = x+ 6. This is

also an equation, but this is not in the standard form. However, this can

be brought to standard form; x− 11 = 0.

Given an equation, it may not have any solution in some system and

may have solution in some other system, as you have noticed earlier. The

equation x2 − 2 = 0 has no solution in the set of all rational numbers, but

it can be solved in the set of real numbers. In fact one of the motivations

to introduce new number systems is that we should be able to solve such

equations. You will learn later that the equation x2 + 1 = 0 also cannot be

solved in the set of all real numbers, but it can be solved in an enlarged

number system called the set of all complex numbers. Thus it is very

important to mention where you are seeking the solution.

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Linear equations 157

An equation in one variable is a statement that an algebraic expression

in one variable is equal to 0. If the degree of the expression is also 1, we

say that the equation is a linear equation. The standard form of a linear

equation in one variable is ax + b = 0, where a 6= 0. Here x is the variable

and a, b are constants(some numbers). For example: x− 9 = 0; 5x− 30 = 0;3

5x+

1

3= 0.

What do we mean by solving an equation?. The statement that some

expression equals to 0 may not be valid for every value of the variable in

it. However, some value(s) of the variable may make the statement true.

In this case, that value of the variable which makes the statement true is

called a solution of the given equation.

The process of finding solution or solutions of a given equation is called

solving the equation.

As you have observed, given a linear equation ax+b = 0, the existence of

solution depends on the number system where you are seeking solution.

If a and b are integers, then there may not be any integer solution unless a

divides b. However, if you are in the system of rational numbers, you can

always get a rational number as a solution. This is the advantage we had

from moving from integers to rational numbers. We are ready to exploit it.

With this back-ground we can now make the statement:

A linear equation with rational coefficients, ax+b = 0, a 6= 0,

has a unique solution in the rational number system.

In fact the solution can be written as x = −ba−1, where a−1 is the multi-

plicative inverse of a. You have seen earlier, while studying rational num-

bers, that any non-zero rational number has its multiplicative inverse in

the system of rational numbers.

You will see later that this statement is also true in the real number

system, provided you look for solution in the real number system. (In fact

this is true in any number system in which every nonzero number has

multiplicative inverse.)

2.3.2 Solving a linear equation in one variable

Consider the equation 5x − 15 = 0. We want to find the value of x for

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158 Unit 3

which 5x− 15 = 0 holds. Here we make use of important axioms. You will

again come across them in Geometry.

1. If equals are added to equals, we again get

equals.

If we know a = b, then for any c, we get a+c = b+c. For example x−5 = 0

is given. We add equal number 5 both sides. What the axioms says is that

equality remains intact. Thus (x−5)+5 = 0+5 which is same as x = 5. The

important point to note is that here a, b, c can be algebraic expressions as

well. Thus 3x+ 2 = 5− x implies that (3x+ 2) + (x− 5) = (5− x) + (x− 5) = 0

or 4x− 3 = 0. This axiom and others are important tools in manipulations

of expressions.

2. If equals are subtracted from equals, we again get equals.

Thus a = b implies that a− c = b− c. Given x+5 = 2x−6, we can subtract

x− 6 from both sides and get (x+ 5)− (x− 6) = (2x− 6)− (x− 6) or 11 = x.

Can you see that there is not much difference between 1 and 2 once

you consider subtraction as addition of additive inverse. Just like you

define additive inverse to numbers, you can do this for polynomials as

well and all the properties which are true for integers are true again for

polynomials.

Example 1. Solve the equation x− 15 = 0.

Solution: We add 15 both sides and invoke axiom 1 to get (x − 15) + 15 =

0 + 15 = 15. This reduces to x = 15.

Example 2. Solve the equation x+ 9 = 20.

Solution: Do you see that this is not in the standard form. We can bring

this to standard form by subtracting 20 both sides. Thus 2 gives (x + 9)−20 = 20 − 20 = 0 or x − 11 = 0. Now this is in standard form. By adding

11 both sides, you will get x − 11 + 11 = 0 + 11 or x = 11. We can combine

both these and reduce the number of steps: subtract 9 both sides. Thus

(x+ 9)− 9 = 20− 9 or x = 11.

Example 3. Solve 2x− 3 = x+ 8.

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Solution: Here you see that both the sides contain algebraic expression.

Suppose you subtract x− 3 from both the sides. You get

(2x− 3)− (x− 3) = (x+ 8)− (x− 3) or equivalently x = 11.

3. If equals are multiplied by equal quantities, we again get

equals.

Thus if a = b, then ac = bc for any c. For example, supposex

2= 1 is

given. You can multiply both the sides from 2 and getx

2×2 = 1×2 or x = 2.

4. If equals are divided by non-zero equal quantities, we get

equals.

This says that if a = b and c 6= 0, thena

c=

b

c. This again is the same as 3

once you think that division is the same as multiplication by multiplicative

inverse.

Example 4. Solvex

3= 9.

Solution: We multiply both the sides from 3 and getx

3× 3 = 9× 3 = 27.

Thus x = 27.

Example 5. Solve2x

9= 5.

Solution: We multiply both the sides by the same number9

2. Note this is

same as multiplying first by 9 and then dividing by 2. We get

2x

9× 9

2= 5× 9

2=

45

2.

We obtain x =45

2.

Example 6. Solve 15x = 120.

Solution: We divide both the sides by 15 and get

15x

15=

120

15= 8.

This gives x = 8.

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160 Unit 3

Example 7. Solve 13y = 100.

We divide both sides by 13 and get

13y

13=

100

13.

This gives y = 100/13.

Another important component of solving an equation is that of veri-

fying the solution. We have to check whether the so called solution we

obtain makes the given statement true. This is done by substituting the

value of the variable we have got in the given equation and by verifying the

truth of the statement.

Example 8. Is 2 a solution of the equation 3x− 5 = 19 ?

Solution: We substitute x = 2, in the given relation. We obtain 3x − 5 =

3(2)− 5 = 6 − 5 = 1, which is the left hand side or LHS. But the right hand

side or RHS is 19. Since 1 6= 19, we see that LHS is not equal to RHS for

x = 2. Hence x = 2 is not a solution.

Example 9. Is 7 a solution of 2x− 4 = 10 ?

Solution: Substitute x = 7 in the given relation. We get LHS = 2x − 4 =

(2 × 7) − 4 = 14 − 4 = 10 and RHS = 10. Thus we see that LHS = RHS.

Therefore, x = 7 is the solution of 2x− 4 = 10.

Example 10. Solve the equation 2x− 3 = 7.

Solution: Adding 3 both the sides, we get 2x − 3 + 3 = 7 + 3. This gives

2x = 10. Now dividing both the sides by 2, we get2x

2=

10

2. Hence x = 5.

Let us verify whether we have got a solution. Putting x = 5, we get 2x− 3 =

2(5)− 3 = 10− 3 = 7. Thus the equation is satisfied.

We can reduce the number of steps in solving an equation. Consider

the equation 2x−3 = 7. We can simply write this as 2x = 7+3. Actually you

are adding 3 both sides. But you drop that as a separate step and do the

addition mentally to write 2x−3+3 = 2x. We say, we have transposed −3 tothe other side. In fact this is a convenient way for doing fast calculations.

Again you can divide both the sides by 2 in mind and write x = 5.

We can also transpose algebraic expressions as well. For example, given

4x− 3 = 3x+ 2, we transpose 3x to the left side and −3 to the write side to

get (4x− 3x) = (3 + 2); i.e., x = 5.

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Example 11. Solve 5x− 12 = 10− 6x.

Solution: By transposing −6x to the right side and −12 to the left side, we

get 5x+ 6x = 10 + 12. Hence 11x = 22 or x = 2.

Rule for transposing: when you transpose an expression

from one side to the other side of an equality, you have

to change the sign of the expression you are transposing.

Thus −6x in the previous example becomes +6x when transposed from

the left side to the right side.

Example 12. Solve 8x− 3 = 9− 2x.

Solution: We transpose the variables on one side and constants on the

other side and get 8x+ 2x = 9+ 3. Thus 10x = 12 or x = 12/10. This reduces

to x =6

5.

Example 13. Solve 8x+ 9 = 3(x− 1) + 7.

Solution: The equation is 8x + 9 = 3x− 3 + 7 = 3x + 4. By transposing, we

get 8x − 3x = 4 − 9 or 5x = −5. We thus get x = −1. We can verify this:

8x+ 9 = 8(−1) + 9 = −8 + 9 = 1 and 3(x− 1) + 7 = 3(−1− 1) + 7 = 3(−2) + 7 =

−6 + 7 = 1. Thus 8x + 9 = 3(x − 1) + 7 is true for x = −1. This checks that

x = −1 is a solution.

Example 14. Solve2

3x =

3

8x+

7

12.

Solution: Here we multiply by the LCM of the denominators: LCM of

3,8,12 is 24. (Which axiom are you using here?) Thus(2

3x

)24 =

(3

8x+

7

12

)24,

and this simplifies to16x = 9x+ 14.

Hence 7x = 14 or x = 2. Check that x = 2 is indeed the solution.

Do you see the advantage of multiplying both the sides by the LCM of all

the denominators?. The new expression becomes an equation with integer

coefficients. You can easily handle this equation. Thus Axiom 3 helps

in transforming an equation with rational coefficients to an equivalent

equation with integer coefficients.

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162 Unit 3

Example 15. Solve the equation2x+ 7

5− 3x+ 11

2=

2x+ 8

3− 5.

Solution: Here 2,3,5 appear in the denominators of various fractions.

Their LCM is 30. We multiply through out by 30. Thus2x+ 7

5× 30− 3x+ 11

2× 30 =

2x+ 8

3× 30− (5× 30).

This reduces to

6(2x+ 7)− 15(3x+ 11) = 10(2x+ 8)− 150.

=⇒ 12x+ 42− 45x− 165 = 20x+ 80− 150.

Transposing appropriate quantities, we get

12x− 45x− 20x = −42 + 165 + 80− 150 =⇒ −53x = 53.

Hence x = −1 is the solution. You may verify this by substituting in the

equation and checking whether LHS and RHS agree.

Example 16. Solve (x+ 4)2 − (x− 5)2 = 9.

Solution: This apparently looks like an equation which is not linear. But

expanding this using identities, we get

x2 + 8x+ 16− x2 + 10x− 25 = 9.

This reduces to 18x−9 = 9. After transposing, we get 18x = 18. Hence x = 1.

It is easy to verify that x = 1 is a solution.

Think it over!

(1) You have learnt how to solve an equation of the form

ax + b = 0, where a 6= 0. You get x = (−b/a), as the unique

solution. Suppose you have two variables x and y; say an

equation ax+by+c = 0, where a 6= 0 and b 6= 0. This is again a

linear equation, but the number of variables is now 2. Can

you solve such an equation? How many solutions (x, y) are

there for such an equation? If a, b, c are integers, is it al-

ways possible to find integers x, y satisfying the equation?

(2) Can you solve (x+ 1)2 = x2 + 2x+ 1 ?

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Exercise 2.3.2

1. Solve the following:

(i) x+ 3 = 11; (ii) y − 9 = 21; (iii) 10 = z + 3; (iv)3

11+ x =

9

11;

(v) 10x = 30; (vi)s

7= 4; (vii)

3x

6= 10; (viii) 1.6 =

x

1.5;

(ix) 8x− 8 = 48; (x)x

3+ 1 =

7

15; (xi)

x

5= 12; (xii)

3x

5= 15;

(xiii) 3(x+ 6) = 24; (xiv)x

4− 8 = 1; (xv) 3(x+ 2)− 2(x− 1) = 7.

2. Solve the equations:

(i) 5x = 3x+ 24; (ii) 8t + 5 = 2t− 31; (iii) 7x− 10 = 4x+ 11;

(iv) 4z + 3 = 6 + 2z; (v) 2x− 1 = 14− x; (vi) 6x+ 1 = 3(x− 1) + 7;

(vii)2x

5− 3

2=

x

2+ 1; (viii)

x− 3

5− 2 =

2x

5; (ix) 3(x+ 1) = 12 + 4(x− 1);

(x) 2x− 5 = 3(x− 5); (xi) 6(1− 4x) + 7(2 + 5x) = 53;

(xii) 3(x+6)+2(x+3) = 64; (xiii)2m

3+8 =

m

2−1; (xiv)

3

4(x−1) = x−3.

2.3.3 Application of linear equations

We shall take up some practical situations leading to linear equations

in one variable.

Example 17. Seven times a number, if increased by 11, is 81. Find the

number.

Solution: We do this in several steps.

Step 1: First we convert the given data to an appropriate equation.

Let the number be x. Apriori, we do not know what this number is.

We formulate a linear equation involving the unknown x, using the

given data. Now seven times the number means 7x. Increasing this

by 11 leads to 7x + 11. The problem says that 7x + 11 = 81. Can you

see now that we have a linear equation in the unknown x?.

Step 2: We now have to solve the equation 7x + 11 = 81. You have

already learnt methods for solving such an equation. Transpose 11

to the other side to get 7x = 81 − 11 = 70. Divide by 7 and you get

x = 10.

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164 Unit 3

Step 3: We have to check whether the number 10 satisfies the state-

ment of our problem. Now seven times the number gives 7 × 10 = 70.

Adding 11 to this gives 81. And that is precisely the data says. Thus

x = 10 is indeed the solution.

Example 18. The present age of Siri’s mother is three times the present

age of Siri. After 5 years, their ages add to 66 years. Find their present

ages.

Solution: Again we go through several steps in the solution of this prob-

lem. Suppose Siri’s present age is x years. Then her mothers age is 3x

years. After 5 years, Siri’s and her mother’s respective ages would be x+5

and 3x + 5 years. The data says that these two numbers would add up to

66. Thus we get the equation

(x+ 5) + (3x+ 5) = 66.

Now it is easy to find the value of x. The equation reduces to 4x + 10 =

66. After transposing 10 to the other side, we obtain 4x = 56 or x = 14.

This means Siri’s present age is 14 years and her mother’s present age is

14× 3 = 42 years.

Let us now verify whether these numbers match with the statement of

the problem. After 5 years, Siri’s age would be 14 + 5 = 19 years. Her

mother’s age would be 42 + 5 = 47 years. Their sum is 19 + 47 = 66 years

which completely matches with the given statement. We conclude that

Siri’s present age is 14 years and her mother’s age is 42 years.

Example 19. The sum of three consecutive even numbers is 252. Find

them.

Solution: Let x be the least number among these three consecutive even

numbers. Then the other numbers are x+2 and x+4. This is because any

two consecutive even numbers differ by 2. The given condition says that

x+ (x+ 2) + (x+ 4) = 252.

Thus we get 3x+ 6 = 252. This reduces to 3x = 246. Solving this, we obtain

x = 82. Hence the numbers are 82, 82 + 2 = 84 and 82 + 4 = 86. We check

that82 + 84 + 86 = 252,

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and this verifies the validity of the solution.

Example 20. If the perimeter of a triangle is 14 cm and the sides are x+4,

3x+ 1 and 4x+ 1, find x.

A B

C

x+4

+1x43 +1x

Solution: You know that the

perimeter of a triangle is the sum

of its three sides. Since the sides

are given to be x + 4, 3x + 1 and

4x + 1, the perimeter is (x + 4) +

(3x + 1) + (4x + 1) = 8x + 6. The

given condition is 8x + 6 = 14. On

solving this, you get x = 1. Hence

the sides are 1+4 = 5, (3×1)+1 = 4

and (4 × 1) + 1 = 5 cm. You get a

triangle with sides 5, 4, 5 cm.

One of the important things to keep in mind is that a mathemati-

cal formulation of a problem may not always be physically feasible.

Suppose you are asked to find the sides of a triangle, given that

they are x, x + 1 and x + 3, and its perimeter is 10 units. You write

x + (x + 1) + (x + 3) = 10 and solve this to get x = 2. You may con-

clude that the sides are 2, 2+ 1 = 3, 2+ 3 = 5. But, there is no triangle

with sides 2, 3, 5, since the sides of a triangle must satisfy the triangle

inequality; the sum of any two sides is greater than the third side.

Thus one has to check whether the solution one obtains is a physi-

cally valid solution. This is a very important part in the solution of a

given problem.

There is nothing strange in this. Mathematics is a game you play

using certain rules. As long as you play it safe confining to rules, you

will always get an end result. Whether that result is right or wrong,

no mathematical law will tell you. You have to go back to the physical

situation and see whether the solution you got is a physically correct

solution.

Example 21. Let P be a point on a line AB such that P lies between A

and B, and AP = 3PB. Given that AB = 10 cm, find the length of AP .

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A P B

Solution: Since P lies between A

and B, we have AB = AP + PB.

Thus 10 = 3PB + PB = 4PB. We

can solve for PB and get PB = 10/4 = 5/2. Hence AP = 3PB = 3 × 5

2=

15

2cm. (Here we are not putting PB = x and get an equation in x. You may

directly treat PB as a variable and get an equation involving PB.)

Activity 1:

In the previous problem you have taken that P lies between A and B. It

may happen that, on the line AB, P may lie to the left of A or to the right

of B. Formulate appropriate equations for both these cases and solve

them. One case gives you negative number. This is not a feasible practical

situation, as length is non-negative number.

Example 22. The sum of the digits of a two digit number is 12. If the new

number formed by reversing the digits is greater than the original number

by 54, find the original number.

Solution: Let x be the digit in units place. Then 12 − x is the digit is ten’s

place. Hence the number is

(12− x)× 10 + x = 120− 9x.

The number obtained by reversing the digits is 10×x+12−x = 12+9x. The

given condition

120− 9x+ 54 = 12 + 9x.

From this we obtain 18x = 120 + 54 − 12 = 162. Hence x = 9. Therefore thedigit in units place 9. The digit is ten’s place is 12 − 9 = 3. The number ishence 39.

Alternate solution: Let the digit in the unit’s place be x and the digit in the ten’s placebe y. Thus the number is 10y + x. We know that x + y = 12. The number obtained byreversing the digits is 10x+ y. The second condition tells that 10x+ y = 10y + x+ 54. Thisreduces to 9(x− y) = 54 or x− y = 6.

Observe that, we have two equations x+y = 12 and x−y = 6. Adding these, you obtain

(x + y) + (x − y) = 12 + 6 = 18. Thus 2x = 18 or x = 9. Since y = 12 − x, you will also get

y = 12− 9 = 3. This shows that the original number is 39.

We verify this. The number obtained by reversing the digits is 93. You

may easily check that 93 = 39 + 54.

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Example 23. The sum of two numbers is 75 and they are in the ratio 3:2.

Find the numbers.

Solution: The numbers which are in the ratio 3:2 are 3x and 2x. We are

given

3x+ 2x = 75.

Thus

5x = 75.

Solving for x, we get x = 15. Hence the numbers are 3x = 45 and 2x = 30

We verify: 45/30 = 3/2 and 45 + 30 = 75.

Exercise 2.3.3

1. If 4 is added to a number and the sum is multiplied by 3, the result is

30. Find the number.

2. Find three consecutive odd numbers whose sum is 219.

3. A rectangle has length which is 5 cm less than twice its breadth. If

the length is decreased by 5 cm and breadth is increased by 2 cm,

the perimeter of the resulting rectangle will be 74 cm. Find the length

and breadth of the original rectangle.

4. A number subtracted by 30 gives 14 subtracted by 3 times the num-

ber. Find the number.

5. Sristi’s salary is same as 4 times Azar’s salary. If together they earn

Rs 3,750 a month, find their individual salaries.

6. Prakruthi’s age is 6 times Sahil’s age. After 15 years, Prakruthi will

be 3 times as old as Sahil. Find their age.

7. In the figure, AB is a straight line. Find

x.

A

D C

B

x

+40x

+20x

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8. If 5 is subtracted from three times a number, the result is 16. Find

the number.

9. Find two numbers such that one of them exceeds the other by 9 and

their sum is 81.

10. The length of a rectangular field is twice its breadth. If the perimeter

of the field is 288 m, find the dimensions of the field.

11. Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will

be twice that of his son. Find their present age.

12. Sanju is 6 years older than his brother Nishu. If the sum of their ages

is 28 years, what are their present age ?

13. Viji is twice as old as his brother Deepu. If the difference of their ages

is 11 years, find their present age.

14. Mrs. Joseph is 27 years older than her daughter Bindu. After 8 years

she will be twice as old as Bindu. Find their present age.

15. After 16 years, Leena will be three times as old as she is now. Find

her present age.

Additional problems on “Linear equations in one variable”

1. Choose the correct answer

(a) The value of x in the equation 5x− 35 = 0 is:

A. 2 B. 7 C. 8 D. 11

(b) If 14 is taken away from one fifth of a number, the result is 20.

The equation expressing this statement is:

A. (x/5)− 14 = 20 B. x− (14/5) = (20/5)

C. x− 14 = (20/5) D. x+ (14/5) = 20

(c) If five times a number increased by 8 is 53, the number is:

A. 12 B. 9 C. 11 D. 2

(d) The value of x in the equation 5(x− 2) = 3(x− 3) is:

A. 2 B. 1/2 C. 3/4 D. 0

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(e) If the sum of two numbers is 84 and their difference is 30, the

numbers are:

A. −57 and 27 B. 57 and 27 C. 57 and −27 D. −57 and −27(f) If the area of a rectangle whose length is twice its breadth is 800

m2, then the length and breadth of the rectangle are:

A. 60 m and 20 m B. 40 m and 20 m

C. 80 m and 10 m D. 100 m and 8 m

(g) If the sum of three consecutive odd numbers is 249, the numbers

are

A. 81,83,85 B. 79,81,83 C. 103,105,107 D. 95,97,99

(h) If (x+ 0.7x)/2 = 0.85, the value of x is:

A. 2 B. 1 C. −1 D. 0

(i) If 2x− (3x− 4) = 3x− 5, then x equals:

A. 4/9 B. 9/4 C. 3/2 D. 2/3

2. Solve: (i) (3x+ 24)/(2x+ 7) = 2: (ii)(1− 9y)/(11− 3y) = (5/8).

3. The sum of two numbers is 45 and their ratio is 7:8. Find the num-

bers.

4. Shona’s mother is four times as old as Shona. After five years, her

mother will be three times as old as Shona(at that time). What are

their present age?

5. The sum of three consecutive even numbers is 336. Find them.

6. Two friends A and B start a joint business with a capital 60, 000. If

A’s share is twice that of B, how much have each invested?

7. Which is the number when 40 is subtracted gives one-third of the

original number?

8. Find the number whose sixth part exceeds its eigth part by 3.

9. A house and a garden together cost 8, 40, 000. The price of the garden

is5

7times the price of the house. Find the price of the house and the

garden.

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170 Unit 3

10. Two farmers A and B together own a stock of grocery. They agree to

divide it by its value. Farmer A takes 72 bags while B takes 92 bags

and gives 8, 000 to A. What is the cost of each bag?.

11. A father’s age is four times that of his son. After 5 years, it will be

three times that of his son. How many more years will take if father’s

age is to be twice that of his son?

12. Find a number which when multiplied by 7 is as much above 132 as

it was originally below it.

13. A person buys 25 pens worth 250, each of equal cost. He wants to

keep 5 pens for himself and sell the remaining to recover his money.

What should be the price of each pen?

14. The sum of the digits of a two-digit number is 12. If the new number

formed by reversing the digits is greater than the original number by

18, find the original number. Check your solution.

15. The distance between two stations is 340 km. Two trains start si-

multaneously from these stations on parallel tracks and cross each

other. The speed of one of the them is greater than that of the other

by 5 km/hr. If the distance between two trains after 2 hours of their

start is 30 km., find the speed of each train.

16. A steamer goes down stream and covers the distance between two

ports in 4 hours while it covers the same distance up stream in 5

hours. If the speed of the steamer upstream is 2 km/hour, find the

speed of steamer in still water.

17. The numerator of the rational number is less than its denominator

by 3. If the numerator becomes three times and the denominator is

increased by 20, the new number becomes 1/8. Find the original

number.

18. The digit at the tens place of a two digit number is three times the

digit at the units’ place. If the sum of this number and the number

formed by reversing its digits is 88, find the number.

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19. The altitude of a triangle is five- thirds the length of its corresponding

base. If the altitude is increased by 4 cm and the base decreased by

2 cm, the area of the triangle would remain the same. Find the base

and altitude of the triangle.

20. One of the angles of a triangle is equal to the sum of the other two

angles. If the ratio of the other two angles of the triangle is 4:5, find

the angles of the triangle.

Glossary

Equation: a statement that a non constant algebraic expression is equal

to 0.

Solution: given an equation, any value of the variable which makes the

statement true.

Linear equation: an equation of degree one.

Transposition: moving a part of the expression to the other side of the

equality; while moving, the sign of the part which has moved changes.

Verification: to check whether the solution obtained satisfies the equa-

tion.

Points to remember

• An equation is valid for a certain set of values of the variable(s) in it;

an identity is valid for all values of the variable(s) in it.

• Given a problem, setting up an equation conforming to the given data

is an important step in the solution of the given problem.

• A mathematical solution may not always be valid physically. One

has to check whether the solution obtained by a valid mathematical

procedure is also correct for the given physical situation.

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CHAPTER 2 UNIT 4

EXPONENTS


• the concept of an integral power to a non-zero base;

• to write large numbers in exponential form;

• about the various laws of exponents and their use in simplifying com-

plicated expressions;

• about the validity of these laws of exponents for algebraic variables.

2.4.1 Introduction

Suppose somebody asks you: how far is the Sun from the Earth? What

is your answer? Perhaps some search in books will give you an idea how

far is the Sun from us. A ray of light travels approximately at the speed of

2,99,792 km per second. It takes roughly eight and a half minutes for a

ray of light to reach the Earth starting from the Sun. Hence the distance

from the Earth to the Sun is about 15, 29,00,000 km. Apart from the Sun,

do you know how far is the nearest star to us? Proxima Centauri is the

closest star to us and it is at a distance of 4.3 light years from us; that is,

the distance a ray of light travels in 4.3 years at the speed of 2,99,792 km

per second. This is equal to

4.3× 365× 24× 60× 60× 299792km,

which you can see is a very huge number.

At the other end what would be the size of an atom ? As you can expect,

it should be very small. The diameter of an atom is roughly 1/100000000000

meters. The subatomic particles have still smaller sizes. Hence there is

a need to represent either big or small numbers in a compact form. The

exponential notation is a very useful and handy notation used for this

purpose. As we shall see there is a methodical way of handling exponen-

tials.

Consider the number 128. An easy factorisation gives

128 = 2× 2× 2× 2× 2× 2× 2.

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Exponents 173

Thus 128 is obtained by multiplying 2 with itself seven times. We write

this as 27. Here 2 is called the base and 7 is called the exponent. We read

this as 2 raised to the power 7. The number 27 is called the exponential

form of 128. Similarly, we observe that

243 = 3× 3× 3× 3× 3 = 35.

Here 3 is the base and 5 is the exponent. We read this as 3 raised to the

power 5.

On the other-hand, consider 72 = 2×2×2×3×3. In the case of 128, the

only prime factor is 2 and we are able to write 128 = 27. Similarly, 343 = 35,

as 3 is the only prime factor of 343. However, 72 has two prime factors,

namely, 2 and 3. We see that 2 occurs 3 times and 3 occurs 2 times. We

write this in the form 72 = 23 · 32. We read this as 2 to the power of 3 times

3 to the power of 2.

Let us consider 81 = 3×3×3×3 = 34. We also observe that 81 = 9×9 = 92.

Thus the same number 81 can be written in two ways: 81 = 34 = 92.

In the first representation, 3 is the base and 4 is the exponent. In the

second representation, 9 is the base and 2 is the exponent. Thus the

same number may be represented using different bases and exponents.

Observe!

9 = 3× 3 = 32. Hence 92 =(32)2. Thus you obtain

(32)2

= 92 = 81 = 34.

Can you recognise some thing?

It is not necessary that we use only numbers. For example, if we have

an algebraic variable a, we write a× a× a× a = a4 and read this as a raised

to the power four. Many times we suppress the word raised, and read a

to the power four or simply a power four. An expression of the form ab4

is read as a times b power four.

Observe:

ab4 = a × b × b × b × b = a(b4), but not ab × ab × ab × ab which is

actually a4b4

To avoid such confusions, it is always better to use braces: instead of

writing ab4, write this as a(b4). If you want 529 = 23× 23, write this as (23)2.

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174 Unit 4

The notation 232 may some time be confused with 2× 32 which is 18. Thus

use proper braces: 2(32) to represent 18 and (23)2 to represent 529.

Another thing you may notice is that it is not necessary to use only

integers as bases. In fact you may use any real number for base:

(0.1)5 = 0.1× 0.1× 0.1× 0.1× 0.1 = 0.00001,

(−1.2)3 = (−1.2)× (−1.2)× (−1.2) = −1.728,(1

5

)4

=1

5× 1

5× 1

5× 1

5=

1

625,

(√2)2 = 2.

Think it over!

Is it possible to take for exponent a real number which is not an

integer?

In general, given a number a or an algebraic variable which we again

denote by a, and a natural number n, we define

an = a× a× a× · · · × a︸︷︷︸n times

.

This is read as a raised to the power n or simply a power n. Here, we say

a is the base and n is the exponent.

Observe:

0n = 0 for any natural number n and a1 = a for any number a.

Moreover, am = an if and only if m = n for any number a 6= 0,

a 6= 1 or a 6= −1

Let us look at more examples:

Number Expanded Exponential Base and

form form exponent

1000 10× 10× 10 103 base 10, exponent 3

m6 m×m×m×m×m×m m6 base m, exponent 6

1

1024

1

2× 1

2× · · · (10 times)

(1

2

)10

base1

2, exponent 10

625 5× 5× 5× 5 54 base 5, exponent 4

625 (−5)× (−5)× (−5)× (−5) (−5)4 base −5, exponent 4.

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Exponents 175

Look at the last two examples. The number 625 is represented as 54

(with base 5 and exponent 4) and also as (−5)4 (with base −5 and exponent

4). Thus the same number has representation in base 5 and base −5.For a number a 6= 0, and a natural number n, we define

a−n =

(1

a

)n

=1

an.

This extends the definition of power to negative integer exponents. Here

are some more examples:

3−4 =

(1

3

)×(1

3

)×(1

3

)×(1

3

); (read as 3 to the power −4.)

(−0.1)−5 =

(1

(−0.1)

)5

= (−10)5 = −100000; (read as (−0.1) to the power −5.)

(4

5

)−6=

(5

4

)6

=5

4× 5

4× 5

4× 5

4× 5

4× 5

4=

56

46.

If a = bn for some integer n 6= 0, 1 and number b 6= 0, we say a is in

exponential form. Here b is called the base and n is called the exponent.

Earlier you have observed that 81 = 34 = 92. Thus the same number may

have different exponential forms.

Caution!

bn is not defined for b = 0, and n = 0 or

n < 0.

Example 1. Express 1024 and 0.1024 using base 2.

Solution: Consider the number 1024 = 210. Here we have base 2 and

exponent 10. However, 1024 is a number in decimal notation: here 4 is in

unit’s place; 2 in 10’s place; 0 in 100’s place; and 1 in 1000’s place. Thus

we have

1024 = (1× 1000) + (0× 100) + (2× 10) + (4× 1)

= (1× 103) + (2× 101) + 4.

Suppose we define b0 = 1 for any number b 6= 0. Then we see that 1024 =

1 ·103+2 ·101+4 ·100. Do you see that we have done this in the unit playing

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176 Unit 4

with numbers? Similarly, consider 0.1024. You observe that

0.1024 = (0.1) + (0.002) + (0.0004)

=1

10+

2

1000+

4

10000

=1

10+

2

103+

4

104

= 1 · 10−1 + 2 · 10−3 + 4 · 10−4.

Example 2. Express 1000 using base 2 and exponents.

Solution: Consider 1000. How do you express this using base 2 and

exponents? You observe that 512 = 29, 256 = 28, 128 = 27 and 26 = 64. Thus

you get the sum

512 + 256 + 128 + 64 = 960.

You are still short by 40 to reach 1000. But 40 = 32+ 8 = 25+23. Thus you

obtain

1000 = 29 + 28 + 27 + 26 + 25 + 23

= 1 · 29 + 1 · 28 + 1 · 27 + 1 · 26 + 1 · 25 + 0 · 24 + 1 · 23 + 0 · 22 + 0 · 21 + 0 · 20.

You observe that this is precisely the binary representation of 1000. We

write this 1000 = (1111101000)2.

Some uses of exponents in daily life

Some examples of how exponents do connect with our everyday lives:

when we speak about square feet, square meters or any such area units;

or about cubic feet, cubic meters, cubic centimeters or any other such

volume units.

The unit square cm is actually 1 cm × 1 cm = 1 cm2. Similarly, a cubic

cm is 1 cm × 1 cm × 1 cm = 1 cm3.

Another kind of indirect example is if you talk about extremely small or

extremely big quantities. For example, the term ‘nanometer’ means 10−9

m. The prefix ‘nano’ means the number 10−9 - an extremely small number.

Or, within computer world you often see megabytes, gigabytes, terabytes.

Mega means 106 or one million; giga means 109, and tera means 1012.

Suppose you have a chemical concentration 0.000442 gm per litre, or

a star having huge mass of 8,290,000,000,000,000,000,000 kg. Those

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Exponents 177

zeroes take up a lot of space. Scientific notation simplifies this by writing

the last number, for example, as 829 × 1019 kg, which means 829 followed

by 19 zeroes. On the other-hand, the chemical concentration is written

with a negative exponent: 4.42× 10−4 gm per litre.

The mass of a proton is 0.000000000000000000000001673 gm. You see how

much space you need to write such a tiny number. If you use exponents,

you may write this in the form 1.673× 10−24 gm.

Another place where exponential notation helps is in calculators. In a

calculator, no display is possible once the number exceeds certain fixed

number of digits. This fixed number varies from calculator to calcula-

tor depending on the capacity of the calculator. Scientific calculators, in

which large or small numbers are calculated, use exponential notation.

For example, a calculator may show only eight significant digits in its dis-

play. So to display a number of the form 234587643214878, which has

15 digits, calculators round it off to 234587640000000 and write the re-

sulting number in the form 23458764×107 ( or some calculators in the form

.23458764× 1015).

Activity 1:

Collect information about many more life situations where the exponential

notation is useful.

Indian contribution to large numbers

The Indians had a passion for high numbers, which is intimately related

to their religious thought. For example, in texts belonging to the Vedic lit-

erature dated from 1200 BC to 500 BC, we find individual Sanskrit names

for each of the powers of 10 up to a trillion and even 1062. One of these

Vedic texts, the Yajur Veda (1200-900 BC), even discusses the concept of

numeric infinity (purna “fullness”), stating that if you subtract purna from

purna, you are still left with purna.

The Lalitavistara Sutra (a Mahayana Buddhist work) recounts a contest

including writing, arithmetic, wrestling and archery, in which the Buddha

was pitted against the great mathematician Arjuna and showed off his

numerical skills by citing the names of the powers of ten up to 1 ‘tallak-

shana’, which equals 1053, but then going on to explain that this is just

one of a series of counting systems that can be expanded geometrically.

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178 Unit 4

The last number at which he arrived after going through nine successive

counting systems was 10421, that is, a 1 followed by 421 zeros.

There is also an analogous system of Sanskrit terms for fractional num-

bers, capable of dealing with both very large and very small numbers.

Larger number in Buddhism seems

107×2122

or 1037218383881977644441306597687849648128,

which appeared as Bodhisattva’s maths in the Avatamsaka Sutra.

Here are a few large numbers used in India by about 5-th century BC

(See Georges Ifrah: A Universal History of Numbers, pp 422-423):

koti - 107, ayuta - 109, niyuta - 1011, kankara - 1013, pakoti - 1014, vivara

- 1015, kshobhya - 1017, vivaha - 1019 , kotippakoti - 1021, bahula - 1023,

nagabala - 1025, and so on. They also had name for 10421 as dhvajagran-

ishamani. (wikipedia.org/wiki/History_of_large_numbers)

Two Indian Legends about Large numbers

1. King Shiraham of India was pleased with his grand Vizier (Chief min-

ister) for inventing the game of Chess and wanted to reward him. Vizier’s

request was simple. He said: “ Your Majesty, give me a grain of wheat

to put on the first square of the chess board, two grains to put on the

second square, four grains to put on the third, eight on the fourth, each

time doubling the number of grains, please give me enough grains to cover

all the 64 squares.” King thought that his minister’s request was very

very modest. He ordered his men to bring the wheat grains and fulfill

his Vizier’s desire. But he soon realised his folly. Do you know how many

grains are needed to comply with the minister’s request? It is 264−1 grains.

This is roughly equal to the world’s wheat production for more than 2000

years.(At the current rate of production.)

2. In the great temple of Beneras(that is Kashi), there is a brass plate

in which three diamond needles are fixed. On one of the needles, at the

creation, Lord Brahma placed sixty four golden discs of decreasing diam-

eters such that each disc sits only on a disc of larger diameter. Day and

night, the priest on duty transfers discs from one diamond needle to the

other according to the immutable laws of Lord Brahma. This requires that

the priest must move only one disc at a time and only in such a way that

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Exponents 179

there is no larger disc on a smaller disc. When all the sixty four discs get

transferred to another needle in accordance with this rule, the world will

come to an end. Can you make a guess of the time required for this? Even

with the assumption that the priests move one disc per second, the total

time needed would be 264 − 1 seconds. If you convert this to years, it will

be more than 58 thousand billion years (58× 1012 years). This is five times

more than the current estimated age of our Universe.

Exercise 2.4.1

1. Express the following numbers in the exponential form:

(i) 1728 (ii)1

512(iii) 0.000169.

2. Write the following numbers using base 10 and exponents:

(i) 12345 (ii) 1010.0101 (iii) 0.1020304

3. Use base 5 and exponents to represent 2010.

4. Express (1234)5 in decimal form

5. Find the value (−0.2)−4.

2.4.2 The first law of exponents

Consider 1024 = 210. Observe

210 = 1024 = 2× 512 = 21 × 29 : 1 + 9 = 10 and 210 = 21+9;

210 = 1024 = 4× 256 = 22 × 28 : 2 + 8 = 10 and 210 = 22+8;

210 = 1024 = 8× 128 = 23 × 27 : 3 + 7 = 10 and 210 = 23+7;

210 = 1024 = 16× 64 = 24 × 26 : 4 + 6 = 10 and 210 = 24+6;

210 = 1024 = 32× 32 = 25 × 25 : 5 + 5 = 10 and 210 = 25+5;

What do you observe? Consider more examples:

54 = 625 = 25× 25 = 52 × 52 : 2 + 2 = 4 and 54 = 52+2;

a7 = (a× a× a× a)× (a× a× a) = a4 × a3 : 4 + 3 = 7 and a7 = a4+3.

Looking at these examples, can you formulate a law?

For any number a and positive integers m,n, am × an =

am+n.

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180 Unit 4

This is also true for an algebraic variable x: xmxn = xm+n. This is called

the first law of exponents. This is useful in simplifying large expressions.

Examples:

3. 25 × 26 = 25+6 = 211;

4. 33 × 36 × 37 = (33 × 36)× 37 = 33+6 × 37 = 39 × 37 = 39+7 = 316.

5. 25 × 52 × 23 × 5 = (25 × 23)× (52 × 5) = 28 × 53.

Exercise 2.4.2

1. Simplify:

(i) 31 × 32 × 33 × 34 × 35 × 36. (ii) 22 × 33 × 24 × 35 × 36.

2. How many zeros are there in 104 × 103 × 102 × 10?

3. Which is larger:(53 × 54 × 55 × 56

)or(57 × 58

)?

2.4.3 The second law of exponents

Look at the following example:

29 = 512 =1024

2=

210

21: 9 = 10− 1 and 29 = 210−1;

28 = 256 =1024

4=

210

22: 8 = 10− 2 and 28 = 210−2;

27 = 128 =1024

8=

210

23: 7 = 10− 3 and 27 = 210−3;

26 = 64 =1024

16=

210

24: 6 = 10− 4 and 26 = 210−4;

26 = 64 =256

4=

28

22: 6 = 8− 2 and 26 = 28−2.

Study one more:

33 = 27 =243

9=

35

32: 3 = 5− 2 and 33 = 35−2.

Can you see some pattern? Is it apparent that some law is followed here

again?

For any number a and positive integers m,n, with m > n,

am

an= am−n.

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Exponents 181

The natural question is: what happens if m = n or m < n. A few exam-

ples will help to clarify the situation.

1

25=

1

32=

2

64=

21

26.

Recall: we have defined a−n = 1an for any a 6= 0 and positive integer n. Thus

2−5 =21

26: −5 = 1− 6.

Similarly,

3−4 =1

34=

1

81=

9

729=

32

36: −4 = 2− 6 and 3−4 = 32−6;

10−1 =1

10=

1000

10000=

103

104; −1 = 3− 4 and 10−1 = 103−4.

The above observations may be put in the following form:

For any number a 6= 0 and positive integers m,n, with m 6= n,

am

an= am−n.

Again recall, we have defined a0 = 1 for all a 6= 0. Observe

25

25= 1 = 20 : 5− 5 = 0 and 25−5 = 20;

34

34= 1 = 30 : 4− 4 = 0 and 34−4 = 30.

This shows that am

am = 1 = a0 for all a 6= 0. We can now reformulate our law:

For any number a 6= 0 and positive integers m,n, not nec-

essarily distinct, am

an= am−n.

The important thing here is that the only condition on m,n is that they are

positive integers. It may happen that m < n, or m = n or m > n. In all these

cases, the above law holds.

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182 Unit 4

There is another way of looking at the second law. Suppose a 6= 0, and

m,n are positive integers. Then

am−n =am

an= am × 1

an= am × a−n,

where we have used a−n =1

an. Thus we obtain

am−n = am × a−n.

Observe that this resembles the first law except that we have negative

integer −n. In this sense, the second law simply extends the first law.

Again observe that for natural numbers m,n,

a−(m+n) =1

am+n =1

am × an(using the first law)

=1

am× 1

an(property of fractions)

= a−m × a−n.

Thus we see that the first law is also valid for negative integers m,n. We

can combine both the first and the second law and state them together:

If a 6= 0 and m,n are integers, then am × an =

am+n.

Example 6. Simplify35 × 22 × 7−3

7−2 × 3−3 × 24.

Solution: The expression is equal to

(35

3−3

)×(22

24

)×(7−3

7−2

)= 35−(−3) × 22−4 × 7−3−(−2)

= 35+3 × 2−2 × 7−3+2

= 38 × 2−2 × 7−1 =38

22 × 7=

6561

28.

Example 7. If 3l × 32 = 35, find the value of l.

Solution: Using the second law of exponents, we obtain 3l+2 = 35. But we

know that for any a 6= 0, 1,−1, if am = an, then m = n. Hence l + 2 = 5 or

l = 3.

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Exponents 183

Exercise 2.4.3

1. Simplify: (i) 10−1 × 102 × 10−3 × 104 × 10−5 × 106; (ii)23 × 32 × 54

33 × 52 × 24.

2. Which is larger:(34 × 23

)or(25 × 32

)?

3. Suppose m and n are distinct integers. Can3m × 2n

2m × 3nbe an integer?

Give reasons.

4. Suppose b is a positive integer such that24

b2is also an integer. What

are the possible values of b?

2.4.4 The third law of exponents

Consider the following examples:

210 = 22+2+2+2+2 = 22 × 22 × 22 × 22 × 22 =(22)5

:

2× 5 = 10 and (22)5 = 210 = 22×5;

210 = 25+5 = 25 × 25 =(25)2

:

5× 2 = 10 and(25)2

= 210 = 25×2;

312 = 32+2+2+2+2+2 = 32 × 32 × 32 × 32 × 32 × 32 =(32)6

:

2× 6 = 12 and(32)6

= 312 = 32×6;

312 = 33+3+3+3 = 33 × 33 × 33 × 33 =(33)4

:

3× 4 = 12 and(33)4

= 312 = 33×4;

312 = 36+6 = 36 × 36 =(36)2

:

6× 2 = 12 and(36)2

= 312 = 36×2.

What do you see? Can you put your observations in to a rule?

If a 6= 0 is a number and m,n are positive integers, then(am)n

= amn.

Study more examples:

22×(−4) = 2−8 =1

28=

1

(22)4=(22)−4

;

3(−5)×2 = 3−10 =1

310=

1

(35)2=

(1

35

)2

=(3−5

)2.

Do you see that negative integer exponents follow similar rule?

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184 Unit 4

If a 6= 0 is a number and m,n are positive integers, then(am)−n

=

a−mn =(a−m

)n.

What do you expect in the case when exponents are negative integers?

Consider the example:

5(−4)×(−3) = 512 =1

5−12=

1(5−4

)3 =(5−4

)−3.

Are you convinced that the negative exponents also follow the same rule:(a−m

)−n= amn for all numbers a 6= 0 and positive integers m,n?

What if one of m,n is equal to 0? You observe that for a 6= 0, a0 = 1 and

hence (a0)n = 1n = 1 = a0 and 0× n = 0.

Thus we can formulate the third law of exponents:

If a 6= 0 is a number and m,n are integers , then(am)n

=

amn.

Example 8. Simplify(1024)3 × (81)4

(243)2 × (128)4.

Solution: Observe 1024 = 210, 81 = 34, 243 = 35, and 128 = 27. Hence the

expression is

(210)3 × (34)4

(35)2 × (27)4=

230 × 316

310 × 228

= 230−28 × 316−10 = 22 × 36 = 2916.

Exercise 2.4.4

1. Simplify: (i)

(25)6 ×

(33)2

(26)5 ×

(32)3 ; (ii)

(5−3

)2 × 34(3−2

)−3 ×(53)−2 .

2. Can you find two integers m,n such that 2m+n = 2mn?

3. If(2m)4

= 46, find the value of m.

2.4.5 The fourth law of exponents

Study the following examples.

(1) 105 = 10× 10× 10× 10× 10

= 2× 5× 2× 5× 2× 5× 2× 5× 2× 5

= (2× 2× 2× 2× 2)× (5× 5× 5× 5× 5)

= 25 × 55.

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Exponents 185

Here you may observe that 10 = 2× 5.

(2) 34 × 24 = (3× 3× 3× 3)× (2× 2× 2× 2)

= (3× 2)× (3× 2)× (3× 2)× (3× 2)

= 6× 6× 6× 6

= 64.

Here again 3× 2 = 6. Consider one more example:

(3) (−2)4 × (44) = (−2)× (−2)× (−2)× (−2)× 4× 4× 4× 4

= (−2× 4)× (−2 × 4)× (−2× 4)× (−2 × 4)

= (−8)× (−8)× (−8)× (−8) = (−8)4.

Once again you find that (−2)× 4 = (−8). Let us see what happens if both

the numbers are negative.

(4) (−3)3 × (−5)3 = (−3)× (−3)× (−3)× (−5)× (−5)× (−5)=

((−3)× (−5)

)×((−3)× (−5)

)×((−3)× (−5)

)

= 15× 15× 15

= 153.

Again you see that expected rule is true: (−3)× (−5) = 15. Now you can

formulate new law:

If a and b are two non-zero numbers and m is any positive integer,

then (a× b)m = am × bm.

Observe that this is also true if m = 0. Since a 6= 0 and b 6= 0, you may

conclude that a × b 6= 0. Thus (a × b)0 = 1. But a0 = 1 and b0 = 1 so that

a0 × b0 = 1. It follows that (a× b)0 = a0 × b0. What happens if m is negative?

If a, b are non-zero numbers and n is a positive integer, you see that

(a× b)−n =1

(a× b)n=

1

an × bn=

1

an× 1

bn= a−n × b−n.

Thus the negative exponents also obey the same rule. We may now refor-

mulate the law:

For any two numbers a 6= 0 and b 6= 0, and integer m,

(a× b)m = am × bm.

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186 Unit 4

Here are more examples .

Example 9. Simplify:154 × 142

213 × 103.

Solution: The numerator is

154 × 142 = (3× 5)4 × (2× 7)2 = 34 × 54 × 22 × 72.

Similarly the denominator is

213 × 103 = (7× 3)3 × (2× 5)3 = 73 × 33 × 23 × 53.

The given expression is therefore

34 × 54 × 22 × 72.

73 × 33 × 23 × 53= 34−3 × 54−3 × 22−3 × 72−3 =

3× 5

2× 7=

15

14.

Example 10. Which is larger: (0.25)4 or (0.35)3?

Solution: We use the simple observation: if a and b are nonzero numbers,

then a < b is equivalent toa

b< 1.

Write 0.25 =1

4and 0.35 =

35

100=

7

20. We have to compare

1

44and

73

203. But

203 = (4× 5)3 = 43 × 53. Thus we see that

(0.25)4

(0.35)3=

43 × 53

44 × 73=

53

4× 73< 1,

since 5 < 7. Thus (0.25)4 < (0.35)3.

Exercise 2.4.5

1. Simplify: (i)68 × 53

103 × 34; (ii)

(15)−3 × (−12)45−6 × (36)2

; (iii)(0.22)4 × (0.222)3

(0.2)5 × (0.2222)2.

2. Is(104)3

513an integer? Justify your answer.

3. Which is larger : (100)4 or (125)3?

2.4.6 The fifth law of exponents

Consider (3

4

)5

=3

4× 3

4× 3

4× 3

4× 3

4=

3× 3× 3× 3× 3

4× 4× 4× 4× 4=

35

45.

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Exponents 187

Similarly, you observe that(−4

5

)3

=

(−45

)×(−4

5

)×(−4

5

)=

(−4)× (−4)× (−4)5× 5× 5

=(−4)353

.

What do you infer from these two examples? Do you see that there is yet

another law of exponent?

If a, b are non-zero numbers and m is a positive integer,(ab

)m=

am

bm.

You may also observe that this is valid if m = 0. In this case(ab

)0= 1 =

1

1=

a0

b0.

What do you think if m happens to be a negative integer? If m < 0, then

n = −m is a positive integer. Hence you can use the law for positive integral

exponent: (ab

)n=

an

bn.

However, we know that an =1

a−n=

1

am. Similarly bn =

1

b−n=

1

bmor

1

bn= bm.

Thus you may obtain(ab

)n=

an

bn= an × 1

bn=

1

am× bm =

bm

am.

But we know 1(ab

)m =(ab

)n=

bm

am.

This implies that am

bm=(ab

)m.

We have obtained law for negative exponent. Thus we may write the fifth

law of exponents:

If a 6= 0 and b 6= 0, and m is an integer, then(ab

)m=

am

bm.

Example 11. Simplify:

(0.2

5

)4

×(

2

0.5

)−3.

Solution: You may simplify the two fractions separately. Thus

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188 Unit 4

(0.2

5

)=

1

52,

(2

0.5

)= 2× 2 = 22.

Thus the expression simplifies to(

1

52

)4

×(22)−3

=1

58 × 26=

1

52 × 106= 0.00000004.

Example 12. Which is larger: (2.5)6 or (1.25)12?

Solution: We write (2.5)6 =

(5

2

)6

=56

26. Similarly, you may obtain

(1.25)12 =

(5

4

)12

=512

412=

512

224.

Thus you have to check which of the numbers56

26,

512

224is larger. Equiva-

lently, you must determine the larger number between 56 and 218 (why?).

However 52 = 25 < 32 = 25. Hence

56 = (52)3 < (25)3 = 215 < 218.

Thus512

224=

56

218× 56

26<

56

26.

Hence (1.25)12 < (2.5)6.

Exercise 2.4.6

1. Simplify: (i)

(2

3

)8

×(6

4

)3

; (ii) (1.8)6 × (4.2)−3; (iii)(0.0006)9

(0.015)−4.

2. What is the least positive integer n such that(30)3

(35)2n is also an integer?

3. Can it happen that for some integer m 6= 0,

(4

25

)m

=

(2

5

)m2

?

4. Find all positive integers m,n such that(3m)n

= 3m × 3n.

Can you see that the fifth law of exponents can be deduced from the

fourth law? You may writea

b= a× b−1. Thus you obtain

(ab

)m= (a× b−1)m = am × (b−1)m = am × b−m =

am

bm.

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Exponents 189

You may notice that you needed (b−1)m = b−m which is a consequence of

the third law. Thus the fifth law is a consequence of the fourth and third

laws.

Important points

All these laws are also valid for algebraic variables. So far you have

seen the validity of these laws for numbers. If x is a variable, we have

xm · xn = xm+n, for all integers m,n. (1)

Here again, we define x0 = 1. Unlike for numbers, where we have defined

a−n =1

anfor a 6= 0, we cannot so easily introduce

1

x. Hence we define x−n,

for a natural number n, as that expression for which xn · x−n = 1 holds.

We write x−n in the form1

xn. Thus we have the result: xn · 1

xn= 1, for all

natural numbers n. Note that xn = 1x−n

also holds for all natural numbers

n. With this understanding, the above law (1) holds. Similarly, you may

write laws:

(xm)n = xmn, for all integers m,n; (2)

(x · y)m = xm · ym, for all integers m and variables x, y. (3)

Additional problems on “Exponents”

1. Mark the correct option:

(a) The value of(3m)n

, for every pair of integers (m,n), is

A. 3m+n B. 3mn C. 3mn

D. 3m + 3n

(b) If x, y, 2x+y

2are nonzero real numbers, then

(2x+

y

2

)−1{(2x)−1 +

(y2

)−1}

equals

A. 1 B. x · y−1 C. x−1 · y D. x−1 · y−1

(c) If 2x − 2x−2 = 192, the value of x is

A. 5 B. 6 C. 7 D. 8

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190 Unit 4

(d) The number

(6

(66))1/6

is equal to

A. 66 B. 666−1 C. 6(6

5) D. 6(56)

(e) The number of pairs positive integers (m,n) such that mn = 25 is

A. 0 B. 1 C. 2 D. more than 2

2. Use the laws of exponents and simplify:

(i)(12)6

162; (ii)

3−4 × 10−5 × (625)

5−3 × 6−4; (iii)

232

(23)2.

3. The diameter of the Sun is 1.4 × 109 meters and that of the Earth is

about 1.2768× 107 meters. Find the approximate ratio of the diameter

of the Sun to that of the Earth.

4. What is the value of(103)2 × 10−4

102?

5. Simplify:

(b−3 · b7 ·

(b−1)2

(−b)2 ·(b2)3

)−2.

6. Find the value of each of the following expressions:

(a)(32)2 −

((−2)3

)2 −(− (52)

)2;

(b)((0.6)2

)0 −((4.5)0

)−2;

(c) (4−1)4 × 25 ×(

1

16

)3

×(8−2

)5 × (642)3;

(d) (−0.75)3 + (0.3)−3 −(−32

)−3;

(e)

(8× (42)4 × 33 × 272

)+(9× 63 × 47 × (32)3

)

(24× (62)4 × (24)2

)+(144× (23)4 × (92)2 × 42

) ;

(f)

(219 × 273

)+(15× 49 × 94

)(69 × 210

)+ 1210

.

7. How many digits are there in the number 23 × 54 × 205?

8. If a7 = 3, find the value of

(a−2

)−3 ×(a3)4 ×

(a−17

)−1

a7.

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Exponents 191

9. If 2m×a2 = 28, where a,m are positive integers, find all possible values

of a +m.

10. Suppose 3k × b2 = 64 for some positive integers k, b. Find all possible

values of k + b.

11. Find the value of(625)6.25 × (25)2.60

(625)7.25 × (5)1.20.

12. A person had some rupees which is a power of 5. He gave a part of it

to his friend which is also a power of 5. He was left with 500. How

much did money he have?

Glossary

Light year: this is the distance travelled by a light ray in one year; it is

equal to 4.3× 365× 24× 60× 60× 299792 km.

Exponential notation: writing a × a × · · · × a, where the product is taken

n times in the form an

Base: in the abbreviation an, a is the base.

Exponent: in the abbreviation an, n is the exponent.

Laws of exponents: am × an = am+n; (am)n = amn; (ab)m = am × bm hold

for non-zero numbers and integers m,n. (These are also true for algebraic

variables a, b and real numbers m,n.)

Points to remember• 0n = 0 for n > 0; 0n is not defined for n ≤ 0.

• am = an if and only if m = n for any a 6= 0, 1 or −1.• For any a 6= 0, and natural number n, a−n =

1

an.

• For any a 6= 0, and integers m,n,

am × an = am+n.

• For any a 6= 0, and integers m,n,(am)n

= amn.

• For any a 6= 0, b 6= 0 and integer m,

(ab)m = am × bm.

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CHAPTER 2 UNIT 5

INTRODUCTION TO GRAPHS

After learning this unit, you learn to:

• identify a point in the plane using a rectangular coordinate system;

• set up a coordinate system and draw graphs of linear curves in such

a system;

• get to know the way the coordinates of a point are related in two

different rectangular coordinate systems with axes parallel to each

other;

• construct the equation of a straight line by looking at the graph of the

straight line on a graph paper.

2.5.1 Introduction

In May, when the summer is at its peak, you may find some days are

hot and some days are not so hot. You may measure the maximum tem-

perature on each day (you may use either Fahrenheit or Celsius to mea-

sure temperature). You get a data for the whole month. Perhaps you will

tabulate it in a straight forward way:

Day 1 2 3 4 5 6 7 8 · · · · · · 30 31

Maximum

Temperature 33◦ 32◦ 34◦ 35◦ 37◦ 32◦ 35◦ 38◦ · · · · · · 36◦ 36◦.

Similarly, the rainfall for each month of a year can be measured in

inches or centimeters and we obtain a data. Recording these data is very

useful for future plans. For example, you can see the rain pattern over

several years and plan the agricultural crops accordingly. You may come

across several such important data in daily life which need to be recorded

for a better future. One easy way is to tabulate as earlier. However, this

is a very cumbersome procedure. People look for efficient way of recording

such data, which also help them to analyse it in a better way. One of the

effective method is the use of graphs.

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Graphs 193

What is a graph? We can define a graph as the visual representa-

tion of numerical data collected during an experiment. By looking at the

graph, one can easily understand the data. Moreover, the graphs also help

us to analyse the data quickly.

There are different types of graphs: Bar graphs; Pie charts; Histograms;

Cartesian graphs. Each of them is useful for describing a particular type

of data. But the principal behind all these graphical representation is the

same; an easy visual comprehension of the collected data.

2.5.2 Bar graphsSuppose a firm makes a profit of 36 lacks during 2006, 48 lakhs

during 2007, 70 lakhs during 2008, 85 lakhs during 2009 and 75

lakhs during 2010. There is an easy visualisation of the firm’s perfor-

mance during this 5-year period, if we represent the profit by a rectangular

vertical bar as shown below.

20072006 2008 2009 2010

70

85

48

36

75

Note that each bar in the above picture represents the profit of the firm

during that year which is given below the bar. You see that the firm has a

good growth, but a set-back in the last year.

Engli

sh

Kann

ada

Socia

l Stu

dies

Mat

hem

atics

Scien

ce

65

8780

9690

It is not necessary that the verti-

cal bars in a bar graph be sepa-

rated. The adjacent graph shows

the marks obtained by Shilpa in

her 8-th standard examination.

It is easy to compare her perfor-

mance in individual subjects.

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194 Unit 5

2.5.3 Pie charts

Rahim is an engineer in an industry and his take home salary is

36,000 per month. His expenditure for a month runs as follows:

How can this data be repre-

sented as a graph? Rahim

would like to see what frac-

tion of his income is spent

on various heads. The graph

must immediately reflect the

House rent —– 6,000

Food —– 6,000

Children’s education —– 3,600

Miscellaneous —– 6,000

Parental care —– 6,000

Savings —– 8,400.

fraction of his expenditure against his total salary income per month.

We may think that 36,000 is represented by a circle. Since the expen-

diture on each item is a part of Rahim’s total earning, such expenditure

must be represented by a part of the circle. The representation must im-

mediately reflect the ratio of the part to the whole For example, Rahim

spends 6,000 on rent and let us find the ratio of his rental expenditure

to that of his income. It is6000

36000=

1

6. Thus his rental expenses is one-

sixth of his earnings and we have to represent rent by one-sixth part of

the whole circle. We can use the fact that the total angle at a point is 360◦.Consider the centre of the circle. Construct a sector of the circle which

measures 60◦, since one-sixth of of 360◦ is 60◦. Thus, this sector repre-

sents exactly1

6part of the whole circle. This we can do for all heads and

find out the angle of the sector needed to represent each of the individual

expenditures. Thus

House rent −→ 6000

36000=

1

6and

1

6× 360◦ = 60◦

Food −→ 6000

36000=

1

6and

1

6× 360◦ = 60◦

Children’s education −→ 3600

36000=

1

10and

1

10× 360◦ = 36◦

Miscellaneous −→ 6000

36000=

1

6and

1

6× 360◦ = 60◦

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Graphs 195

Parental care −→ 6000

36000=

1

6and

1

6× 360◦ = 60◦

Savings −→ 8400

36000=

7

30and

7

30× 360◦ = 84◦

The graph we have generated is as follows:

Educatio

n

Miscellaneous

ParentsRent

Food

Savings

1/6

1/6

1/10

1/6

1/67/30

Observe!1

6+1

6+

1

10+1

6+1

6+

7

30= 1.

2.5.4 Coordinate system

Another extremely useful and important method of representing data

is the use of Cartesian graphs. These are also called coordinate graphs

as the basic principle depends on the use of a coordinate system. A

coordinate system is a useful device for representing points on the plane.

If you take a line, then the points on the line correspond to real numbers.

In a coordinate system in the plane, two lines, one perpendicular to the

other are used. This will help us to identify the points on the plane in a

definite manner.

To get an idea how a coordinate system works, suppose there is a rect-

angular grazing yard and you are standing at one corner of the yard. A

cow is grazing at some point in the yard. You have to identify its location

by some numerical data. What do you do?

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196 Unit 5

An easy way is: walk in the direction of the cow from where you stand

and measure the distance you move. You may measure it using some

convenient measurement. But when you describe the situation to some

one else, you have to tell the direction in which you had moved and the

distance you had moved. Describing the direction is not so easy. You have

to use a different mechanism.

A B

CD

F(you)

GE (cow)

300

100

The best way is to use the bound-

aries of the grazing yard. Suppose

you are at the point A and the cow

is grazing at E.(Look at the adja-

cent figure.) You may walk along

the boundary AB to a point F

where EF is parallel to AD. Thenmake a 90◦ turn anti clock-wise and move in the direction of the cow;

remember EF is parallel to AD. If AF = 300 m and FE = 100 m, you may

say the cow is located at 300 m in the direction of AB and 100 m in the

direction of AD.

Briefly we say E has coordinates (300, 100) with respect to the point A and

the system (AB,AD). Note that 300 is the distance in the direction of AB

and 100 is the distance in the direction of AD.

You may also walk along AD first to a point G where AG = 100 m, and

then walk along GE(which is parallel to AB) to the point E where GE = 300

m. Thus if you use the point A and the system (AD,AB), then E has

coordinates (100, 300).

Observe!

The point E is described by the ordered pair (300, 100) with re-

spect to the point A and the system (AB,AD). The same point E

is described by the ordered pair (100, 300) now with respect to the

point A and the system (AD,AC). Thus the same point E may

have different coordinates with respect to different systems.

Caution!!

With respect to the point A and the system (AB,AD), the points

(300, 100) and (100, 300) are totally different.

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Graphs 197

B

CD

F(you)

P Q

100 200

50

RA

E(cow)

50

50

Instead being at the point A, sup-

pose you are at the point P (See

the figure). How do you describe

the location of the cow with re-

spect to new position? Again, you

may walk along PQ to the point Q,

where PQ is parallel to AB and QE

is parallel to AD. Then you may move along QE to reach E. Now you

can describe E with respect to the point P and the system (AB,AD). Note

that PQ is parallel to AB and QE is parallel to AD; we are using the same

system (AB,AD) but a different starting point P . If you know PQ and

QE, you can easily describe E. Draw PR parallel to AD with R on AB.

Suppose AR = 100 m. Then PQ = RF = 200 m. Similarly, if RP = 50 m,

then QE = FE−FQ = FE−RP = 100− 50 = 50 m. Thus with respect to the

point P and the system (AB,AD), the point E has coordinates (200, 50)

Observe!

The point P can be described with respect to A and the

system (AB,AD) by (100, 50), since AR = 100 and RP = 50

units. Using the point P and the system (AB,AD), the

point E is described by (200, 50). But E is described by

(300, 100) using the point A and the system (AB,AD). Can

you see that 300 = 100 + 200 and 100 = 50 + 50? What do you

conclude?

Now you imagine a situation of a huge grazing yard which is endless

and having no boundaries. Again a cow is grazing at some point in the

yard and you are standing at some other point. How do you describe

the position of the cow? Earlier, you had a rectangular yard and you

had a rectangular frame given by the boundaries. Now you are on your

own and you do not have any such frame. You may consider two lines

perpendicular to each other and passing through the point where you are

standing. Let O be the point where you are standing and let X ′OX, Y ′OY

be two line perpendicular to each other and passing through O. Let the

cow be grazing at E(See the figure).

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198 Unit 5

F

K

X’ X

Y’

G E

P

Q O

Y

L

Again you may move along−−→OX to F , where EF is par-

allel to Y ′OY . Then you

may move from F to E

along FE. With respect to

the point O and the system

(X ′OX, Y ′OY ), you can de-

scribe E using the lengths

OF and FE = OG(here G is

on the line Y ′OY such that

GE is parallel to X ′OX).

But there is a difficulty. If the point E is below the line X ′OX, as shown

in the figure, then you have to move along the ray−−→OX to the point F , and

then you have to move from F to E in the direction of the ray−−→OY ′. If some

other cow is located at the point K, then you have to move from O to L in

the direction of the ray−−→OX ′ and then you have to move from L to K in the

direction of the ray−−→OY ′. Similarly, you can reach the point P from O, by

moving first along−−→OX ′ and then from Q to P in the direction of

−−→OY .

Now you may observe that there are four different directions to move

starting from O;−−→OX;

−−→OX ′;

−−→OY ; and

−−→OY ′. You may think the line X ′OX

as the real number line and O corresponding to the number zero. Thus

all points to the left of O (along−−→OX ′) correspond to negative numbers and

all points to the right of O (along−−→OX) correspond to the positive num-

bers. Similarly, we can view the line Y ′OY as another copy of number line

and O corresponding to zero; all points above O corresponding to positive

numbers and all points below O corresponding to negative numbers. Note

that O has the position 0 with respect to both the lines. We say O has

coordinates (0, 0).Y

Y’

X’ XOL

K

P

−5

−8

−7

−6

−3

−1

−2

0

1

2

3

4

5

6

7

1 2 3 4 5−1−2

Q−3−4

−4

−5−6−7−8−9−108 9 10 76

F

E

For reaching E, you have to move 5

units to F in the direction−−→OX and 4

units in the direction−−→OY ′, from F to E.

Since−−→OY ′ corresponds to the negative

part of the number line, we describe E

by (5,−4). We say E has coordinates

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Graphs 199

(5,−4) with respect to the rectangular coordinate system X ′OX ←→Y ′OY . (Here X ′OX ←→ Y ′OY is a symbol used to represent the coordi-

nate system.)

Here the line X ′OX is called the x-axis and Y ′OY is called the y-axis.

The point O is called the origin of the coordinate system. The ray−−→OX

is called as the positive x-axis and−−→OX ′ is called as the negative x-

axis. Similarly,−−→OY is the positive y-axis and

−−→OY ′ is the negative y-axis.

The point E has coordinates (5,−4) in this system. We say 5 is the x-

coordinate of E and −4 is the y-coordinate of E. Note that the first num-

ber is always identifier along the horizontal axis and the second number

is always the identifier along the vertical axis. Similarly, the point K has

−6 as its x-coordinate and −3 as its y-coordinate. We say K has coordi-

nates (−6,−4) in this system. Similarly, P has coordinates (−4, 4).

X ’

Y ’

OX

YThe lines X ′OX and Y ′OY di-

vide the plane in to 4 regions.

The region bounded by the rays−−→OX and

−−→OY is called the first

quadrant. Similarly, you have

the second quadrant, the third

quadrant and the fourth quad-

rant. If you take any point in

the first quadrant, you have to move in the directions of−−→OX and

−−→OY to

reach that point. Thus you use the positive x and y-axes.

You will now recognise that all points in the first quadrant has non-

negative x- and y- coordinates; each point on the ray−−→OX has its y-coordinate

equal to 0; and each point on the ray−−→OY has its x-coordinate 0. In the

second quadrant, you have to use the rays−−→OX ′ and

−−→OY ; hence each point

in the second quadrant has non-positive x-coordinate and non-negative

y-coordinate. Similarly, the points in the third quadrant are described by

the coordinates (x, y), where x ≤ 0 and y ≤ 0. Each point in the fourth

quadrant is represented by the ordered pair (x, y) where x ≥ 0 and y ≤ 0.

The x-coordinate of a point is also called its the abscissa and the y-

coordinate is called the ordinate of that point.

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200 Unit 5

Observe that the lines X ′OX and Y ′OY are completely left to

your choice. The only condition is that they must be perpendic-

ular to each other for obtaining a rectangular coordinate system.

René Descartes was born on

March 31st, 1596 in the town

of La Haye in the south of

France. In 1606, at the age

of 8, René Descartes started

studying literature, grammar,

science, and mathematics. In

1616, he received his bac-

calaureate and licentiate de-

grees in Law. Aside from his

Law degrees, Descartes also

spent time studying philoso-

phy, theology, and medicine.

After a short stay in the military, Descartes went on to lead a quiet life,

continuing his intellectual pursuits, writing philosophical essays, and ex-

ploring the world of science and mathematics. In 1637, he published “Ge-

ometry”, in which his combination of algebra and geometry gave birth to

analytical geometry, better known as Cartesian geometry.

We use only rectangular coordinate system throughout this chap-

ter. However, it is possible to have a coordinate system in which

the two axes need not be perpendicular to each other. Such a system

called oblique coordinate system is also useful in solving many practical

problems.

The idea of introducing coordinates to identify the points on the plane

is due to René Descartes, a french mathematician and philosopher. This

has revolutionised the thinking as it helps to convert geometrical problems

to equivalent algebraic problems. To commemorate his contribution, the

system we study is also known as Cartesian coordinate system. This

has developed in to a new branch of mathematics known as Analytic ge-

ometry.

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Graphs 201

Activity 1:

Locating a given point on a graph paper

Suppose you are given a graph paper. You also fix up your coordinate

system in the graph paper, say X ′OX and Y ′OY which are perpendicular

to each other. How do you locate the point P with coordinates (−4, 6)?

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 87 9 10

1

2

3

4

5

6

7

8

−1

−3

−4

−5

−6

−2

−7

−8

Q

P(−4,6)

Y’

Y

X’ X

Starting from the origin O, move 4 units in the direction of the ray−−→OX ′.

Since−−→OX ′ represents negative x-axis, when −4 is given as x-coordinate,

you have to move 4 units in the direction of−−→OX ′. You end up with the point

Q on−−→OX ′. Now move 6 units up in the direction parallel to

−−→OY . Remember−−→

OY represents positive y-axis. Hence the number 6(with positive sign)tells

you that you have to move 6 units upwards parallel to−−→OY . You then reach

a point P whose coordinates are (−4, 6).Activity 2:

Take a sheet of graph paper. Now set up two rectangular coordinate axes:

X ′OX ←→ Y ′OY and X ′1O1X1 ←→ Y ′1O1Y1, where X ′OX ‖ X ′1O1X1. Locate

a point P with coordinates (10, 5) with respect to the system X ′1O1X1 ←→Y ′1O1Y1. What are the coordinates of P with respect to the systemX ′OX ←→Y ′OY ? Find the coordinates of O1 with respect to the system X ′OX−Y ′OY .

Now can you find the relation among the coordinates of P in the system

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202 Unit 5

X ′OX ←→ Y ′OY , the coordinates of O1 in the system X ′OX ←→ Y ′OY and

the coordinates of P in the system X ′1O1X1 ←→ Y ′1O1Y1?

−6−7−8−13 −12 −11 −10 −9 0−2−3−4−5 1 2 3 4 65 7 10 11 12 13 14

3

4

5

1

2

6

7

−1

−3

−4

−2

−6

−5

−7

−8

−9

−10

−11

−12

−13

−14

−15

−16

1 20

9

−1−2−3−4

−1

−2

−3

−4

−5

1

3

2

7

6

5

4

8

9

8

10

11

12

13

14

16

15

17

−1

Y1

X 1’ X 1

Y 1’

XX ’ O

Y

Y ’

O1

P

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

In the above figure, you see that O1 has coordinates (11, 11) in the sys-

tem X ′OX ←→ Y ′OY . The point P has the coordinates (21, 16) in the sys-

tem X ′OX ←→ Y ′OY and (10, 5) in the system X ′1O1X1 ←→ Y ′1O1Y1. Observe

that21 = 11 + 10, 16 = 11 + 5.

We write this in the form (21, 16) = (11, 11)+(10, 5). Repeat this with various

positions of P and different systems X ′OX ←→ Y ′OY , X ′1O1X1 ←→ Y ′1O1Y1.

What is the conclusion you draw from these activities?

Observe!

Suppose X ′OX ←→ Y ′OY and X ′1O1X1 ←→ Y ′1O1Y1 are two system

of coordinates such that X ′OX ‖ X ′1O1X1. Let a point P have

coordinates (x, y) in the system X ′OX ←→ Y ′OY and (x′, y′) in the

second system X ′1O1X1 ←→ Y ′1O1Y1. If O1 has coordinates (a, b)

with respect to the system X ′OX − Y OY , then x = a + x′ andy = b+ y′.

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Graphs 203

Activity 3:

Take a sheet of graph paper and set up your coordinate system X ′OX ←→Y ′OY . Locate a point P with coordinates (−5, 8). Trace the point with

coordinates (8,−5). Do you get the same point P?

Example 1.

Consider the number 3. Let us tabulate the multiples of 3, some positive

and some negative.

x −3 −2 −1 0 1 2 3

3× (−3) 3× (−2) 3× (−1) 3× 0 3× 1 3× 2 3× 3

y −9 −6 −3 0 3 6 9

(x, y) (−3,−9) (−2,−6) (−1,−3) (0, 0) (1, 3) (2, 6) (3, 9)

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 87 9 10

1

2

3

4

5

6

7

8

−1

−3

−4

−6

−2

−7

12

11

10

9

−9

−11

−10−11−12−13 131211

−8

−5

−10

−12

Y ’

Y

XX ’

(1,3)

(2,6)

(3,9)

(4,12)

(−1,−3)

(−2,−6)

(−3,−9)

(−4,−12)

(0,0)

O

Take a sheet of graph paper, and set up your own coordinate system,

X ′OX ←→ Y ′OY . Locate the points (x, y) tabulated in the last row. Can

you see that all these lie on a straight-line?

Example 2. (Area versus perimeter of a square)

Consider a square of side length 1 unit. What is its area? You know that

the area of a square is l2, where l is the length of the square. And its

perimeter is l+ l+ l+ l = 4l. Hence the area of a square of side length 1 unit

is 1 square unit and its perimeter is 4 units. What are these for a square

of length 2 units? You see that they are respectively 4 square units and 8

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204 Unit 5

units. Let us tabulate the perimeter and area of squares of different side

lengths.

l 1 2 3 4 5 6

x = 4l 4 8 12 16 20 24

y = l2 1 4 9 16 25 36

(x, y) (4,1) (8,4) (12,9) (16,16) (20,25) (24,36)

(4,1)

(8,4)

(12,9)

(16,16)

(20,25)

(24,36)

X’

Y’

Y

XO

1

5

15

30

40

35

20

25

10

0 5 10 15 20 25 30

Again set up your own coordinate system on a graph paper and plot

these points (x, y) on the graph paper. Do you think that these points lie

on a straight-line?

Example 3. (Simple interest versus number of years)

Suppose Shiva deposits 1,000 in a bank for 5 years for simple interest

at the rate of 8% per annum. You can easily calculate what he earns as

interest over the years and tabulate it.

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Graphs 205

Year=x 1 2 3 4 5

Interest=y 80 160 240 320 400

(x, y) (1, 80) (2, 160) (3, 240) (4, 320) (5, 400)

X’

Y’

X

Y

O0−1−2−4 −3−5 3 421 5 6 7 8 9 10 11 12 13

160

400

80

120

200

240

280

320

360

(1,80)

(2,160)

(3,240)

(4,320)

(5,400)

40

Again you are required

to plot these points on

a graph paper. Here

you face a practical prob-

lem. The interest Shiva

obtains are big numbers

and it is difficult to lo-

cate numbers like 240,

320, 400 on a graph pa-

per, unless the graph pa-

per you take is very huge.However, this difficulty may be avoided using different scaling. For exam-

ple, you may use 1 unit=40 along y-axis. Here again you notice that all

the points lie on a straight-line.

Activity 4:

Suppose a car moves with a constant speed of 40 km per hour. Make a

table of the distance covered by the car at the end of 1-st, 2-nd, 3-rd, 4-th,

5-th, 6-th ,7-th and 8-th hour. Trace the points on a graph paper where

x-axis represents the time and y-axis represents the distance traveled by

the car.

Exercise 2.5.4

1. Fix up your own coordinate system on a graph paper and locate the

following points on the sheet:

(i) P (−3, 5); (ii) Q(0,−8); (iii) R(4, 0); (iv) S(−4,−9).2. Suppose you are given a coordinate system. Determine the quadrant

in which the following points lie:

(i) A(4, 5); (ii)B(−4,−5); (iii)C(4,−5).3. Suppose P is a point with coordinates (−8, 3) with respect to a coor-

dinate system X ′OX ←→ Y ′OY . Let X ′1O1X1 ←→ Y ′1O1Y1 be another

system with X ′OX ‖ X ′1O1X1 and suppose O1 has coordinates (9, 5)

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206 Unit 5

with respect to X ′OX ←→ Y ′OY . What are the coordinates of P in the

system X ′1O1X1 ←→ Y ′1O1Y1?

4. Suppose P has coordinates (10, 2) in a coordinate system X ′OX−Y ′OY

and (−3,−6) in another coordinate system X ′1O1X1 ←→ Y ′1O1Y1 with

X ′OX ‖ X ′1O1X1. Determine the coordinates of O with respect to the

system X ′1O1X1 ←→ Y ′1O1Y1.

2.5.5 Linear graphs

Take a re-look at Example 1, where you have plotted integral multiples

of 3. Instead of taking integer multiples of 3, suppose you take real multi-

ples of 3: for each real number x, consider y = 3x(obtained by multiplying

x by 3).

x 0 1 2 3 −1 −2 1

2

−12

1

3

y 0 3 6 9 −3 −6 3

2

−32

1

As you assign different real values for x, you get back different real

numbers y = 3x. Thus for each real number x, you get a point whose

coordinates are (x, 3x). Using a coordinate system, we can trace these

points on a graph sheet.

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 87 9 10

1

2

3

4

5

6

7

8

−1

−3

−4

−5

−6

−2

−7

−8

12

11

10

9

−9

−10

−11

−12

−10−11−12−13 131211

Y’

Y

XX’

O

(3,9)

(2,6)

(1,3)

(−1,−3)

(−2,−6)

(1/3, 1)

(1/2,3/2)

(−1/2, −3/2)

If you take more and more values of x, you get more and more points

(x, 3x). As you go on tracing them on graph paper, you see that these points

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Graphs 207

get more and more clustered. Of course, you have to get finer and finer

graph paper to trace these points. However, you see that all these points

lie on a straight-line. Now it is clear that as x exhaust all real numbers,

the points (x, y), where y = 3x, trace a straight-line in the plane. We say

the equation y = 3x represents a straight-line on the coordinate plane.

Activity 5:

Use the scaling 1 unit =1

3on the x-axis. Calculate more values of y = 3x

as x varies over real numbers and trace the points (x, y) on a graph sheet

using the suggested scaling.

The relation y = 3x is of the first degree. Such a relation always repre-

sents a straight-line. The general form of the equation for a straight-line

is y = ax + b.(Here a and b are real numbers.) If a = 0, then y = b is a

straight-line parallel to y-axis.

Example 4. Suppose a person deposits 10,000 in a bank for simple

interest at the rate of 7% per annum. Obtain a relation for accrued interest

in terms of years. Draw the graph of this relation. Use this to read the

interest at the end of 10 years.

Solution: Let y be the interest obtained in x years. The interest for x years

is7

100× 10000 × x = 700x. Thus you get y = 700x as the required relation.

Now use the scaling: 1 unit= 700 along y-axis. Note that

for x = 1, you get y = 700 : (x, y) = (1, 700),

for x = 2, you get y = 1400 : (x, y) = (2, 1400).

6543210 87 9 11 12 13 14 15

700

1400

2100

3500

2800

4200

4900

5600

6300

7000

7700

OX’X

Y’

Y

P(1,700)

Q(2,1400)

R

T

S

10

Take a sheet of graph pa-

per and set up coordinate

axes X ′OX ←→ Y ′OY . Lo-

cate points P (1, 700) and

Q(2, 1400). Remember you are

using the scale 1 unit =700

along y-axis. Now the equation is a straight-line and hence P , Q determine

this straight-line.

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208 Unit 5

Observe!

To draw a straight-line, it is enough to know only two points on

that line; given any two distinct points in the plane, you can

use the scale to draw the unique line passing through those two

points.

Use a straight-edge to join PQ and extend it to possible extent. This

straight-line represents the curve of the interest versus years for this

problem. Now look at the point 10 on x-axis, say T . Draw a perpen-

dicular to x-axis at T to meet the straight-line y = 700x at R. From R draw

a perpendicular to y-axis. Record the point where it cuts y-axis. That gives

the interest at the end of 10 years, which you read as 7,000.

Example 5. Draw the graph of y = 3x+ 5.

Solution: Give different values for x and get values for y. Tabulate them.

x 0 1 2 3 −1 −2y 5 8 11 14 2 −1

0−2 −1 1 2 3 4 5

(−2, −1)

(−1,2)

(0,5)

(1,8)

(3,14)

(2,11)

15

10

5

20

Let us trace these on a graph paper. Fix x and y axes and locate the

points

(x, y) = (0, 5), (1, 8), (2, 11), (3, 14), (−1, 2), (−2,−1).

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Graphs 209

Do you see that all these points lie on a straight line? Use a straight-

edge and join all these points by a straight line segment.

Activity 6: Draw the graphs of y = x+4, y = 2x− 3, y = 3, x = 2y+1, x = 2.

All these are straight lines. Do you see that y = ax + b is always a straight

line?

Example 6. Determine the equation of the line in the following given

graph.

X’

Y’

X

Y

O3 421 5 6 7 8 9 10 11 12 130−1−2−3−4−5

1

2

4

3

5

6

7

8

9

10

−1

−2

−3

P

Q

Solution: Recall that

the general equation of a

straight-line is y = ax + b,

where a, b are real constants.

The given graph shows that

a 6= 0. (If a = 0, then y = b

represents a straight-line

parallel to y-axis.) If x = 0,

then y = b. Thus (0, b) is a

point on the straight-line.

Similarly, taking y = 0, you get ax + b = 0 or x = − b

a. Thus

(− b

a, 0

)is also

a point on the line. Looking at the graph, we see that it cuts y-axis at (0, 4)

and the x-axis at (3, 0). We conclude that

(0, b) = (0, 4),

(− b

a, 0

)= (3, 0).

Thus b = 4 and − b

a= 3, which gives a = −4

3. Hence the equation of the

given graph is y = −43x+ 4. This may also be written as 3y + 4x = 12(why?)

Example 7. Determine the equation of the line in the following given

graph.

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210 Unit 5

X’

Y’

X

Y

O3 421 5 6 7 8 9 10 11 12 130−1−2−3−4−5

2

4

3

5

6

7

8

9

10

−1

−2

−3

1

P

Q

Solution: We may take

the equation in the form

y = ax + b. Observe that

the straight-line passes

through (0, 0). This

means that y = 0 when

x = 0. But the substitu-

tion x = 0 in the equa-

tion gives 0 = y = b. Thus

b = 0 and so that y = ax.

We also observe that the line passes through (1,−3); that is y = −3, when-

ever x = 1. This gives −3 = a × 1 or a = −3. The equation of the line is

therefore y = −3x.

Exercise 2.5.5

1. Draw the graphs of the following straight-lines:

(i) y=3-x; (ii) y=x-3; (iii) y=3x-2; (iv) y=5-3x ;

(v) 4y = −x+ 3; (vi) 3y = 4x+ 1; (vii) x = 4; (viii) 3y = 1.

2. Draw the graph ofy

x=

y + 1

x+ 2.

3. Determine the equation of the line in each of the following graphs:

X’

Y’

X

Y

O3 421 5 6 7 8 9 10 11 12 130−1−2−3−4−5

2

4

3

5

6

7

8

9

10

−1

−2

−3

1

X’

Y’

X

Y

O3 421 5 6 7 8 9 10 11 12 130−1−2−3−4−5

2

4

3

5

6

7

8

9

10

−1

−2

−3

1

4. A boat is moving in a river, down stream, whose stream has speed 8

km per hour. The speed of the motor of the boat is 22 km per hour.

Draw the graph of the distance covered by the boat versus hour.

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Graphs 211

5. Find the point of intersection of the straight-lines 3y + 4x = 7 and

4y + 3x = 7, by drawing their graphs and looking for the point where

they meet.

Additional problems on “Introduction to graphs”


(a) The point (4, 0) lie on the line ——————–

A. y − x = 0 B. y = 0 C. x = 0 D. y + x = 0

(b) The point (−5, 4) lie in ——————–

A. the first quadrant B. the second quadrant C. the third

quadrant D. the fourth quadrant

(c) If a straight-line pass through (0, 0) and (1, 5), then its equation is

——————–

A. y = x B. y = 5x C. 5y = x D. y = x+ 5

(d) If a point P has coordinates (3, 4) in a coordinate systemX ′OX ←→Y ′OY , and if O has coordinates (4, 3) in another systemX ′1O1X1 ←→Y ′1O1Y1 with X ′OX ‖ X ′1O1X1, then the coordinates of P in the new

system X ′1O1X1 ←→ Y ′1O1Y1 is ——————–

A. (3, 4) B. (1,−1) C. (7, 7) D. (−1, 1)(e) The coordinates of a point P in a systemX ′OX ←→ Y ′OY are (5, 8).

The coordinates of the same point in the system Y ′OY ←→ XOX ′

are ——————–

A. (−8, 5) B. (8, 5) C. (8,−5) D. (−8,−5)(f) The signs of the coordinates of a point in the third quadrant are

——————–

A. (+,−) B. (−,+) C. (+,+) D. (−,−)(g) If a person moves either 1 unit in the direction of positive x-axis or

1 unit in the direction of positive y-axis per step, then the number

of steps he requires to reach (10, 12) starting from the origin (0, 0)

is ——————–

A. 10 B. 12 C. 22 D. 120

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212 Unit 5

(h) The y-coordinate of the point of intersection of the line y = 3x + 4

with x = 3 is ——————–

A. 4 B. 7 C. 10 D. 13

(i) The equation of the line which passes through (0, 0) and (1, 1) is

——————–

A. y = x B. y = −x C. y = 1 D. x = 1

2. Find the quadrant in which the following points lie:

(i) (5, 10); (ii) (−8, 9); (iii) (−800,−3000); (iv) (8,−100).

3. Match the following:

(A) On the x-axis (i) x coordinate is negative

(B) In the second quadrant (ii) cuts the y-axis at (0, 4)

(C) The line y = 3x+ 4 (iii) coordinates of a point are of

the form (a, 0).

4. Fill in the blanks:

(a) The y-coordinate of a point on the x-axis is —————–.

(b) The x- coordinate is called as —————–.

(c) The x-axis and y-axis intersect at —————–.

(d) If a point (x, y) 6= (0, 0) is in the third quadrant, then x + y has

—————– sign.

(e) If a point (x, y) lies above horizontal axis, then y is always ———

——–.

(f) The point of intersection of x = y and x = −y is —————–.

(g) The line y = 4x+ 5 intersects y-axis at the point —————–.

5. True or false?

(a) The equation of the x-axis is x = 0.

(b) The line x = 4 is parallel to y-axis.

(c) The line y = 8 is perpendicular to x-axis.

(d) The lines x = y and x = −y are perpendicular to each other.

(e) The lines x = 9 and y = 9 are perpendicular to each other.

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Graphs 213

(f) The graph of y = x2 is a straight line.

(g) The line y = 3x+ 4 does not intersect x-axis.

(h) In a rectangular coordinate system, the coordinate axes are cho-

sen such that they form a pair of perpendicular lines.

6. Determine the equation of the line which passes through the points

(0,−8) and (7, 0).

7. Determine the equation of the line in each of the following graph:

(i)

O3 421 50−1−2−3−4−5

2

4

3

5

6

7

8

9

10

−1

−2

−3

1

Y’

X’ X

Y

(ii)

X’

Y’

X

Y

O3 421 5 6 7 8 9 10 11 12 130−1−2−3−4−5

2

4

3

5

6

7

8

9

10

−1

−2

−3

1

8. A point P has coordinates (7, 10) in a coordinate system X ′OX ←→Y ′OY . Suppose it has coordinates (10, 7) in another coordinate system

X ′1O1X1 ←→ Y ′1O1Y1 with X ′OX ‖ X ′1O1X1. Find the coordinates of O1

in the system X ′OX ←→ Y ′OY .

9. Sketch the region {(x, y) : x ≥ 0, y ≥ 0, x + 2y ≤ 4} in a coordinate

system set up by you.

10. Draw the graphs of lines 3y = 4x− 4 and 2x = 3y+4 and determine the

point at which these lines meet.

11. If a ⋆ b = ab+ a + b, draw the graph of y = 3 ⋆ x+ 1 ⋆ 2.

Glossary

Graph: a visual representation of numerical data.

Bar graph: graphs in which data is represented by rectangular bars.

Pie chart: a graph in which data is represented by sectors of a circle.

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214 Unit 5

Rectangular frame: two perpendicular lines in a plane which helps to lo-

cate points on the plane.

Rectangular coordinate system: a system which making use of rectan-

gular frames locates points as an ordered pair of real numbers.

x-axis: the horizontal line in a rectangular coordinate system.

y-axis: the vertical line in a rectangular coordinate system.

Quadrant: the four parts that a plane gets divided by a rectangular coor-

dinate system.

Abscissa: the x-coordinate of a point.

Ordinate: the y-coordinate of a point.

Cartesian coordinate system: a coordinate system in which every point

in the plane is determined by a pair of real numbers called the coordinates

of the point.

Analytic geometry: the geometry in which all the geometrical concepts

are studied using the coordinate system.

Points to remember

• You can set up your own coordinate system; there is no sacred coor-

dinate system.

• The coordinates of a point depends on the coordinate system you

choose.

• A rectangular coordinate system is only a convenient coordinate sys-

tem; it is not necessary that one should always use a rectangular

system.

• A graph is visual representation of the numerical data.


Exercise 2.1.1

1. Constants: 15,−37,√3, 7; Variables: 12+ z,

−x5

,√x,

2

3xy;

5xy

2, 7− x,

6x+ 4y, −7z, 8yz

4x, y + 4,

y

4,

2x

8yz.

2. Monomials: 7xyz,8x

y,

8

5x2y2; Binomials: 9− 4y, 4y2 − xz, 7x+ z2;

Trinomials: x− 2y + 3z, 4 + 5y − 6z.

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Answers 215

Exercise 2.1.3

1. {4x2, 3x2}, {xy, 8xy}, {−8x3, 6x3,−74x3},{1

3x

}, {7xyz}.

2. (i) 4x−14y+11; (ii) 4x2−3xy−9y2. 3. (i) 9a+6b; (ii) 2x2y+5xy2+2y3.

4. (i) 10x2y − 3xy2; (ii) 3a+ 7b.

Exercise 2.1.41.

First → 3x −6y 4x2 −8xy 9x2y −11x3y2Second ↓

3x 9x2 −18xy 12x3 −24x2y 27x3y −33x4y2−6y −18xy 36y2 −24x2y 48xy2 −54x2y2 66x3y3

4x2 12x3 −24x2y 16x4 −32x3y 36x4y −44x6y2−8xy −24x2y 48xy2 −32x3y 64x2y2 −72x3y2 88x4y3

9x2y 27x3y −54x2y2 36x4y −72x3y2 8x4y2 −99x5y3−11x3y2 −33x4y2 66x3y3 −44x5y2 88x4y3 −99x5y3 121x6y4

2. (i) 15x2 + 24x; (ii) 45p4q3 + 3pq4; (iii)6

5a3x− 6

5b3x; (iv) −x3 + 15x.

3. (i) 6x3y2 − 10x2y − 3x2y2 + 5xy; (ii) 12x3y3 − 18x3y4 + 4xy − 6xy2;

(iii) 6x4 + 4x3 + 9x2 + 6x; (iv) 10m3 + 12m2 − 3m.

Exercise 2.1.5

1. (i) a2+8a+15; (ii) 9t2+15t+4; (iii) a2−6a−16; (iv) a2−8a+12. 2. (i) 2915;

(ii) 10812; (iii) 1224; (iv) 9888. 3. x3+ (a+ b+ c)x2+ (ab+ bc+ ca)x+ abc.

4. (i) a2+12a+36; (ii) 9x2+12xy+4y2; (iii) 4p2+12pq+9q2; (iv) x4+10x2+25.

5. (i) 1156; (ii) 104.04; (iii) 2809; (iv) 1681. 6. x2 − 12x+ 36;

(ii) 9x2 − 30xy + 25y2; (iii) 25a2 − 40ab+ 16b2; (iv) p4 − 2p2q2 + q4.

7. (i) 2401; (ii) 96.04; (iii) 3481; (iv) 39204. 8. x2 − 36; (ii) 9x2 − 25;

(iii) 4a2 − 16b2; (iv)

(4x2

9− 1

). 9. (i) 2475; (ii) 851; (iii) 80.75;

(iv) 9996. 10. (i) x4 − 81; (ii) 16a4 − 81; (iii) p4 − 16; (iv)

(1

16m4 − 1

81

);

(v) 16x4 − y4; (vi) 16x4 − 81y4.

Additional problems on “Algebraic expressions”

1. (a) B. (b) B. (c) C. (d) A. (e) A. (f) C. (g) C. (h) B. (i) C.

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216 Answers

2. 2x+17y+p−q. 3. (i) 16x2+24x+9; (ii) x2+4xy+4y2; (iii) x2+(1/x2)+2;

(iv) x2 + (1/x2)− 2. 4. (i) 4t2 − 25; (ii) x2y2 − 25; (iii) 4x2 − 9y2.

5. (i) n4 − 1; (ii) n4 − (1/n4); (iii) x8 − 1; (iv) 16x4 − y4. 6. (i) 10609; (ii)

9216; (iii) 9951; (iv) 999936; (v) 21000. 7. 25. 8. 80. 9. ±8.10. 29 11. 7 and 18. 12. 34 and 1154. 13. (i) 2(x2 + y2);

(ii) x4− 2x2y2+ y4. 14. (i)

(3(x+ z)

2

)2

−((z − x)

2

)2

; (ii) (x+2y)2− (2x+ y)2;

(iii) (x+ 100)2 − 12; (iv) 5002 − 52.

15. (i) 13x− 4y; (ii) −28x − 11y. 16. 4x2 − 22x+ 6. 17. 2x2 − 2xy + 4z2.

18. (18x2 − 9xy − 20y2)/2. 19. 5x2 + 11xy + 5y2.

Exercise 2.2.2

1. (i) x(x+ y); (ii) 3x(x− 2); (iii) (0.8)a(2a− 1); (iv) 5(1− 2m− 4n).

2. (i) (a+x)(a+b); (ii) (3a+7b)(c−d); (iii) (x−2z)(3y−3t); (iv) (y−3)(y2+2−x).3. (i) (2a+ 5)(2a− 5); (ii)

(x+

3

4

)(x− 3

4

); (iii) (x2 + y2)(x+ y)(x− y);

(iv)1222

25; (v) 0.4; (vi) (7a− 3b)(3a− b).

Exercise 2.2.3

1. (i) p = 9, q = 2; (ii) p = −8, q = −4; (iii) p = 6, q = −4; (iv) p = 12, q = −1;(v) p = −6, q = 1; (vi) p = −11’ q = 4. 2. (i) (x+ 4)(x+ 2); (ii) x+ 3)(x+ 1);

(iii) (a+ 3)(a+ 2); (iv) (a− 3)(a− 2); (v) (a− 8)(a+ 5); (vi) (x− 9)(x+ 8).

3. (i) (x+7)(x+7); (ii) (2x+1)(2x+1); (iii) (a− 5)(a− 5); (iv) 2(x− 6)(x− 6);

(v) (p− 12)(p− 12); (vi) x(x− 6)(x− 6).

Additional problems on “Factorisation”

1. (a) C. (b) B. (c) D. (d) B. (e) D. (f) A.

2. (i) (x+ 3)2; (ii) (1− 4x)2; (iii) (2x+ 9y)(2x− 9y); (iv) (2a+ b)2;

(v) (a2 − d2)(b2 − c2).

3. (i) (x+ 3)(x+ 4); (ii) (x+ 4)(x− 3); (iii) (x− 6)(x+ 3); (iv) (x+ 7)(x− 3);

(v) (x− 16)(x+ 12); (vi) (x− 1)(x+ 1)(x− 2)(x+ 2);

(vii) (x− 2y)(x+ 2y)(x− 3y)(x+ 3y).

4. (i) (2x−3)(x+2); (ii) (x−4)(3x−5); (iii) (x−2)(6x+7); (iv) (2x+y)(2x+5y);

(v) (2x− 1)(2x+ 1)(x− 1)(x+ 1). 5. (i) (x− 4)(x+ 4)(x2 + y2)(x4 + y4);

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Answers 217

(ii) a4x4(a− x)(a+ x)(a2 + x2)(a4 + x4); (iii) (x2 + x+ 1)(x2 − x+ 1);

(iv) (x2 + x+ 3)(x2 − x+ 3). 6. (x2 + 2xy + 2y2)(x2 − 2xy + 2y2).

Exercise 2.3.2

1. (i) x = 8; (ii) y = 30; (iii) z = 7; (iv) x =6

11; (v) x = 3; (vi) s = 28;

(vii) x = 20; (viii) x = 2.4; (ix) x = 7; (x) x =−85; (xi) x = 60; (xii) x = 25;

(xiii) x = 2; (xiv) x = 36; (xv) x = −1. 2. (i) x = 12; (ii) t = −6; (iii) x = 7;

(iv) z =3

2; (v) x = 5; (vi) x = 1; (vii) x = −25; (viii) x = −13; (ix) x = −5;

(x) x = 10; (xi) x = 3; (xii) x = 8; (xiii) x = −54; (xiv) x = 9.

Exercise 2.3.3

1 6. 2. 71, 73, 75. 3. 25 cm and 15 cm. 4. 11. 5. 3,000 and

750. 6. 60 and 10. 7. x = 40. 8. 7. 9. 45 and 36. 10. 96 m and

48 m. 11. Ahmed’s age 12 and his father’s age 36 years. 12. Nishu’s

age 11 and Sanju’s age 17 years. 13. Deepu’s age 11 and Viji’s age 22

years. 14. Bindu’s age 19 and Mrs. Joseph’s age 46 years. 15. 8 years.

Additional problems on “Linear equations in one variable”

1 (a) B. (b) A. (c) B. (d) B. (e) B. (f) B. (g) A. (h) B.

2. (i) 10; (ii) −4757

. 3. 21 and 24. 4. 10 and 40. 5. 110, 112, 114.

6. A’ share 40,000 and B’s share 20,000. 7. 60. 8. 72. 9. garden

35,000 and house 49,000. 10. 800. 11. 15 years. 12. 33.

13. 12.50 14. 57. 15. 90 km and 95 km. 16. 2.25 km. 17. 1/4.

18. 62. 19. altitude 20 cm and base 12 cm. 20. 40◦, 50◦ and 90◦.

Exercise 2.4.1

1. (i) 123; (ii) 2−9; (iii) (0.013)2. 2. (i) 104 + 2 · 103 + 3 · 102 + 4 · 10 + 5;

(ii) 103 + 101 +1

102+

1

104; (iii)

1

10+

2

102+

3

105+

4

107. 3. 3 · 54 + 1 · 53 + 2 · 5.

4. 194. 5. 625.

Exercise 2.4.2

1. (i) 321; (ii) 26 · 314. 2. 10 zeros. 3. 53 × 54 × 55 × 56 > 57 × 58.

Exercise 2.4.3

1. (i) 103; (ii)25

6. 2. 34 × 23 > 25 × 32. 4 b = 1, 2 or 4.

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218 Answers

Exercise 2.4.4

1. (i) 1; (ii)1

9. 2. m = 2, n = 2. 3. m = 3.

Exercise 2.4.5

1. (i) 2592; (ii)2000

27; (iii)

(1.11)3 × (1.1)2

(1.01)2. 2. No. 3. 1004 > 1253.

Exercise 2.4.6

1. (i)32

243; (ii)

19683

42875; (iii)

313

239 × 544. 2. n = 49. 3. can happen for

m = 2.

4. (m,n) = (2, 2).

Additional problems on “Exponents”

1. (i) B. (ii) D. (iii) D. (iv) C. (v) B.

2. (i) 1152; (ii) 25/2; (iii) 8.

3. 109.65. 4. 1. 5. b12. 6. (a) 642; (b) 0; (c) 2−9; (d) 944/27; (e) 1/2.

7. 11. 8. 81. 9. a+m = 8. 10. k + 2 = 8 or 14. 11. 1. 12. 625.

Exercise 2.5.4

2. (i) first quadrant; (ii) third quadrant; (iii) fourth quadrant.

3. (−17,−2). 4. (−13,−8).Exercise 2.5.5

3. (i) 3x− 2y = 6; (ii) x+ y = 5. 5. (1, 1).

Additional problems on “Introduction to graphs”

1. (a) B. (b) B. (c) C. (d) C. (e) C. (f) D. (g) C. (h) B. (i) A.

2. (a) First; (b) second; (c) third; (d) fourth.

3. (A) −→ (iii); (B) −→ (i); (C) −→ (ii).

4. (a) zero; (b) abscissa; (c) (0, 0); (d) negative; (e) positive; (f) (0, 0);

(g) (0, 5).

5. (a) false; (b) true; (c) false; (d) true; (e) true; (f) false; (g) false;

(h) true.

6. 7y = 8(x− 7); 7y = 8x. 7. (i) 6y = 11x+ 15; (ii) 16y = −5x+ 12.

8. (−3, 3).

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CHAPTER 3 UNIT 1

AXIOMS, POSTULATES AND THEOREMS


• the meaning of undefined objects, axioms, postulates and hypothesis;

• that the lines, points, plane, space are undefined objects in Euclidean

geometry;

• various types of angles and relation among these angles;

• the properties of parallel lines and about Euclid’s fifth postulate.

3.1.1 Introduction

In earlier classes, you have studied many geometrical objects: straight

lines, triangles, quadrilaterals and circles. You have also studied some

geometrical properties of these objects: angles and different types of an-

gles; triangle inequality(the sum of two sides of a triangle is greater than

the third side); medians; altitudes; area of a triangle and a circle. Most of

these are taught to you through observations. You may wonder that these

were developed by our ancestors more than 2000 years ago.

Indeed, the concept of geometry is very old. Egyptian civilisation devel-

oped the early geometrical methods and measurements. In fact, geometry

is derived from two Greek words: Geo to mean Earth and metron meaning

measurement. When the Nile river flooded the whole region, the cultivated

land used to submerge in water erasing all the boundaries. Hence Egyp-

tians developed certain geometrical methods to demarcate the boundaries

afresh. They also introduced area of plane figures and volume of some

three dimensional objects which were used as granaries. Perhaps Pyra-

mids, which still occupy a place among seven wonders of the world and

whose construction is certainly one of the greatest human achievements,

will give you an idea how much Egyptians were advanced in the use of

geometry.

Thus the ancient geometry developed through the practical requirement

of measuring land. However, a systematic treatment of geometry started

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220 Unit 1

with ancient Greek around 2500 years ago. They are the first one to realise

the need to conceptualise the geometrical ideas.

The practical geometry used point, line and plane without bothering

much what they mean. But, the Greek Philosophers and Mathematicians

were more interested in proving statements by deductive reasoning. It

was, perhaps, Thales(640 BC- 546 BC) who first introduced the concept of

proof. He realised the need for proving a statement by logical reasoning.

Many more Greeks, like Appolonius, Plato, Pythagoras, Diophantus and

Ptolemy made enormous contributions to the systematic development of

Geometry and other areas of Mathematics and laid the firm foundation to

make Mathematics a science of logical reasoning.

However, it was Euclid who collected all these contributions to Geome-

try and other branches of mathematics in to thirteen volumes of a book

called the Elements along with his own original ideas.

Euclid(around 300 BC) was a Greek

mathematician often called the Father

of Geometry. He was a contemporary

of Ptolemy(323 BC-283 BC), another fa-

mous Greek mathematician of antiq-

uity. Euclid’s work Elements is one of

the most influential work in the whole

history of mathematics which changed

the face of Mathematics laying the foun-

dation for the future development.

Euclid deduced his results, what is now known as Euclidean Geometry,

from a small number of principles called Axioms and Geometrical postu-

lates. Euclid has also contributed to other branches of Mathematics. His

proof that there are infinitely many prime numbers is a classic example of

the deductive reasoning that Euclid employed in his works. Nothing much

is known about Euclid’s life. The date and place of his birth are unknown.

Like-wise, the date and circumstances of his death are also unknown. All

we know about him is through the references made by other people in their

work. The picture of Euclid we have today is also through the imagination of

an artist.

In ancient India, Sulva Sutras are perhaps the first record of Math-

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Axioms 221

ematical lore, especially geometry(600 BC to 300 BC). These are records

of the mathematical principles developed during Vedic period and subse-

quent time. The Sulva Sutras contain several geometrical principles.

The Indian geometry developed out of religious needs of constructing

sacrificial altars to propitiate Gods and later for the study of Eclipses.

The Baudhayana Sutra, the most ancient among the Sulva Sutras says

the diagonals of a rectangle bisect each other. This also contained the idea

of the celebrated Pythagoras’s theorem, but unfortunately no proof was

given.

The Sulva Sutras also gave methods of constructing a square whose

area equals the area of a given circle. The construction involved approx-

imating π and the approximation used was 3.088, which is quite close to

the present day approximations to π.

With the passage of time, others also made significant contributions:

Aryabhata I, Bhaskara I, Varahamihara, Brahmagupta, Mahaviracharya,

Bhaskaracarya II, Madhava, Nilakantha Somayaji, and many others made

innumerable contribution to the advancement of Mathematics.

3.1.2 Axioms and Postulates

You have seen that a straight angle is defined as the angle measuring

180 degrees. You use a protractor to measure angle. However, the cali-

bration of protractor is such that when you place it on a straight line, you

read 180 degrees. Thus the protractor is designed to measure 180 degrees

when a straight angle is given. Can you see that you have to go around

from protractor to straight angle to protractor?

This was the major difficulty faced by Greek mathematicians while de-

veloping Geometry as a pure deductive science. They had to depend on

certain primitive notions like points, straight lines and planes and space.

But this was not enough to deduce every thing. They had to set up cer-

tain statements, whose validity was accepted unquestionably, applicable

to geometry alone. They had to depend some more statements applicable

to all of Mathematics and science in general and Geometry in particular.

The general statements which are accepted without question and which

are applicable to all branches of science are commonly referred as Axioms.

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222 Unit 1

The statements which are particular to Geometry and accepted without

question are called Geometrical Postulates. Any result you further prove

depends on these axioms and postulates.

Another problem with the deductive method is how to define some ge-

ometrical terms. For example, you all have an intuitive idea what a point

is. But can you define a point? When you define some thing, you must

do it so using what you know already. Hence you have to depend on some

undefined terms.

In Euclid’s geometry, the undefined terms are point, line, plane. They

are only certain abstract ideas. Thus you cannot see a point. If you take

a sharp pencil and make a dot on a paper, that approximately resembles

a point. Similarly, you cannot see a line. When a point moves in both the

directions, it produces a straight line. A line is endless.

If A and B are points on a line, we denote the straight line by←→AB. If a

straight line is cut, you get two pieces: each piece is called a ray. Thus a

ray has an initial point and extends indefinitely in one direction. If A is the

initial point of a ray and B is any other point on a ray, we denote the ray

by−→AB. Take a line and choose any two points A and B on it. The part of

the line between A and B is called a line segment and is denoted by AB.

BA

Straight Line

A BRay

A BSegment

Likewise, you cannot define a plane. Intuitively, a plane is flat infinite

surface without any thickness. A black board or the surface of still water in

a big tank resemble finite part of a plane. Thus, there are undefined terms

in Euclid’s geometry. With suitable axioms and geometrical postulates,

Euclid’s geometry tells you what edifice can be built. Let us study these

axioms and postulates.

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Axioms 223

I. Axioms

There are certain elementary statements, which are self evident and

which are accepted without any questions. These are called axioms.

These statements are also applicable to other areas of mathematics and

science. Euclid used the following statements which he called Common

Notions.

Axiom 1: Things which are

equal to the same thing are

equal to one another.

10 cmAB =

CD = AB

A B

DCimplies CD= 10 cm

Take a sheet of paper. Draw a

line segment AB of length 10 cm.

Draw a second line CD having

length equal to that of AB, using

a compass. Measure the length

of CD. Do you observe that CD

has length equal to 10 cm? You

may write this as: CD = AB and

AB = 10 cm implies CD = 10 cm.

Suppose you have three baskets A, B and C having mangoes, oranges

and bananas. Suppose A, B have equal number of fruits and B, C also

have equal number of fruits. Can you conclude that A and C have equal

number of fruits?

Axiom 2: If equals are added to

equals, the wholes are equal.

A B C

D E F

AB=DE and BC=EF implies AC=DF

Suppose you have two line seg-

ments AB and DE of equal length.

Add BC to AB and add EF to DE.

If BC = EF , then AC = DF .

Take a basket A of 10 mangoes

and a basket B of 10 oranges.

Add 5 apples to both of these bas-

kets. Do you see that the number

of fruits in both the baskets are

equal?(equal to 15)

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224 Unit 1

Axiom 3: If equals are sub-

tracted from equals, then the

remainders are equal.

A B C

D E F

AC=DF and BC=EF implies AB=DE

Suppose you have two line seg-

ments AC and DF of equal length.

Remove BC from AC and EF from

DF respectively. If BC = EF , then

AB = DE.

Take a basket A of 10 mangoes

and B of 10 oranges. Remove 2

fruits from each basket. Then A

and B again have equal number

of fruits.

Axiom 4: Things which coincide with one another must be equal to

one another.

This means that if two geometric figures can fit completely one in to other,

then they are essentially the same.

Axiom 5: The whole is greater than the part.

Take a container of water. Remove some water from it. Will the remaining

volume of water the same as the original volume?

Euclid’s common notions are these first five axioms. The first three

concern “equals” or “equal things”; the fourth is interpreted now to

mean that if two figures, such as linesegments, angles, triangles, or

circles, are such that one can be moved to coincide with the other,

the figures are equal. In modern language they are called congruent.

Two ideas run through these common notions: (1) that geometrical

figures can be treated as magnitudes, and (2) that, if one figure is

visibly part of another (perhaps after a motion), then the magnitude

of the part is less than the magnitude of the whole. For adding

and comparing, you should have magnitudes of the same kind. For

example, you cannot add area to length. Or you cannot compare a

triangle with a point. Axiom 5 can be used to define greater than.

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Axioms 225

If b is a part of a, then a is greater than b. Again this comparison

is between magnitudes of the same kind. (For example, you cannot

take out a point from a line and say line is greater than a point which

is clearly meaningless.) Since point has no magnitude of any kind,

you cannot compare two points. But you can say some line segment

is larger than some other line segment as you can compare their

lengths.

II. Geometrical postulates

Apart from these common notions, Euclid also made the following

geometrical postulates to deduce new propositions.

Postulate 1. A straight line segment can be drawn joining any two

points.

Postulate 2. Any straight line segment can be extended indefinitely

in a straight line.

Postulate 3. Given any straight line segment, a circle can be drawn

having the segment as radius and one endpoint as center.

Postulate 4. All right angles are congruent.

Postulate 5. If a straight line meets two other lines, so as to make

the two interior angles on one side of it together less than two right

angles, the other straight lines will meet if produced on that side on

which the angles are less than two right angles.

The Postulate 5 is the famous Euclid’s parallel postulate. What it

asserts is that two distinct straight lines in a plane are either parallel

or meet exactly in one point. This postulate cannot be proved as

a theorem, although this was attempted by many people. Euclid

himself used only the first four postulates (“absolute geometry”) for

the first 28 propositions of the Elements, but was forced to invoke

the parallel postulate on the 29-th.

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226 Unit 1

In 1823, Janos Bolyai and Nicolai Lobachevsky independently re-

alized that entirely self-consistent “non-Euclidean geometries” could

be created in which the parallel postulate did not hold. The paral-

lel postulate is actually equivalent to: any two straight lines in the

plane either do not meet at all or meet in one point.

While proving his propositions, Euclid made several tacit assumptions.

For example, Postulate 1 says that there is a line passing through any two

given points. But what Euclid had in his mind seems to be: given any two

distinct point in the plane, there is a unique line passing through these two

points. On the other hand, you can draw infinitely many lines through a

given point in the plane.

Postulate 2 says that given a line segment in the plane, this can be

extended to a unique straight line. Postulate 4 is concerned about right

angles. But this is not defined any where by Euclid. What he seems to

have thought was that angle by a straight line is made up of two right

angles.

By present day standards, there are several inconsistencies in Euclid’s

Elements. Nevertheless, it is definitely the first book, based on rigorous

mathematical principles. In recent years, many attempts have been made

to introduce new set of undefined objects, axioms and postulates, so that

the terms hitherto undefined could be defined using these new objects,

axioms and postulates. The advancement of set theory and axioms on

number systems make it possible to define a point, a line or a plane using

coordinate systems.

Exercise 3.1.2

1. What are undefined objects in Euclid’s geometry?

2. What is the difference between an axiom and a postulate?

3. Give an example for the following axioms from your experience:

(a) If equals are added to equals, the wholes are equal.

(b) The whole is greater than the part.

4. What is the need of introducing axioms?

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Axioms 227

5. You have seen earlier that the set of all natural numbers is closed

under addition(closure property). Is this an axiom or some thing you

can prove?

3.1.3 Lines and Angles

B

O A

AOB

Suppose you have a ray−→OA on a plane with

end point O. With the same end point O,

consider another ray−−→OB in the same plane.

You observe that−−→OB is obtained from

−→OA

through suitable rotation around the point

O.We say−−→OB subtends an angle with

−→OA.

The amount of rotation is the measure of

this angle.

We use a numerical measurement called degree to measure angles. We

use the notation a◦ to denote a degrees. The rays−→OA and

−−→OB are called

the sides of the angle and O is called the vertex of the angle. The angle

subtended by the rays−→OA and

−−→OB is denoted by ∠AOB or AOB.

Note: Consider two rays−→OA and

−−→OB. If X is any point on

−→OA, then

the rays−→OA and

−−→OX are the same. Similarly, for any point Y on the

ray−−→OB, the rays

−−→OY and

−−→OB are the same. Thus ∠AOB = ∠XOY .

There is a geometrical instrument called protractor which is used for

measuring angles.

Activity 1:

Construct an angle which measures 40 degrees using protractor.

Activity 2:

Take two rays−→OA and

−−→OB and measure the angle between them using a

protractor.

Warning! Even with the best protractor, scale and pencil, you may

not be able to produce an angle which measures exactly 40◦. Your

eyes also play an important role, as parallax error normally creep

in. Nevertheless your construction is good enough for all practical

purposes.

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228 Unit 1

Recall what you have studied about different types of angles: straight

angle, right angle, acute angle, obtuse angle, reflex angle, complete

angle, adjacent angles, complimentary angles and supplementary an-

gles.

Consider a straight line and let O be point on the straight line. Then O

divides the straight line in to two rays. If B is to the left of O on the line

and A to the right of O, then there are two rays−→OA and

−−→OB. The angle

between these two rays is called a straight angle.

If you set your protractor such that its centre coincides with O, then you

see that the angle between−→OA and

−−→OB is 180◦. However, your protractor

is so calibrated that it measures any straight angle exactly 180◦. Thus you

cannot define a straight angle using a measuring device.

We will see that in Euclid’s geometry, this is taken as one of the postu-

lates. But once you know a straight angle, you can define all other types

of angles. For example, a right angle is that angle which measures 90◦

or you need half of a protractor. Similarly, an acute angle is one having

measure less than 90◦ and an obtuse angle is that angle which measures

more than 90◦ but less than a straight angle. A reflex angle is an angle

measuring more than 180◦ but less than 360◦. Finally, a complete angle

is an angle measuring 360◦. This corresponds to a complete revolution of

a ray−→OA around the initial point O.

A O B O

B

A

Straight angle Right angle

O A O A

B

B

Acute angle Obtuse angle

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Axioms 229

B

AO

A

O B

Reflex angle

Complete angle

B AO

O A

B

Right angle

Straight angle

We say two rays−→OA and

−−→OB are per-

pendicular to each other if the an-

gle between them is 90◦ and we write−→OA ⊥ −−→OB. We say two rays

−→OA and−−→

OB are supplementary rays if the an-

gle between them is 180◦. Observe that

in this case−→OA and

−−→OB are in opposite

directions.

Suppose you have a straight line and O is a point on this line. Then O

divides the straight line in to two rays: if B is to the left of O and A to the

right of O, then the straight line is made up of two rays−→OA and

−−→OB, the

angle between them being one straight angle or 180◦.

Two angles are said to be supplementary angles if their sum is 180◦.Similarly, two angles are said to be complementary if they add up to 90◦.

Two angles are said to be adjacent angles, if both the angles have

a common vertex and a common side. If two straight lines←→AB and

←→CD

intersect at a point O, then you see that four angles are formed at O: if

the first line is divided by O in to two rays−→OA and

−−→OB and if the second

line is divided in to two rays−→OC and

−−→OD, then you get four angles ∠AOC,

∠COB, ∠BOD and ∠DOA. The pair of angles ∠AOC and ∠BOD are called

vertically opposite angles. Observe the other pair ∠COB and ∠DOA is

also a pair of vertically opposite angles.

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230 Unit 1

OA

X

B

C

AD

B

O

Adjacent angles

Vertically opposite angles

While measuring lengths of line segments and angles, we observe the

following rules. These were not stated separately by Euclid, but tacitly

assumed by him in the derivation of new propositions. We take them as

additional postulates.

Rule 1. Every line segment has a positive length. (The length of the

line segment AB is denoted by AB or |AB|.)

Rule 2. If a point C lies on a line segment AB, then the length of AB

is equal to the sum of the lengths of AC and CB; that is AB = AC+CB.

Rule 3. Every angle has a certain magnitude. A straight angle mea-

sures 180◦.

Rule 4. If−→OA,

−−→OB and

−→OC are such that

−→OC lies between

−→OA and

−−→OB,

then ∠AOB = ∠AOC + ∠COB.

Rule 5. If the angle between two rays is zero then they coincide.

Conversely, if two rays coincide, the angle between them is either

zero or an integral multiple of 360◦.

Note: While measuring angles, we use the following convention: if

the angle is measured anti-clock-wise, it is positive. If it is measured

clock-wise, then it is negative.

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Axioms 231

Note: What is the minimum number of axioms needed to develop a

self-consistent geometry is a question in Mathematical Philosophy.

However, here we do not bother about the minimum number re-

quired nor about retaining original axioms and postulates made by

Euclid. In fact there were several gaps in the set of axioms used by

Euclid and later mathematicians added some more to those used by

Euclid.

Activity 3:

On a sheet of paper draw a straight line←→AB. Choose a point O on it.

Draw a ray−→OC on it. Measure angles ∠BOC and ∠COA using a protractor.

What is the sum ∠BOC + ∠COA? Repeat this with different positions of−→OC. What do you find?

You will always find that the sum of these two angles is 180◦. Can you

prove this using axioms and postulates.?

Note: Perhaps, you may realise here the need for a logical proof. No

matter which configuration you take with a line and a ray standing

on it, you see that the sum of the two adjacent angles always add

up to 180◦. However, this does not deny that there is a case of a line

and a ray on it such that the sum of the two adjacent angles is dif-

ferent from 180◦. This is inherent in the structure itself as there are

infinitely many possibilities of a line and a ray on it, and you cannot

verify your finding with all of them. This is the reason why Mathe-

maticians look for logical proofs based on axioms and postulates or

on the propositions which have already been proved.

Let us see what is the statement we need to prove. We have to take

an arbitrary line and an arbitrary ray standing on it. Then there are two

adjacent angles formed by the line and the ray. We have to show that the

sum of these adjacent angles is 180◦. We put this as a proposition. A

Proposition is a statement which is to be proved using the axioms and

postulates. It also depends on certain assumptions called hypotheses

which are used in the proof of the statement.

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232 Unit 1

Proposition 1. Let←→AB be a straight line and

−→OC be a ray standing

on the line←→AB. Then ∠BOC + ∠COA = 180◦.

A O B

C

Fig . 1

Before getting on to the proof, let us

see what are given and what we need

to prove.

Given (or hypothesis): A ray−→OC

stands on a straight-line←→AB forming

two adjacent angles ∠BOC and ∠COA.

To prove: ∠BOC + ∠COA = 180◦.Proof: We have ∠BOC +∠COA = ∠BOA

by rule 4. But ∠BOA is a straight an-

gle determined by the line←→AB. By rule

3, ∠BOA = 180◦. Now we can invoke

Axiom 1: ∠BOC + ∠COA and 180◦ are

both equal to the same thing, ∠BOA.

We conclude that ∠BOC+∠COA = 180◦.Look at the proposition once again. It says: if a ray stands on a straight

line, then the sum of two adjacent angles formed is 180◦. That is, the two

adjacent angles are supplementary. This is the general nature of all our

new propositions: given that a certain statement S is true, some other

statement R is true. We say S is the hypothesis and R is the conclu-

sion.(Here a ray stands on a straight line is the statement S which is the

hypothesis and the sum of adjacent angles is 180◦ is the statement R, which

is the conclusion.) Hence if R is not true, then S cannot be true. Formally

we say “S implies R” is equivalent to “not R implies not S.”

This was one of the methods frequently employed by Euclid and also

by later mathematicians to prove new propositions. If you want to prove

that “S implies R”, it is enough to prove “not R implies not F .” This is

called method of reductio ad absurdum. This is a latin word with meaning

“reduction to the absurdity.”

What is the converse of a proposition? Naturally, hypothesis and con-

clusion must change their places. If our original proposition has S as

hypothesis and R as conclusion, the converse proposition must have R

as hypothesis and S as conclusion. In the present context the converse is:

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Axioms 233

If there are three rays−→OA,

−−→OB and

−→OC such that ∠BOC and ∠COA are adja-

cent angles(that is−→OC is between

−→OA and

−−→OB) and if ∠BOC+∠COA = 180◦,

then A,O,B all lie on the same straight line. We say A,O,B are collinear

if all of them lie on the same straight line. We put forth this as another

proposition.

Proposition 2. Let−→OA,

−−→OB and

−→OC be three rays such that

−→OC is

between−→OA and

−−→OB. Suppose ∠BOC + ∠COA = 180◦. Then A,O,B are

collinear,that is, they lie on the same straight-line.

A O

C

B

DFig . 2

A O

C

Fig . 3

D

B

Given: Three rays−→OA,

−−→OB and

−→OC are such that ∠BOC and ∠COA are

adjacent angles which add up to 180◦.To prove: A,O,B all lie on the same line..

Proof: Here we do some construction. Extend AO to D such that A,O,D

all lie on the same line,←→AD. Using proposition 1, we have ∠DOC+∠COA =

180◦. But we are given that ∠BOC+∠COA = 180◦. Using Axiom 1, it follows

that∠DOC + ∠COA = ∠BOC + ∠COA.

Now use Axiom 3 to get ∠DOC = ∠BOC. There are two possibilities:−−→OB

lies between−−→OD and

−→OC(see Fig.2) or

−−→OD lies between

−−→OB and

−→OC(see Fig.

3). In the first case, using rule 4, we get

∠BOC = ∠DOC = ∠DOB + ∠BOC,

and Axiom 3 implies that ∠DOB = 0. In the second case, rule 4 gives

∠DOC = ∠BOC = ∠BOD + ∠DOC,

and Axiom 3 implies that ∠BOD = 0. Thus the angle between the rays−−→OB and

−−→OD is zero. Using rule 5, we conclude that the rays

−−→OB and

−−→OD coincide. This means B and O are on the line

←→AD. Thus A,O,B are

collinear.

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234 Unit 1

Note: Proposition 1 and proposition 2 are two geometrical state-

ments which are converses of each other. In geometry, if some

statement is true, then many times its converse is also true. How-

ever, this is not universally valid. There may be statements which

are true, but whose converses may fail. Later you will see that an

equilateral triangle is isosceles, but an isosceles triangle need not be

equilateral.

A O B

C

Fig . 4

Example 1. In the adjoining figure, if

∠COA− ∠BOC = 50◦, find these angles.

Solution: We know by proposition 1,

∠BOC + ∠COA = 180◦.

Adding, two relations, we get

2∠COA = 230◦.(Which axiom is used here?) This implies that ∠COA = 115◦. (Which axiom

is needed here?) Now

∠BOC = 180◦ − ∠COA = 180◦ − 115◦ = 65◦.

Example 2. In the adjoining figure, if the angles ∠AOB, ∠BOC, ∠COD are

in the ratio 1:2:3 and←→AD is a straight line, find the measures of all the

angles.

OD A

C

B

1

23

Fig. 5

Solution: Using the proposition

1, we see that

∠AOB + ∠BOC + ∠COD = 180◦.

But the given hypothesis is

∠BOC = 2∠AOB and ∠COD =

3∠AOB. Thus we get 6∠AOB =

180◦. It follows that ∠AOB = 30◦.This gives ∠BOC = 2 × 30◦ = 60◦

and ∠COD = 3× 30◦ = 90◦.

Definition: Suppose ∠AOB is an angle formed by two rays−→OA and

−−→OB. If−→

OP is another ray between−→OA and

−−→OB such that ∠AOP = ∠POB, we say

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Axioms 235

−→OP bisects ∠AOB or

−→OP is the angle bisector of ∠AOB. We observe that

in this case ∠AOP = ∠POB = 12∠AOB.

Activity 4:

Draw a straight line←→AB and a ray

−→OC on

←→AB. Measure ∠BOC and ∠COA.

Construct ray−→OP such that it bisects ∠BOC. Similarly, construct ray

−→OQ

such that, it bisects ∠COA. Measure ∠POQ. Do you see that ∠POQ = 90◦?.

Repeat this taking different rays−→OC on

←→AB. Do you always find that

∠POQ = 90◦? Can you formulate this as a proposition?

Proposition 3. Let←→AB be a straight line and let

−→OC be a ray standing

on it. Let−→OP be the bisector of ∠BOC, and let

−→OQ be the bisector of

∠COA. Then ∠POQ = 90◦.

A O B

C

Q

P

Fig. 6

Given:−→OP bisects ∠BOC and

−→OQ bi-

sects ∠COA.

To prove: ∠POQ = 90◦.

Proof:

Since−→OP bisects ∠BOC, we have

∠POC =1

2∠BOC. (1)

Since−→OQ bisects ∠COA, we also have

∠COQ =1

2∠COA. (2).

Adding these two and using rule 4, we obtain

∠POQ =1

2(∠BOC + ∠COA).

By proposition 1, ∠BOC + ∠COA = 180◦. Thus we obtain

∠POQ =1

2× 180◦ = 90◦.

Activity 5: Take two lines←→AB and

←→CD intersecting at a point O. Measure

∠BOD, ∠DOA, ∠AOC, ∠COB. Compare ∠BOD and ∠AOC. Similarly com-

pare ∠DOA and ∠COB. Do you observe some thing pertinent? Repeat this

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236 Unit 1

with different positions of←→CD with respect to

←→AB. Can you formulate a

new geometrical proposition?

Proposition 4. If two straight lines intersect at a point, then the

vertically opposite angles are equal.

C

A B O

D

Fig. 7

Given:←→AB and

←→CD intersecting at a

point O.

To prove: ∠BOD = ∠AOC and ∠DOA =

∠COB.

Proof: Consider the straight line←→AB

and the ray−−→OD standing on it. Then

∠BOD and ∠DOA are adjacent angles.

By proposition 1, we have

∠BOD + ∠DOA = 180◦. (1)

Similarly, considering the straight line←→CD and the ray

−→OA, we see that

∠DOA and ∠AOC are adjacent angles. Thus proposition 1 again gives

∠DOA+ ∠AOC = 180◦. (2)

Using Axiom 1, we can compare (1) and (2) and get

∠BOD + ∠DOA = ∠DOA+ ∠AOC. (3)

Since we can remove ∠DOA using Axiom 3, we obtain

∠BOD = ∠AOC.

A similar argument gives ∠DOA = ∠COB.

Example 3. Let←→AB and

←→CD be straight lines intersecting at O. Let

−→OP be

the bisector of ∠BOD and−→OQ be the bisector of ∠AOC. Prove that Q,O, P

are collinear.

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Axioms 237

C

A B O

D

Q

P

Fig. 8

Solution: We have to show that

∠POQ = 180◦. Since−→OP bisects ∠BOD,

we have

∠POD =1

2∠BOD. (1)

Similarly, using the given hypothesis

that−→OQ bisects ∠AOC, we also obtain

∠AOQ =1

2∠AOC. (2)

However, we have

∠POQ = ∠POD + ∠DOA+ ∠AOQ (using rule 4)

= ∠DOA+1

2(∠BOD + ∠AOC) (from (1) and (2))

= ∠DOA+1

2× 2∠AOC (∠BOD = ∠AOC as vertical opposite angles)

= ∠DOA+ ∠AOC

= 180◦ (using proposition 1).

It follows that P,O,Q are collinear.

Exercise 3.1.3

1. Draw diagrams illustrating each of the following situation:

(a) Three straight lines which do not pass through a fixed point.(b) A point and rays emanating from that point such that the angle

between any two adjacent rays is an acute angle.(c) Two angles which are not adjacent angles, but still supplemen-

tary.(d) Three points in the plane which are equidistant from each other.

2. Recognise the type of angles in the following figures:

PQ

ABO

OB A

X

X

Y

R

A

OB

(i)(ii) (iii)

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238 Unit 1

3. Find the value of x in each of the following diagrams:

(i)

O

C

BA

x2x

(ii)

A O B

D

C

x4x

90

(iii)

A B

C

D

x

x90

(iv)

O

x+yx−y

A B

C

(v)

A B

E

F

O

C

D

x

3x

x +30

(vi)

C

A B

F

E

D x

y

y

65

4. Which pair of angles are supplementary in the following diagram? Are

there supplementary rays?

130

12050

O

D

A

B

C

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Axioms 239

5. Suppose two adjacent angles are supplementary. Show that if one of

them is an obtuse angle, then the other angle must be acute.

3.1.4 Parallel lines and Euclid’s fifth postulate

Take any two distinct straight lines←→AB and

←→CD. We first show that

there is at most one point common to them. Suppose the contrary and

assume that P and Q are two distinct points which lie on both the lines.

But we know that by postulate 1 that there is a unique line←→PQ passing

through P and Q. Since P and Q are on←→AB, we must have

←→AB=

←→PQ. A

similar argument shows that←→CD=

←→PQ. Using Axiom 1, we obtain

←→AB=

←→CD,

which contradicts the assumption that←→AB and

←→CD are distinct. Thus

given two distinct lines, either they do not have any point in common or

there is one point common to them. In the latter case, the two straight

line intersect at this common point.

We say two straight lines are parallel to each other, if either they are

identical or they do not intersect. Thus two distinct lines←→AB and

←→CD are

parallel to each other if and only if they do not share any common point.

Let us get back to Euclid’s Postulate 5:

Postulate 5. If a straight line meets two other lines, so as to make

the two interior angles on one side of it together less than two right

angles, the other straight lines will meet if produced on that side on

which the angles are less than two right angles.

This is one of the most complicated postulate made by Euclid. In later

years, many attempts have been made to arrive at a simpler equivalent

versions of Euclid’s fifth postulate.

Suppose←→AB and

←→CD are two straight lines and let

←→PQ be a line which

meets←→AB in L and

←→CD in M .(see Fig. 9) If a line intersects two or more

lines, it is called a transversal to those set of lines. Here←→PQ is a transversal

to←→AB and

←→CD. There are eight angles formed by these three lines: ∠1, ∠2,

∠3, ∠4, ∠5, ∠6, ∠7 and ∠8 as shown in figure. In this ∠1, ∠4, ∠7 and ∠6 are

called exterior angles; ∠3, ∠2, ∠5 and ∠8 are called interior angles. The

angles ∠3 and ∠5 are called a pair interior alternate angles. Observe that

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240 Unit 1

∠2 and ∠8 are also a pair of interior alternate angles. Similarly, the pairs

∠1,∠7 and ∠4,∠6 are called pairs of exterior alternate angles. Angles

∠1 and ∠5 are called a pair corresponding angles. There are three more

pairs of corresponding angles: ∠2, ∠6; ∠3, ∠7; and ∠4, ∠8.

A

B

C D

P

Q

L

M

1

23

4

5

67

8

Fig. 9

Look at ∠3 and ∠8. They are the interior angles on the same side of

the line←→PQ. According Postulate 5, if ∠3 + ∠8 < 180◦, then

←→AB and

←→CD

meet on the left side of←→PQ. (If it happens that ∠2 + ∠5 < 180◦, then they

must meet on the right side of←→PQ.) Let us explore more on the condition

∠3 + ∠8 6= 180◦. We have the following proposition.

Proposition 5. If a transversal cuts two parallel lines, then the sum

of two internal angles on the same side of the transversal is equal to

180◦.

Given: a transversal intersecting two parallel lines.

To prove: the sum of internal angles on the same side of trnasversal is

equal to 180◦.

Proof: Suppose the result is not true. (see fig 9.) If ∠3 + ∠8 6= 180◦, then

either you must have ∠3 +∠8 < 180◦ or ∠3 +∠8 > 180◦. In the first case←→AB

and←→CD meet on the left side of

←→PQ. Suppose ∠3 + ∠8 > 180◦. We observe

that

∠3 + ∠8 + ∠2 + ∠5 = (∠3 + ∠2) + (∠8 + ∠5)

= ∠ALB + ∠CMD

= 180◦ + 180◦ = 360◦.

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Axioms 241

Thus

∠2 + ∠5 = 360◦ − (∠3 + ∠8) < 360◦ − 180◦ = 180◦.

=⇒ ∠2+∠5 < 180◦. Hence Postulate 5 tells us that←→AB and

←→CD meet on the

right side of←→PQ.

We conclude that: if the sum of the interior angles on the same side of←→PQ is not equal to 180◦, then

←→AB and

←→CD meet at some point(either to the

left or to the right of←→PQ). Thus if

←→AB and

←→CD are parallel, then the sum of

the interior angles on the same side of any transversal←→PQ is equal to 180◦.

This completes the proof of the proposition.

Thus Postulate 5 implies that given a pair of parallel lines and a transver-

sal, the sum of the internal angles on the same side of the transversal is

equal to 180◦.Is the converse true? Given two straight lines and a transversal such

that the sum of two internal angles on the same side of the transversal is

equal to 180◦, does it follow that the two lines are parallel?. Here we make

use of a fairly simple equivalent version of parallel postulate of Euclid.

This was first given by a Scottish mathematician called Playfair.

Playfair’s postulate: Given a line in a plane and a point outside the

line in the same plane, there is a unique line passing through the

given point and parallel to the given line.

We have the following statement.

Proposition 6. If a transversal cuts two distinct straight lines in such

a way that the sum of two internal angles on the same side of the

transversal is equal to 180◦, then the two lines are parallel to each

other.

C M

P

Q

B

D

LA

X

Y

S

Fig. 10

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242 Unit 1

Given: two straight lines←→AB and

←→CD and a transversal

←→PQ intersecting

←→AB in L and

←→CD in M respectively; and ∠ALM + ∠LMC = 180◦.

To prove:←→AB‖

←→CD.

Proof: Assume that←→AB and

←→CD are not parallel. Then they must meet at

some point, say S. (see Fig.10) By Playfair’s postulate, there is a unique

line←→XY passing through S and parallel to

←→PQ.

Since←→XY ‖

←→PQ, we have ∠QLS + ∠LSY = 180◦(they are internal angles

on the same side of the transversal←→SB to the parallel lines

←→XY and

←→PQ).

But ∠QLS + ∠ALM = 180◦( they are adjacent angles formed by the ray−→LA

standing on the line←→PQ). Hence it follows that ∠LSY = ∠ALM . (Which

axioms are needed here?) But ∠ALM + ∠LMC = 180◦(given data). We also

have ∠LMC + ∠MSY = 180◦(since they are the sum of the internal angles

on the same side of the transversal−→SD cutting the parallel lines

←→XY and

←→PQ). Thus we get ∠ALM = ∠MSY . It now follows that ∠LSY = ∠MSY . But

∠MSY = ∠MSL+∠LSY . We obtain ∠MSL = 0. Hence←→SB and

←→SD coincide.

This forces that the straight lines←→AB and

←→CD are the same, contradicting

that they are distinct lines. We conclude that←→AB ‖

←→CD.

Activity 6:

Draw two parallel lines. Draw a transversal and measure different angles

formed by intersections. You will see that:

1. Any pair of alternate angles are equal.

2. Any pair of corresponding angles are equal.

Repeat the same with different positions of the transversal. You will see

that the same results repeat. We formulate this as a theorem.

Theorem 1. If two parallel lines are cut by a transversal, then

(i) each pair of alternate angles are equal;

(ii) each pair of corresponding angles are equal.

Given:←→AB and

←→CD two distinct parallel lines and

←→PQ a transversal inter-

secting←→AB in L and

←→CD in M . (see Fig. 11)

To prove: ∠3 = ∠5 and ∠1 = ∠5.

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Axioms 243

C D

P

Q

L

M

5

67

8

4 1

23

A B

Fig. 11

Proof: Since ∠3 and ∠8 are two

internal angles on the same side

of the transversal←→PQ cutting the

parallel lines←→AB and

←→CD, we

know that

∠3 + ∠8 = 180◦.But ∠8 and ∠5 are the adjacent angles formed by the ray

−−→MP standing

on the line←→CD. Hence we also know that

∠8 + ∠5 = 180◦.

Comparing, we see that ∠3 = ∠5.

Again observe that ∠2 + ∠3 = 180◦ = ∠8 + ∠5. Using ∠3 = ∠5, we get

∠2 = ∠8.

We also observe that ∠1 = ∠3, since they are vertically opposite angles.

Using this with ∠3 = ∠5, we conclude that ∠1 = ∠5.

Thus the pair ∠1,∠5 of corresponding angles are equal.

Similarly we can prove ∠2 = ∠6, ∠4 = ∠8 and ∠3 = ∠7, ∠1 = ∠7, ∠4 = ∠6.

Think it over!

There are two statements about parallel lines and a transversal:

(i) if a transversal cuts two distinct parallel lines, then any pair

of alternate angles are equal; (ii) if a transversal cuts two dis-

tinct parallel lines, then any pair of corresponding angles are

equal. But these two are not independent statements. You can

easily prove any of them assuming the other and using proposi-

tions 1 and 4.

What is the converse of theorem 1. If there are two distinct straight

lines and a transversal such that any pair of alternate angles are equal,

can we prove that the two lines are parallel to each other. Note that if we

are able to prove this result, you can also prove that: given two distinct

lines and a transversal such that any pair of corresponding angles are

equal, then the lines are parallel(use the previous observation). Thus we

have the following theorem.

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244 Unit 1

Theorem 2. Suppose a transversal cuts two distinct straight lines

such that a pair of alternate angles are equal. Then the two lines are

parallel to each other.

Given: two straight lines←→AB and

←→CD and a transversal

←→PQ (see Fig. 11)

and ∠3 = ∠5.

To prove:←→AB ‖

←→CD.

Proof: We know that ∠8 +∠5 = 180◦(since they are supplementary angles).

By the given hypothesis, we know that ∠3 = ∠5. We thus obtain ∠3 + ∠8 =

180◦. However, ∠3 and ∠8 are the internal angles on the same side of the

transversal to the lines←→AB and

←→CD. By proposition 6, we conclude that

←→AB ‖

←→CD.

Corollary: If a transversal cuts a pair of straight lines in such a way

that a pair of corresponding angles are equal, then the two lines are

parallel to each other.

Proof: Referring to Fig. 11, we are given, say, ∠1 = ∠5. But ∠1 = ∠3, since

they are vertically opposite angles. We obtain ∠3 = ∠5. Hence by theorem

2,←→AB ‖

←→CD.

Activity 7:

Draw two parallel lines←→AB and

←→CD. Draw a line

←→XY parallel to

←→CD. Draw

a transversal←→PQ. Let it cut

←→AB in L,

←→CD in M and

←→XY in N . Measure

∠BLQ and ∠Y NQ. Do you see that they measure the same. Repeat this

with different positions of←→PQ.

Example 4. Two lines which are parallel to a common line are parallel to

each other.

PX Y

DC

A B

Q

L

M

N

Fig. 12

Solution: Suppose←→AB and

←→XY

are two lines which are parallel

to a common line←→CD. We show

that←→AB ‖

←→XY . Draw a transversal

←→PQ, cutting

←→AB at L,

←→CD at M and

←→XY at N , respectively. We observe

that

∠BLP = ∠DMP,

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Axioms 245

as they are corresponding angles made by transversal←→PQ with the parallel

lines←→AB and

←→CD. Similarly,

∠DMP = ∠Y NP (Why?).

Using Axiom 1, we obtain ∠BLP = ∠Y NP . Now we can use corollary to

theorem 2 and conclude that←→AB ‖

←→XY .

Example 5. Let←→AB be a straight line. Let

←→CD and

←→EF be two straight lines

such that each of them is perpendicular to←→AB. Prove that

←→CD ‖

←→EF .

A

C E

D F

BML

Fig. 13

Solution: Let←→AB intersect

←→CD

and←→EF at L and M respectively.

Since←→CD⊥

←→AB, we have ∠DLA =

90◦. Using←→EF⊥

←→AB, we also get

∠FMA = 90◦. Thus ∠DLA =

∠FMA. But these are corre-

sponding angles made by the

transversal←→AB with the lines

←→CD and

←→EF . Hence by corollary to theorem

2, we conclude that←→CD ‖

←→EF .

Example 6. Show that the angle bisectors of a pair of alternate angles

made by the transversal with two parallel lines are parallel to each other.

A B

C D

P

Q

L

M

X

S

R

Y

Fig. 14

Solution: We are given two

parallel lines←→AB and

←→CD, and

transversal←→PQ. Consider the pair

of alternate angles ∠ALM and

∠LMD. Let−−→LX be the bisector of

∠ALM ; let−−→MY be the bisector of

∠LMD. Extend the ray−−→LX to the

straight line←→XR and the ray

−−→MY

to the straight line←→SY as shown

in the figure(see Fig. 14). We

have to show that−−→XR‖−→SY . Con-

sider the lines←→XR and

←→SY with

transversal←→PQ.

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246 Unit 1

We have

∠XLM =1

2∠ALM, ∠LMY =

1

2∠LMD.

However, ∠ALM = ∠LMD(why?). We hence obtain ∠XLM = ∠LMY . But

∠XLM and ∠LMY are a pair of alternate angles made by the transversal←→PQ with the lines

←→XR and

←→SY . It follows, by theorem 2, that

←→XR ‖

←→SY .

Exercise 3.1.41. Find all the angles in the follow-

ing figure.

AL

M

Q

B

C D

P

135

2. Find the value of x in the diagram

below.

x

130

90

3. Show that if a straight line is perpendicular to one of the two or more

parallel lines, then it is also perpendicular to the remaining lines.

4. Let←→AB and

←→CD be two parallel lines and

←→PQ be a transversal. Show

that the angle bisectors of a pair of two internal angles on the same

side of the transversal are perpendicular to each other.

Additional problems on “Axioms, postulates and theorems


(i) If a = 60 and b = a, then b = 60 by ————

A. Axiom 1 B. Axiom 2 C. Axiom 3 D. Axiom 4(ii) Given a point on the plane, one can draw ————

A. unique B. two C. finite number of D. infinitely many

lines through that point.(iii) Given two points in a plane, the number of lines which can be

drawn to pass through these two points is ————

A. zero B. exactly one C. at most one D. more than one(iv) If two angles are supplementary, then their sum is ————

A. 90◦ B. 180◦ C. 270◦ D. 360◦

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Axioms 247

(v) The measure of an angle which is 5 times its supplement is

————

A. 30◦ B. 60◦ C. 120◦ D. 150◦

2. What is the difference between a pair of supplementary angles and a

pair of complimentary angles?

3. What is the least number of non-collinear points required to deter-

mine a plane?

4. When do you say two angles are adjacent?

5. Let AB be a segment with C and D between them such that the order

of points on the segment is A,C,D,B. Suppose AD = BC. Prove that

AC = DB.

6. Let←→AB and

←→CD be two straight lines intersecting at O. Let

−−→OX be the

bisector of ∠BOD. Draw−−→OY between

−−→OD and

−→OA such that

−−→OY ⊥ −−→OX.

Prove that−−→OY bisects ∠DOA.

7. Let←→AB and

←→CD be two parallel lines and

←→PQ be a transversal. Let

←→PQ

intersect←→AB in L. Suppose the bisector of ∠ALP intersect

←→CD in R

and the bisector of ∠PLB intersect←→CD in S. Prove that

∠LRS + ∠RSL = 90◦.

8. In the adjoining figure,←→AB and

←→CD

are parallel lines. The transversals←→PQ and

←→RS intersect at U on the line

←→AB. Given that ∠DWU = 110◦ and

∠CV P = 70◦, find the measure of

∠QUS.

WV

UA B

DC

PR

QS

x

110

70

9. What is the angle between the hour’s hand and minute’s hand of a

clock at (i) 1.40 hours, (ii) 2.15 hours? (Use 1◦ = 60 minutes.)

10. How much would hour’s hand have moved from its position at 12

noon when the time is 4.24 p.m.?

11. Let AB be a line segment and let C be the midpoint of AB. Extend AB

to D such that B lies between A and D. Prove that AD +BD = 2CD.

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248 Unit 1

12. Let←→AB and

←→CD be two lines intersecting at a point O. Let

−−→OX be a

ray bisecting ∠BOD. Prove that the extension of−−→OX to the left of O

bisects ∠AOC.

13. Let−−→OX be a ray and let

−→OA and

−−→OB be two rays on the same side of−−→

OX, with−→OA between

−−→OX and

−−→OB. Let

−→OC be the bisector of ∠AOB.

Prove that

∠XOA+ ∠XOB = 2∠XOC.

14. Let−→OA and

−−→OB be two rays and let

−−→OX be a ray between

−→OA and

−−→OB

such that ∠AOX > ∠XOB. Let−→OC be the bisector of ∠AOB. Prove

that

∠AOX − ∠XOB = 2∠COX.

15. Let−→OA,

−−→OB,

−→OC be three rays such that

−→OC lies between

−→OA and

−−→OB.

Suppose the bisectors of ∠AOC and ∠COB are perpendicular to each

other. Prove that B,O,A are collinear.

16. In the adjoining figure,←→AB ‖

←→DE.

Prove that

∠ABC − ∠DCB + ∠CDE = 180◦.

A B

C

ED

17. Consider two parallel lines and a transversal. Among the measures of

8 angles formed, how many distinct numbers are there?

Glossary

Undefined objects: those objects in mathematics which cannot be de-

fined using the terms already known.

Axioms: certain statements which are valid in all branches of mathe-

matics whose validity is taken for granted without seeking mathematical

proofs.

Postulates: some statements which are taken for granted in a particular

branch of mathematics.

Hypothesis: certain conditions assumed while proving a proposition.

Adjacent angles: a pair of angles made by a ray standing on a line.

Complementary angles: a pair of angles which add up to 90◦.Supplementary angles: a pair of angles which add up to 180◦.

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Axioms 249

Straight angle: an angle formed by a straight line; equal to 180◦.Complete angle: an angle which measures 360◦.Reflex angle: an angle which measures more than 180◦, but less than

360◦.Linear pair: a pair of angles which make a starlight line.

Vertically opposite angles: when two straight lines intersect each other,

a pair of angles which do not form a linear pair are vertically opposite an-

gles.

Collinear: points all lying on the same straight line.

Parallel lines: a collection of lines which do not intersect pairwise.

Alternate angles: when a transversal cuts a pair of lines, the angles

formed by the transversal which is not a linear pair and lying on both

sides of the transversal.

Corresponding angles: when a transversal cuts a pair of lines, the angles

formed by the transversal which lie on the same side of the transversal

and also on the similar side of the two lines.

Points to remember

• Mathematics is a game in which certain objects are given and you

have to play the game according to certain pre-laid rules.

• In geometry, the objects are points, lines plane; the rules are axioms

and postulates of geometry.

• Euclid’s fifth postulate– equivalent formulation is that given any line

and a point out side that line, there is a unique line passing through

the given point and parallel to the given line– is the one which gives

Euclidean geometry. Changing this postulate will lead to different

geometry.

Purity, patience, and perseverance are three essentials to sucess and

above all, love. —–Swamy Vivekananda

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CHAPTER 3 UNIT 2

THEOREMS ON TRIANGLESAfter studying this unit, you learn to:

• identify a triangle in a collection of figures;

• classify different types of triangles based on sides and angles;

• recognise the angle sum property of a triangle;

• identify the interior and exterior angles of a triangle;

• establish the relationship between the exterior angle and interior op-

posite angles;

• prove logically angle sum property of the triangle;

• solve problems based on the angles of a triangle.

3.2.1 Introduction

In the previous chapter you have studied th properties lines and angles.

You have seen how the axioms of Euclidean geometry helps you to build

some nice relations between angles and lines. In this chapter, you shall

study about a closed plane figure formed by three non parallel lines, a

triangle.

A plane figure bound by three non concurrent line segments in plane

is called a triangle.

This needs an explanation. When we say a plane figure, we actually

mean the linear figure, not the two-dimensional figure. Let A,B,C be three

points such that they are not on the same line; we say A,B,C are non-

collinear. Join AB, BC and CA. You get a linear figure which consists

of three line segments which meet only at their end points. Such a linear

figure is called a triangle. We say A, B and C are the vertices of the

triangle ABC. The segments AB, BC, and CA are called the sides of the

triangle; and the angles ∠BAC, ∠ABC and ∠ACB are called the angles of

the triangle ABC (or the interior angles of ABC).A triangle consists 9 elements:

Vertices Sides Angles

A AB ∠BAC or ∠A

B BC ∠ABC or ∠B

C AC ∠ACB or ∠C

A

B C

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Triangles 251

Note: When there is no confusion, the sides of a triangle ABC are

also denoted by AB, BC and AC. These symbols are also used to

denote their respective lengths. The context tells whether the side or

the length to be taken.

Points to ponder:

(i) A triangle cut in a plane sheet of paper is a triangular sheet,

but not a triangle. Only the three line segments constitute a

triangle. (ii) Triangular sheet as a plane figure has only area,

but has no thickness.

Triangles are classified based on the measure of sides and angles.

classification based on sides:

(i) Equilateral triangle; (ii) Isosceles triangle; (iii) Scalene triangle.

B C

A

Equilateral triangle:

A triangle in which

all sides are of equal

length is called an

equilateral triangle.

In triangle ABC,

AB = BC = CA.

B C

A

Isosceles triangle:

A triangle in which

two sides are of equal

length is called an

isosceles triangle. In

the above triangle

ABC,

AB = BC.

B C

A

Scalene triangle:

A triangle in which

all sides are of differ-

ent lengths is called

a scalene triangle In

the above triangle

ABC,

AB 6= BC 6= CA 6= AB.

Do you see that an equilateral triangle is also isosceles? But an

isosceles triangle need not be equilateral.

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252 Unit 2

classification based on angles :

(i) Acute angled triangle; (ii) Right angled triangle; (iii) Obtuse angled

triangle.

B C

A

Acute angled triangle:

A triangle in which all

the angles are smaller

than 90◦ is called an

acute angled triangle. In

the above triangle ABC,

∠ABC < 90◦,∠BCA < 90◦,∠CAB < 90◦.

B C

A

Right angled trian-

gle: A triangle hav-

ing an angle equal

to 90◦ is called a

right angled triangle.

In the above triangle

ABC,

∠ABC = 90◦.

B C

A

Obtuse angled tri-

angle: If a triangle

has an angle greater

than 90◦, it is called

an obtuse angled tri-

angle. In the above

triangle ABC,

∠ABC > 90◦.

Activity 1:

Name the types of triangles given below :

100

130

(i)(ii) (iii)

(iv) (v)(vi)

(vii)(viii)

(ix)

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Triangles 253

(x) (xi) (xii)

Exercise 3.2.1

1. Match the following:

(1)

(a) Equilateral triangle

(2)

(b) Acute angled triangle

(3)

(c) Right angled triangle

(4)

(d) Obtuse angled triangle

2. Based on the sides, classify the following triangles (figures not drawn

to the scales):

4 cm

3 cm5 cm 7 cm

4 cm

5 cm

4.5 cm

6 cm

3.5cm

(i) (ii) (iii)

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254 Unit 2

4.3 cm4 cm 3.2 cm

4.1 cm 2.5 cm

(iv) (v) (vi)

5.6 cm

3 cm

6.5 cm6.5 cm

5 cm 9 cm

6 cm

5 cm

3 cm 3 cm

3 cm 5 cm

3.5 cm 8 cm

6 cm 6 cm

(vii) (viii) (ix) (x)

3.2.2 Sum of interior angles

Let us do some paper activity before guessing a geometrical result.

Activity 2:

Take a sheet of paper, fold it into four folds. On one of its folds, draw a

triangle using a scale and a pencil. Then cut the triangle from a pair of

scissors. Now you have four identical triangular sheets. Select three of

them and mark identical angles as 1, 2, and 3 on each sheet of the paper

as shown below.

1 13 3

2

1 3

2 2

Draw a straight line on a sheet of your note book. Arrange the triangles

such that angle 1 of the first triangle, 2 of the second and 3 of the third as

shown in the figure.

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Triangles 255

21 2

32

3 11

3

You can find that all the three angles together form a straight angle. But

you see that these three angles are precisely the angles of a triangle. Thus

you can guess that the sum of three angles of a triangle is 180◦.

Activity 3:

Draw a triangle ABC on a sheet as shown in the figure. Cut the remaining

part of the paper. Fold the triangular sheet such that the vertex A touches

the base line of the paper at M . Now fold the vertex B and C to meet the

point M . You will find that they make a straight angle. (See the figure

below.)

C

B C B M C

AA

BA

Draw right angled triangle such that angle ∠2 = 90◦. (See the figure below.)

2

1

3 2

1

3

4

We want to find the sum of the three angles: ∠1+∠2+∠3. Draw the parallel

line to base of the triangle and passing through the vertex at the top.

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256 Unit 2

Now we find another angle ∠4 equal to ∠3, because they are alternate

interior angles between two parallel lines. Therefore ∠1+∠3 = ∠1+∠4 = 90◦.This implies that

∠1 + ∠2 + ∠3 = ∠2 + (∠1 + ∠3) = 90◦ + 90◦ = 180◦.

Thus the sum of interior angles of a right triangle is 180◦. Now we make

use of this to show that the sum of three interior angles of a triangle is

180◦.

31 5

42

6

Take an arbitrary triangle. This

can be split into two right trian-

gles, by drawing a perpendicular

to the base. We know that,

∠1 + ∠2 + ∠3 = 180◦, ∠4 + ∠5 + ∠6 = 180◦.

Adding these we obtain

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360◦.

But angles ∠3 and ∠5 are supplementary angles and they make a straight

line. Therefore ∠3 + ∠5 = 180◦. Thus we get

∠1 + ∠2 + ∠4 + ∠6 = 360◦ − (∠3 + ∠5) = 360◦ − 180◦ = 180◦.

But can you see that ∠2+∠4 is the angle at one of the vertex of the triangle?

Hence the sum of the three angles of triangle is 180◦

If you observe the proof given above, it consists of two parts. In the first

part you prove that the sum of three interior angles of a right triangle is

180◦, using a construction; drawing a line parallel to the base line through

the top vertex. A general triangle is split in to two right angled triangles

and we use the result for right triangle for getting on to a general triangle.

Can’t we construct a parallel to the base through the top vertex for a

general triangle and proceed with the proof? We take up this approach

below.

Theorem 1. In any triangle, the sum of the three interior angles is

180◦. (Interior angle theorem)

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Triangles 257

B C

AFE

Given: ABC is a triangle.

To prove: ∠ABC + ∠BCA +

∠CAB = 180◦.Construction: Through point A,

draw the line EF ‖ BC.

Proof: Below we give several statements and the reason for the truth of

each statement. Finally we arrive at the desired conclusion.

Statement Reason

∠ABC = ∠EAB alternate angles by the transversal AB

with the parallel lines BC and EF

∠BCA = ∠FAC alternate angles by the transversal AC

with the parallel lines

∠EAB + ∠BAC + ∠FAC = 180◦ sum of the linear angles at A.

By substituting ∠EAB = ∠ABC and ∠BCA = ∠FAC, we finally get

∠ABC + ∠BAC + ∠BCA = 180◦.

This completes the proof.

Example 1 In a triangle ABC, it is given that ∠B = 105◦ and ∠C = 50◦.Find ∠A.

10550

CB

ASolution: We have, in triangle

ABC, (by Theorem 1),

∠A+ ∠B + ∠C = 180◦

=⇒ ∠A+ 105◦ + 50◦ = 180◦

=⇒ ∠A+ 155◦ = 180◦

=⇒ ∠A = 180◦ − 155◦

=⇒ ∠A = 25◦.

Thus ∠A measures 25◦.Example 2 In the given figure, find all the angles.

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258 Unit 2

B

A

C3

5

2x

x

x

Solution: In triangle ABC, if we

make use of theorem 1, we get

∠A+ ∠B + ∠C = 180◦. Hence

5x+ 3x+ 2x = 180◦ =⇒ 10x = 180◦

=⇒ x = 180◦/10 =⇒ x = 18◦.

Hence,∠A = 5x = 90◦;∠B = 3x = 54◦;∠C = 2x = 36◦.

Example 3 If the bisectors of the angles ∠ABC and ∠ACB of a triangle

ABC meet at a point O, then Prove that ∠BOC = 90◦ +1

2∠BAC.

Solution:

Given: A triangle ABC and the bisectors of ∠ABC and ∠ACB meeting at

point O.

To prove: ∠BOC = 90◦ +1

2∠BAC.

B C

O

A

1 2

Proof: In triangle BOC we have

∠1 + ∠2 + ∠BOC = 180◦ (1)

In triangle ABC, we have ∠A +

∠B+∠C = 180◦. Since BO and CO

are bisectors of ∠ABC and ∠ACB

respectively, we have

∠B = 2∠1 and ∠C = 2∠2.

We therefore get ∠A+2(∠1)+ 2(∠2) = 180◦. Dividing by 2, we get∠A

2+∠1+

∠2 = 90◦. This gives

∠1 + ∠2 = 90◦ − ∠A

2(2)

From (1) and (2), we get

90◦ − ∠A

2+ ∠BOC = 180◦.

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Triangles 259

Hence

∠BOC = 90◦ +1

2∠BAC.

Exercise 3.2.2

1. In a triangle ABC, if ∠A = 55◦ and ∠B = 40◦, find ∠C.

2. In a right angled triangle, if one of the other two angles is 35◦, find the

remaining angle.

3. If the vertex angle of an isosceles triangle is 50◦, find the other angles.

4. The angles of a triangle are in the ratio 1:2:3. Determine the three

angles.

5. In the adjacent triangle ABC, find

the value of x and calculate the mea-

sure of all the angles of the triangle.

B C

A

x x

x

−15

+15

+30

6. The angles of a triangle are arranged in ascending order of their mag-

nitude. If the difference between two consecutive angles is 10◦, findthe three angles.

3.2.3 Exterior angles

DCB

AConsider a triangle ABC. If the

side BC is produced externally

to form a ray−−→BD, then ∠ACD is

called an exterior angle of trian-

gle ABC at C and is denoted by

Ext∠C.

Note: If you produce AC to a point E (instead of BC), you get an

angle ∠BCE. But ∠ACD = ∠BCE as they are vertically opposite

angles. Thus Ext∠C is the same whether you use the side BC

or the side AC; it depends only on ∠C of the triangle ABC.

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260 Unit 2

With respect to Ext∠C of triangle ABC, ∠A and ∠B are called interior

opposite angles.

F

D

E

A

BC

Now in triangle ABC side CA,

BC, and AB are produced to form

rays−−→CE,

−−→BD and

−→AF . Then

∠BAE, ∠ACD and ∠CBF are ex-

terior angles of the triangle ABC.

B C

A

3

1 2

Activity 4: Take three sheets of

paper of size 8 × 10 cm. Place

one above the other and cut three

right angled triangles such that

one of the corners of each sheet

becomes the right angle of the tri-

angle. Now you will have threeidentical triangles. Mark the angles of each triangle as 1,2,3 as shown in

the figure.

B C

A

3

P Q1 2

Draw straight line PQ on an-

other sheet of paper and place

one of the triangular sheet on the

line as shown in figure such that

∠ACQ forms an exterior angle of

the triangle ABC.

B C

A

3

P Q1 23

1

12

2

3

Now place the remaining two tri-

angular sheets with angles 3 and

1 as shown in the adjacent fig-

ure. Can you see that ∠ACQ =

∠3 + ∠1? Thus, the measure of

an exterior angle of a triangle

is equal to the sum of the cor-

responding two interior opposite

angles.

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Triangles 261

Theorem 2. If a side of triangle is produced, the exterior angle so

formed is equal to the sum of the corresponding interior opposite

angles. (Exterior angle theorem.)

SRQ

P Given: In triangle PQR, produce

QR to S. Then ∠PRS is an ex-

terior angle and the correspond-

ing interior opposite angles are

∠PQR and ∠QPR.

To prove:∠PRS = ∠QPR+∠PQR.

Proof:

Statement Reason

∠QPR+ ∠PQR + ∠PRQ = 180◦; interior angle theorem;

∠PRQ+ ∠PRS = 180◦; linear pair

∠QPR + ∠PQR + ∠PRQ = ∠PRQ+ ∠PRS; Axiom 1(see Chap 3, unit 1);

∠QPR + ∠PQR = ∠PRS; Axiom 3(see Chap 3, unit 1).


Example 4 An exterior angle of a triangle is 100◦ and one of the interior

opposite angles is 45◦. Find the other two angles of the triangle.

Solution: Let ABC be a triangle whose side BC is produced to form an

exterior angle ∠ACD such that Ext∠C = 100◦. Let ∠B = 45◦. By exterior

angle theorem we have

B CD

A

100

∠ACD = ∠B + ∠A

=⇒ 100◦ = 45◦ + ∠A

=⇒ ∠A = 100◦ − 45◦

Hence ∠A = 55◦. Thus we get

∠C = 180◦ − (∠A+ ∠B)

= 180◦ − (55◦ + 45◦) = 80◦. Hence

∠C = 80◦.

Example 5 In the given figure, sides QP and RQ of a triangle PQR are

produced to the points S and T respectively. If ∠SPR = 135◦ and ∠PQT =

110◦, find ∠PRQ.

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262 Unit 2

110

135

T

S

Q R

P

Solution: Since Q, P and S all lie

on the same line,

∠QPR+ ∠SPR = 180◦.Hence

∠QPR + 135◦ = 180◦

or

∠QPR = 180◦ − 135◦ = 45◦.Using exterior angle property in

triangle PQR, we have

∠PQT = ∠QPR + ∠PRQ.

This gives 110◦ = 45◦ + ∠PRQ.

Solving for ∠PRQ, we get ∠PRQ = 110◦ − 45◦ = 65◦.

Example 6 The side BC of a triangle ABC is produced on both sides. Show

that the sum of the exterior angles so formed is greater than ∠A by two

right angles.

E FB C

A

24 3 5

1

Solution: Draw a triangle ABC

and produce BC on both sides to

points D and F . Denote the an-

gles as shown in the figure. We

have to show that

∠4 + ∠5 = ∠1 + 180◦.By exterior angle theorem, we

have

∠4 = ∠1 + ∠3 and ∠5 = ∠1 + ∠2.

Adding these two we get

∠4 + ∠5 = (∠1 + ∠3) + (∠1 + ∠2) = ∠1 + (∠1 + ∠2 + ∠3) = ∠1 + 180◦,since the sum of all the interior angles of a triangle is 180◦.

Exercise 3.2.3

1. The exterior angles obtained on producing the base of a triangle both

ways are 104◦ and 136◦. Find the angles of the triangle.

2. Sides BC, CA and AB of a triangle ABC are produced in an order,

forming exterior angles ∠ACD, ∠BAE and ∠CBF . Show that ∠ACD+

∠BAE + ∠CBF = 360◦.

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Triangles 263

3. Compute the value of x in each of the following figures:

50 106

130

100

65

112

120

x x

(i)(ii) (iii)

(iv) (v)

x x

x

20

4. In the figure, QT ⊥ PR, ∠TQR = 40◦

and ∠SPR = 30◦. Find ∠TRS and

∠PSQ.

Q S R

P

30

40

T

5. An exterior angle of a triangle is 120◦ and one of the interior opposite

angles is 30◦. Find the other angles of the triangle.

6. Find the sum of all the angles at the

five vertices of the adjoining star.

Additional problems on “Theorems on triangles”

1. Fill up the blanks to make the following statements true:

(a) Sum of the angles of a triangle is————

(b) An exterior angle of a triangle is equal to the sum of ——————

——— opposite angles.

(c) An exterior angle of a triangle is always ————— than either of

the interior opposite angles.

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264 Unit 2

(d) A triangle cannot have more than ————-right angle.

(e) A triangle cannot have more than———————– obtuse angle.

2. Choose the correct answer from the given alternatives:

(a) In a triangle ABC, ∠A = 80◦ and AB = AC, then ∠B is ———

A. 50◦ B. 60◦ C. 40◦ D. 70◦

(b) In right angled triangle, ∠A is right angle and ∠B = 35◦, then ∠C

is ———

A. 65◦ B. 55◦ C. 75◦ D. 45◦

(c) In a triangle ABC, ∠B = ∠C = 45◦, then the triangle is ———

A. right triangle B. acute angled triangle C. obtuse angle

triangle D. equilateral triangle

(d) In an equilateral triangle, each exterior angle is ———

A. 60◦ B. 90◦ C. 120◦ D. 150◦

(e) Sum of the three exterior angles of a triangle is ———

A. two right angles B. three right angles C. one right angle

D. four right angles

3. In a triangle ABC, ∠B = 70◦. Find ∠A + ∠C.

4. In a triangle ABC, ∠A = 110◦ and AB = AC. Find ∠B and ∠C.

5. If three angles of a triangle are in the ratio 2: 3: 5, determine three

angles.

6. The angles of triangle are arranged in ascending order of magnitude.

If the difference between two consecutive angles is 15◦, find the three

angles.

7. The sum of two angles of a triangle is equal to its third angle. Deter-

mine the measure of the third angle.

8. In a triangle ABC, if 2∠A = 3∠B = 6∠C, determine ∠A, ∠B and ∠C.

9. The angles of triangle are x− 40◦ , x− 20◦ and1

2x+15◦. Find the value

of x.

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Triangles 265

10. In triangle ABC, ∠A − ∠B = 15◦ and ∠B − ∠C = 30◦, find ∠A,∠B and

∠C.

11. The sum of two angles of a triangle is 80◦ and their difference is 20◦.Find the angles of the triangle.

12. In a triangle ABC, ∠B = 60◦ and ∠C = 80◦. Suppose the bisector of

∠B and ∠C meet at I. Find ∠BIC.

13. In a triangle, each of the smaller angles is half the largest angle. Find

the angles.

14. In a triangle, each of the bigger angles is twice the third angle. Find

the angles.

15. In a triangle ABC, ∠B = 50◦ and ∠A = 60◦. Suppose BC is extended

to D. Find ∠ACD.

16. In an isosceles triangle, the vertex angle is twice the sum of the base

angles. Find the angles of the triangle.

Glossary

Exterior angle: the angle subtended externally when a side of a triangle

is extended.

Interior opposite angles: angles opposite to the interior angle which is

supplementary to the exterior angle.

Points to remember

• Triangles are classified according to their angles and their sides.

• The sum of the three interior angles of a triangle is 180◦.

• An exterior angle of a triangle is the sum of two interior opposite

angles.

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CHAPTER 3 UNIT 3

CONGRUENCY OF TRIANGLES


• identify the congruent figures;

• identify the congruent triangles;

• identify the corresponding sides and corresponding angles of congru-

ent triangles;

• state the postulates for congruency of triangles;

• understand that particular triples of elements determine the congru-

ency of triangles;

• deduce logical methods for proving theorems;

• solve problems based on different postulates of congruency;

• appreciate the use of congruency of triangles in solving practical day

to day problems.

3.3.1 Introduction

Suppose you have two equilateral triangles, each of side length, say 1

cm. Can you place one on the other so that all the three sides collace?

That is all the three sides of one triangle sit exactly on three sides of the

other triangle. Take equilateral triangles of side lengths 1 cm and 2 cm.

Can you put one on the other so that one exactly fits the other? No matter

how you adjust them, you see that it is impossible to superpose one on

the other. The exact geometrical idea which needed to understand these

is congruency.

Congruency is one of the fundamental concepts in geometry. This con-

cept is used to classify the geometrical figures on the basis of their shapes.

Two geometrical figures are said to be congruent, if they have same shape

and size. For example:

1. Two line segments are congruent

if they have same length.

C

A B

D6 cm

6 cm

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Congruency 267

2. Two angles are congruent if they

have same measure.

P

RQ Y Z

X

28o o28

3. Two circles are congruent if they

have same radii. 2 cm 2 cm

4. Two squares are congruent if they

have sides of same length.

M

N P

Q

V U2.2 cm2.2 cm

TS

Note: 1. Two geometrical congruent figures can be made to superimpose

one on the other so that one exactly covers the other.

2. Two congruent geometrical figures have same parameters.

Congruency of triangles

Two triangles are said to be congruent if all the sides and angles of one

triangle are equal to the corresponding sides and angles of the other

triangle.

CB

A

E F

D In triangles ABC and DEF , you

observe AB = DE, AC = DF and

BC = EF ; ∠A = ∠D, ∠B = ∠E,

and ∠C = ∠F .

Therefore, triangles ABC and

DEF are congruent.

We write this as

∆ABC ∼= ∆DEF.

A few words about the use of this notation. When we write

∆ABC ∼= ∆DEF , the important thing is to observe that the vertices A,B,C

correspond to the verticesD,E, F in that order. If we write∆ABC ∼= ∆EFD,

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268 Unit 3

this gives a different meaning. This means ∠A = ∠E, ∠B = ∠F and

∠C = ∠D; and AB = EF , BC = FD and CA = DE. Can you see the

difference? So while using the notation for congruency, keep the order of

the vertices in proper way.

Recall that there are six elements associated with a triangle. Two tri-

angles are congruent if and only if these six elements match in a suitable

sense. This will be made clear by understanding corresponding sides and

angles.

Corresponding sides and angles

Let us say that, on superposition, triangle ABC covers triangle DEF

exactly in such way that

1. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F ;

2. AB = DE, BC = EF , AC = DF .

Then ∆ABC ∼= ∆DEF . Angles which coincide on superposition are called

corresponding angles. Sides which coincide on superposition are called

corresponding sides. Generally it is not always possible to superimpose

one triangle over other triangle to know which angles are the correspond-

ing angles and which are the corresponding sides. Can you see that if two

triangles are congruent, then there are corresponding vertices as well.

However, vertices are points and do not have any numerical quantities as-

sociated with them. Hence we use only sides and angles for determining

congruency.

Geometrically, in two congruent triangles ABC and DEF , angles oppo-

site to equal sides are corresponding angles and so they are equal. There-

fore, BC = EF implies that angle opposite BC= angle opposite to EF .

Symbolically BC = EF =⇒ ∠A = ∠D. Similarly, AC = DF =⇒ ∠B = ∠E

and AB = DE =⇒ ∠C = ∠F .

Exercise 3.3.1

1. Identify the corresponding sides and corresponding angles in the fol-

lowing congruent triangles:

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Congruency 269

(a)(b)

RA B

PC

Q2. Pair of congruent triangles and incomplete statements related to them

are given below. Observe the figures carefully and fill up the blanks:

(a) In the adjoining figure

if ∠C = ∠F , then AB = − −−−− and BC = −−−−−.

D

F

E

BA

C

(b) In the adjoining figure

if BC = EF , then ∠C = − −−−− and ∠A = −−−−−.

B C E F

DA

(c) In the adjoining figure,

if AC = CE and ∆ABC ∼= ∆DEC,

then ∠D = − − − − − and ∠A =

−−−−−.

E

B

A

D

C

3.3.2 SAS postulate for the congruency of triangles

Least number of conditions for congruency

Now we know that two triangles are congruent if and only if the cor-

responding sides and corresponding angles of two triangles are equal (six

elements), i.e., three for corresponding angles and three for corresponding

sides. Now the natural question is; what is the least number of conditions

required to ensure the congruency of two triangles? Do we need the corre-

sponding equality of all the six elements or a lesser number of conditions

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270 Unit 3

will suffice to ensure the congruency of two triangles? In this and subse-

quent sections, we shall see that if three properly chosen elements out of

six of a triangle are equal to the corresponding there elements of another

triangle, then other three elements of two triangles coincide in order and

we have the congruency of two triangles. Let us discuss those three con-

ditions which will ensure the congruency of two triangles. One has to be

careful in choosing three conditions. For example three angles will not do.

(Draw two non-congruent triangles which have same set of angles.)

Side-angle-side postulate [SAS postulate]

If the two sides and included angle of one triangle are equal to the

corresponding two sides and the included angle of the other triangle,

then the two triangles are congruent.

B EC F

DA In triangles ABC and DEF , you

observe that AB = DE, AC = DF

and ∠A = ∠D. Hence SAS postu-

late tells

∆ABC ∼= ∆DEF.

Q R Y Z

XP

60

60

Look at the following trian-

gles PQR and XY Z. You

observe that PQ = XY and

QR = Y Z. Moreover ∠P =

60◦ = ∠Y . Still the triangles

PQR and XY Z need not becongruent because the included angles ∠Q and ∠Y , which are cor-

responding angles, need not be equal.

Example 1. In the figure O is the midpoint of AB and CD. Prove that

(i) ∆AOC ∼= ∆BOD; (ii) AC = BD.

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Congruency 271

CB

DA

O

Solution: In triangles AOC and BOD, we

haveAO = BO, (O, the midpoint of AB);

∠AOC = ∠BOD, (vertically opposite angles);

CO = OD, (O, the midpoint of CD).So by SAS postulate we have

∆AOC ∼= ∆BOD.

Hence AC = BD, as they are corresponding parts of congruent triangles.

Example 2. In the figure, it is given that AE = AD and BD = CE. Prove

that ∆AEB is congruent ∆ADC.

C B

DE

A

Solution: We have AE = AD

and CE = BD. Adding, we

get AE + CE = AD + BD

=⇒ AC = AB. In trian-

gles AEB and ADC, we have

AE = AD, (given);

AB = AC, (proved);

∠EAB = ∠DAC, (common angle).

By SAS postulate ∆AEB ∼= ∆ADC.

Example 3. In a quadrilateral ACBD, AC = AD and AB bisect ∠A . Show

that ∆ABC is congruent to ∆ABD.

A B

D

C Solution: In triangles ABC and

ABD, we haveAC = AD, (given);

∠CAB = ∠DAB, (AB bisects ∠A);

AB = AB, (common side).Hence

∆ABC ∼= ∆ABD.(Can you see that BA bisects ∠CBD?)

Exercise 3.3.2

1. In the adjoining figure PQRS is a

rectangle. Identify the congruent tri-

angles formed by the diagonals.

S R

QP

O

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272 Unit 3

2. In the figure ABCD is a square,

M,N,O and P are the midpoints of

sides AB, BC, CD and DA respec-

tively. Identify the congruent trian-

gles.

3. In a triangle ABC, AB = AC. Points E on AB and D on AC are such

that AE = AD. Prove that triangles BCD and CBE are congruent.

4. In the adjoining figure, the sides BA

and CA have been produced such

that BA = AD and CA = AE. Prove

that DE ‖ BC. [ hint:-use the con-

cept of alternate angles.]CB

DE

A

3.3.3 Consequences of SAS postulate

Now you have learnt how to compare two triangles using SAS condition.

This comparison will lead to some very interesting consequences about the

properties of triangles. We study a few of them here.

Theorem 1. In a triangle, the angles opposite to equal sides are equal.

B CD

AGiven: a triangle ABC in which

AB = AC.

To prove: ∠C = ∠B.

Construction: Draw the angle bi-

sector of ∠A. Let it cut BC at D.

Let us compare triangles ABD and

ACD:Proof:

Statement Reasons

AB = AC; given;

AD = AD; common side;

∠BAD = ∠CAD; by construction.

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Congruency 273

We can use SAS postulate to conclude that ∆ADB ∼= ∆ADC. Hence

∠ABC = ∠ACB, since these are corresponding angles of congruent tri-

angles. Thus the theorem is proved.

Corollary 1: In an isosceles triangle, the angle bisector of the apex angle

is the perpendicular bisector of the base.

Given: ABC is a triangle in which AB = AC and apex angle ∠A.

Construction: draw the angle bisector AD from A on BC. (Refer to the

figure of the previous theorem.)

To prove: AD is the perpendicular bisector of BC. Equivalently, we have

to show that BD = DC and ∠ADB = ∠ADC = 90◦.

Proof:

∆ADB ∼= ∆ADC, by (theorem 1)

=⇒ DB = DC and ∠ADB = ∠ADC.

But ∠ADB + ∠ADC = 180◦ (linear pair)

=⇒ ∠ADB + ∠ADC = 180◦

=⇒ 2∠ADB = 180◦

=⇒ ∠ADB = 180◦/2 = 90◦.

=⇒ ∠ADC = 180◦ − ∠ADB = 90◦.

We also observe that ∆ABD ∼= ∆ACD by SAS postulate. Hence

BD = CD.


Now you may wonder whether the converse of the theorem is true: in

any triangle, the sides opposite to the equal angles are equal. It is also

true, but its proof needs a different condition of congruency, which you

will study later.

Example 4. In the figure AB = AC and DB = DC. Prove that

∠ABD = ∠ACD.

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274 Unit 3

B C

A

D

Solution: In ∆ABC, we have

AB = AC =⇒ ∠ABC = ∠ACB

(angles opposite to equal sides).

Again in ∆DBC, we have

DB = DC (given) =⇒ ∠DBC = ∠DCB (angles opposite to equal sides).

Hence we obtain

∠ABC − ∠DBC = ∠ACB − ∠DCB.

This gives

∠ABD = ∠ACD.

Exercise 3.3.3

1. In a ∆ABC, AB = AC and ∠A = 50◦. Find ∠B and ∠C.

2. In ∆ABC, AB = BC and ∠B = 64◦. Find ∠C.

3. In each of the following figure, find the value of x:

B C D

A

x

AB=ACB C

A

D

AC=CD

x

40

65

30

55 75

x

AB=AC

50

x

CCB

A

A

B

BD=DC=AD

D

4. Suppose ABC is an equilateral triangle. Its base BC is produced to D

such that BC = CD. Calculate (i) ∠ACD and (ii) ∠ADC.

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Congruency 275

5. Show that the perpendiculars drawn from the vertices of the base of

an isosceles triangle to the opposite sides are equal.

6. Prove that a ∆ABC is an isosceles triangle if the altitude AD from A

on BC bisects BC.

7. Suppose a triangle is equilateral. Prove that it is equiangular.

3.3.4 ASA postulate for congruency

If two angles and the common side of one triangle are equal to the

corresponding two angles and the common side of the other triangle,

then the two triangles are congruent.

B EC F

DA Given two triangles ABC

and DEF such that

∠B = ∠E, ∠C = ∠F and

BC = EF , the ASA postu-

lates tells

∆ABC ∼= ∆DEF .

Earlier you have seen that the angles opposite to equal sides of a trian-

gle are equal. Now we are ready to prove the converse of this result.

Theorem 2. If in a triangle two angles are equal, then the sides oppo-

site to them are equal. (Converse of Theorem 1.)

B CD

A

Given: triangle ABC in which

∠B = ∠C.

To prove: AC = AB.

Construction: draw AD ⊥ BC.

Proof: Then ∠ADB = ∠ADC = 90◦. We are given ∠DBA = ∠DCA. Consider

triangles ADB and ADC. We have

∠ADB + ∠DBA+ ∠BAD = 180◦ = ∠ADC + ∠DCA+ ∠CAD.

It follows that ∠BAD = ∠CAD(why?). Consider triangles ADB and ADC.

We have

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276 Unit 3

∠BAD = ∠CAD (proved);

∠ADB = ∠ADC (both are right angles);

AD = AD (common side.)

=⇒ ∆ADB ∼= ∆ADC, by ASA condition. We conclude that AB = AC, by

property of congruency. This completes the proof of the converse of theo-

rem 1.

Example 5. In a triangle ABC, AB = AC and the bisectors of angles B

and C intersect at O. Prove that BO = CO and AO is the bisector of angle

∠BAC.

B C

A

O

Solution: Since the angles oppo-

site to equal sides are equal,

AB = AC

=⇒ ∠C = ∠B

=⇒ ∠B

2=

∠C

2.

Since BO and CO are bisectors of ∠B and ∠C, we also have

∠ABO = ∠B/2 and ∠ACO = ∠C/2.

Hence

∠ABO =∠B

2=

∠C

2= ∠ACO.

Consider ∆BCO:

∠OBC = ∠OCB (Why?)

=⇒ BO = CO, (Theorem 2).

Finally, consider triangles ABO and ACO.

BA = CA (given);

BO = CO (proved);

∠ABO = ∠ACO (proved).

Hence by SAS postulate

∆ABO ∼= ∆ACO

=⇒ ∠BAO = ∠CAO =⇒ AO bisects ∠A.

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Congruency 277

Example 6. Diagonal AC of a quadrilateral ABCD bisects the angles ∠A

and ∠C. Prove that AB = AD and CB = CD.

A

D

C

B Solution: Since diagonal AC

bisects the angles ∠A and ∠C,

we have ∠BAC = ∠DAC and

∠BCA = ∠DCA. In triangles ABC

and ADC, we have∠BAC = ∠DAC (given);

∠BCA = ∠DCA (given);

AC = AC (common side).

So, by ASA postulate, we have

∆BAC ∼= ∆DAC

=⇒ BA = AD and CB = CD (Corresponding parts of congruent triangle).

Example 7. Two parallel lines l and m are intersected by another pair of

parallel lines p and q as in the figure. Show that triangles ABC and CDA

are congruent.

34

21

A D

B Cm

l

qp

Solution: Since l and m are par-

allel lines and AC is a transver-

sal, ∠1 = ∠4. Similarly transversal

AC cuts parallel lines p and q, so

that ∠2 = ∠3. In triangles ABC

and CDA, we have∠1 = ∠4 (proved);

∠2 = ∠3 (proved);

AC = AC (common side).By ASA postulate,

∆ABC ∼= ∆CDA.

Example 8. In the figure, ∠BCD = ∠ADC and ∠ACB = ∠BDA. Prove that

AD = BC and ∠A = ∠B.

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278 Unit 3

D

BA

C

31 2

4

Solution: We have ∠1 = ∠2 and

∠3 = ∠4

=⇒ ∠1 + ∠3 = ∠2 + ∠4

=⇒ ∠ACD = ∠BDC.

Thus in triangles ACD and BDC,

we have,∠ADC = ∠BCD (given);

CD = CD (common);

∠ACD = ∠BDC (proved.)

By ASA condition ∆ACD ∼= ∆BDC. Therefore

AD = BC and ∠A = ∠B.

Think it over!

We have taken ASA condition as a postulate. But actually, it

can be proved as a theorem based on SAS postulate. Try to con-

struct a proof. In deductive geometry, one takes only a mini-

mum number of postulates and try to construct as many results

as possible using these minimum number of postulates. One

can also take ASA as a postulate and obtain SAS condition as a

theorem.

Exercise 3.3.4

1. In the given figure, If AB ‖ DC and P

is the midpoint of BD, prove that P

is also the midpoint of AC.

D C

BA

P

2. In the adjacent figure, CD and BE

are altitudes of an isosceles triangle

ABC with AC = AB. Prove that AE =

AD.

B C

ED

A

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Congruency 279

3. In figure, AP and BQ are perpen-

diculars to the line segment AB and

AP = BQ. Prove that O is the mid-

point of line segment AB as well as

PQ.

P

A

B

Q

O

4. Suppose ABC is an isosceles triangle with AB = AC; BD and CE are

bisectors of ∠B and ∠C. Prove that BD = CE.

5. Suppose ABC is an equiangular triangle. Prove that it is equilat-

eral. (You have seen earlier that an equilateral triangle is equiangular.

Thus for triangles equiangularity is equivalent to equilaterality.)

3.3.5 SSS postulate for congruency

We will study one more condition for congruency of two triangles. It

depends only on the sides.

If three sides of one triangle are equal to the three corresponding

sides of the other triangle, then the triangles are congruent.

Activity 1:

Take three sheets of papers; one in the shape of a square, another rect-

angular and a third-one in the shape of a parallelogram (for this you may

have to draw a parallelogram on a sheet of paper and cut it along the

boundary). Draw diagonals as shown in the figure. Cut the sheets along

the diagonals.

A B

CD1

S R

QP

2Z Y

XW

3

You will get two triangles from each sheet. Now you place one of the

triangular sheet obtained from each figure on the other triangular sheet

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280 Unit 3

of the same figure such that it exactly covers the other triangular sheet.

You will notice that each pair of triangles are congruent and in each case

three sides of one triangle are equal to the corresponding sides of the other

triangle.

Now we look for the converse of this. If three sides of one triangle are

equal to the three corresponding sides of another triangle, can we put one

on the other such that each covers the other exactly? SSS congruency

condition says that it is indeed the case.

Think it over!

Since there is SSS congruence postulate, can we have AAA

postulate and SSA postulate?

Example 9. In the figure, it is given that AB = CD and AD = BC. Prove

that triangles ADC and CBA are congruent.

B

C

D

A

Solution: In triangles ADC and

CBA, we have

AB = CD (given);

AD = BC (given);

AC = AC (common side.)By SSS congruency condition, ∆ADC ∼= ∆CBA.

Example 10. In the figure AD = BC and BD = CA. Prove that

∠ADB = ∠BCA and ∠DAB = ∠CBA.BA

CD

Solution: In triangles ABD and

BAC, we have

AD = BC given;

AB = AB (common);

AC = BD (given.)We can use SSS condition to conclude that ∆ABD ∼= ∆BAC. From this we

conclude that

∠ADB = ∠BCA and ∠DAB = ∠CBA.

Example 11. In the adjoining figure, AB = CD and AD = BC. Show that

∠1 = ∠2.

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Congruency 281

1

2

D C

BA

Solution: In triangles ABD and

CDB, we haveAB = CD (given);

AD = BC (given);

BD = DB (common side.)Hence ∆ABD ∼= ∆CDB, by SSS

postulate. Comparing the angles,

we get

∠1 = ∠2.

(Later you will see that, under the given conditions, ABCD is a parallelo-

gram so that AD ‖ BC, and ∠1 and ∠2 are alternate angles formed by the

transversal BD.)

Think it over!

We have taken SSS condition as a postulate. But this is

also a consequence of SAS postulate. You can get SSS as a

theorem from SAS postulate.

Exercise 3.3.5

1. In a triangle ABC, AC = AB and the altitude AD bisects BC. Prove

that ∆ADC ∼= ∆ADB.

2. In a square PQRS, diagonals bisect each other at O. Prove that

∆POQ ∼= ∆QOR ∼= ∆ROS ∼= ∆SOP .

3. In the figure, two sides AB, BC and

the median AD of ∆ABC are respec-

tively equal to two sides PQ, QR and

median PS of ∆PQR. Prove that

(i) ∆ADB ∼= ∆PSQ;

(ii) ∆ADC ∼= ∆PSR.

Does it follow that triangles ABC and

PQR are congruent?

P

Q S R

DB C

A

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282 Unit 3

4. In △PQR, PQ = QR; L,M and N

are the midpoints of the sides of PQ,

QR and RP respectively. Prove that

LN = MN .

P N R

M

Q

L

3.3.6 RHS theorem

CDB

A

Activity 2:

Draw an equilateral triangle ABC

on a sheet of paper. From the ver-

tex A, draw the perpendicular AD

to the base the BC. Cut the sheet

of paper along the triangle. Now

fold it along the perpendicular line. You will notice that two right angled

triangles superpose one on the other. So the two triangles are congruent.

D C

BAActivity 3:

Take a square sheet of paper. Fold

the sheet of the paper along one of

its diagonal. You will notice that

two triangles so formed by fold-

ing the sheet are right angled tri-

angles and they superpose one on

the other.

D C

BA Activity 4:

Take a rectangular sheet of paper,

such that one of its side length is

equal to the length of the square

you had taken earlier and the

other is different from the lengthof the square . Draw one of its diagonals. Cut the sheet along the diagonal,

to obtain two right triangles. Can you see that here also you can superpose

one right triangle on the other?

Now place one of the triangular sheets from the square and another

from the rectangular sheet. You will notice that even though the two tri-

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Congruency 283

angles are right angled and one set of corresponding sides are equal, they

do not superpose one on the other.

In SAS postulate, You have noticed earlier that you need the equality of

two corresponding sides and the included angle for congruency. However

for a right angled triangle, you can relax this a bit to get what is known as

RHS condition. We have the following theorem on right angled triangles.

Theorem 3. Two right-angled triangles are congruent if the hypotenuse

and a side of one triangle are equal to the hypotenuse and the corre-

sponding side of the other triangle. (RHS theorem.)

B C G E F

DA

Given: two right-angled triangles ABC and DEF such that

(i) ∠B = ∠E = 90◦;(ii) Hypotenuse AC = Hypotenuse DF ; and

(iii) AB = DE.

To prove: ∆ABC ∼= ∆DEF .

Construction: Produce FE to G so that EG = BC. Join DG.

Proof: In triangles ABC and DEG, observe that

AB = DE (given);

BC = EG (by construction);

∠ABC = ∠DEG (each equal to 90◦.)

Hence by SAS, ∆ABC ∼= ∆DEG =⇒ ∠ACB = ∠DGE and AC = DG. But

AC = DF , by the given hypothesis. We thus get

DG = AC = DF .

In triangle DGF , we have got DG = DF (just proved). This implies that

∠G = ∠F (angles opposite to equal sides are equal).

In triangles DEF and DEG,

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284 Unit 3

∠G = ∠F (proved);

∠DEG = ∠DEF (both equal to 90◦.)Hence

∠GDE = 180◦ − (∠G+ ∠DEG) = 180◦ − (∠F + ∠DEF ) = ∠FDE.

Consider triangles DEG and DEF . We have

DG = DF (proved);

DE = DE (common);

∠GDE = ∠FDE (proved).

Hence by SAS condition,

∆DEG ∼= ∆DEF .

But we have already proved that

∆ABC ∼= ∆DEG.

It follows that

∆ABC ∼= ∆DEF .

Note: We have used an important result: If there are three triangles ABC,

DEF and JKL such that ABC is congruent to DEF and DEF is congruent

to JKL, then ABC is congruent to JKL. This is precisely Axiom 3 in the

unit 1 of chapter 3.

Think it over!

If two sides of a right triangle are respectively equal to the corre-

sponding sides of another right triangle, can you conclude that

the two triangles are congruent? (We are not demanding that

hypotenuse of two triangles be equal.)

Example 12. Suppose ABC is an isosceles triangle such that AB = AC

and AD is the altitude from A on BC. Prove that (i) AD bisects ∠A, (ii) AD

bisects BC.

CDB

A Solution: We have to show that

∠BAD = ∠CAD and BD = DC.

In right triangles ADB and ADC, we

haveAB = AC (given);

AD = AD (common side).

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Congruency 285

So by RHS congruency of triangles, we have ∆ABD ∼= ∆ACD. Hence

∠BAD = ∠CAD and BD = DC.

Example 13. Suppose the altitudes AD, BE and CF of triangle ABC are

equal. Prove that ABC is an equilateral triangle.

F

A

E

CDB

Solution: In right triangles BCE

and CBF , we have,BC = BC (common hypotenuse);

BE = CF (given).Hence BCE and CBF are congruent,

by RHS theorem. Comparing the tri-

angles, we get ∠B = ∠C.

This implies that

AC = AB (sides opposite to equal angles).

Similarly,

AD = BE =⇒ ∠B = ∠A

=⇒ AC = BC.

Together, we get AB = BC = AC or ABC is equilateral.

Exercise 3.3.6

1. Suppose ABCD is rectangle. Using RHS theorem, prove that triangles

ABC and ADC are congruent.

2. Suppose ABC is a triangle and D is the midpoint of BC. Assume that

the perpendiculars from D to AB and AC are of equal length. Prove

that ABC is isosceles.

3. Suppose ABC is a triangle in which BE and CF are respectively the

perpendiculars to the sides AC and AB. If BE = CF , prove that

triangle ABC is isosceles.

3.3.7 Some consequences

You have seen earlier that in a triangle the angles opposite to equal

sides are equal. And conversely, sides opposite to equal angles are equal.

So the natural question is: if angles are of different measures, can we

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286 Unit 3

compare the sides opposite to them? Can we say some thing about angles

if sides are of different lengths?

Proposition 1. Suppose two sides of a triangle are not equal. Then

the angle opposite to a larger side is greater than the angle opposite

to the smaller side.

B C

D

A

Given: a triangle ABC in which

AC > AB.

To prove: ∠B > ∠C.

Construction: take a point D on

AC such that AB = AD. (This is

possible since AC > AB.)

Join BD.

Proof: In triangle ABD, we have

AB = AD(by construction) =⇒ ∠ABD = ∠ADB (angles opposite to equal sides).

Now ∠BDC is an exterior angle for triangle BCD. Hence it is larger than

interior opposite angle ∠BCD. We thus get

∠C < ∠BDA = ∠ABD < ∠ABC = ∠B.


Proposition 2. In a triangle, if two angles are unequal, then the side

opposite to the larger angle is greater than the side opposite to the

smaller angle.

B C

A

Given: a triangle ABC in which

∠B > ∠C.

To prove: AC > AB.

Proof: Observe that

∠B > ∠C =⇒ AC 6= AB.

For, AC = AB implies that ∠B =

∠C(angles opposite to equal sides

are equal).Thus either

AC < AB or AC > AB.

If AC < AB, then by previous proposition ∠B < ∠C; but this contradicts

the given hypothesis. The only possibility left out is

AC > AB.

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Congruency 287

Note: Here we are using law of trichotomy. Given any two real numbers a

and b, you have only one of the possibilities: a < b; a = b; or a > b.

Proposition 3. In a triangle, the sum of any two sides is greater than

the third side.

C

A

D

B

Given: a triangle ABC.

To prove: AB + AC > BC.

Construction: Extend BA to D such that

AD = AC and join DC.

Proof: Then

BD = BA + AD = BA+ AC.

Since AD = AC, we have

∠ADC = ∠ACD (angle opposite to equal sides).Hence we obtain

∠BCD > ∠ACD = ∠ADC = ∠BDC.

In triangle BCD, we have

∠BCD > ∠BDC =⇒ BD > BC(by proposition 2).

But BD = BA+ AC, as we have observed earlier. We thus get

BA+ AC > BC.

You can similarly prove CA < AB +BC and AB < BC + CA.

Note: The inequalities BC < CA+AB, CA < AB+BC and AB < BC+CA are

called triangle inequalities. They are necessary conditions for the existence

of a triangle with sides AB, BC and CA. This tells that the straight line

is the shortest path between any two point. Given three numbers a, b, c,

necessary conditions for the existence of a triangle with sides a, b, c are that

a < b+ c, b < c + a and c < a + b. These conditions are also sufficient which

you will see while constructing a triangle with three given sides.

Example 14. Show that in a right angled triangle, hypotenuse is larger

than any of the remaining sides.

Solution: Suppose ABC is a right angled triangle with ∠B = 90◦. Then

AC is the hypotenuse. Observe that ∠BAC < ∠B and ∠BCA < ∠B. Now

the side opposite to ∠BAC is BC and side opposite ∠BCA is AB. Hence by

proposition 2, BC < AC and AB < AC.

Example 15. In the adjoining figure, ∠B = 70◦, ∠C = 50◦ and AD is the

bisector of ∠A. Prove that AB > AD > CD.

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288 Unit 3

B D C

A

70 50

Solution: Observe that

∠A = 180◦ − (∠B + ∠C)

= 180◦ − (70◦ + 50◦) = 60◦.

Hence ∠BAD = ∠DAC = 30◦.

Consider triangle BAD. We can compute ∠ADB:

∠ADB = 180◦ − (70◦ + 30◦) = 80◦ > ∠ABD

Hence AB > AD, by proposition 2. In triangle ADC, we have

∠DAC = 30◦ < 50◦ = ∠ACD.

Again proposition 2 gives CD < AD.

Exercise 3.3.7

1. In a triangle ABC, ∠B = 28◦ and ∠C = 56◦. Find the largest and the

smallest sides.

2. In a triangle ABC, we have AB = 4cm, BC = 5.6cm and CA = 7.6cm.

Write the angles of the triangle in ascending order of measures.

3. Let ABC be a triangle such that ∠B = 70◦ and ∠C = 40◦. Suppose D

is a point on BC such that AB = AD. Prove that AB > CD.

4. Let ABCD be a quadrilateral in which AD is the largest side and BC

is the smallest side. Prove that ∠A < ∠C. (Hint: Join AC)

5. Let ABC be a triangle and P be an interior point. Prove that AB +

BC + CA < 2(PA+ PB + PC).

Additinal problems on “Congruency of triangles”

1. Fill in the blanks to make the statements true .

(a) In right triangle the hypotenuse is the ————— side.

(b) The sum of three altitudes of a triangle is ————— than its

perimeter.

(c) The sum of any two sides of a triangle is —————- than the

third side.

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Congruency 289

(d) If two angles of a triangle are unequal , then the smaller angle

has the ————– side opposite to it.

(e) Difference of any two sides of a triangle is —————- than the

third side.

(f) If two sides of a triangle are unequal, then the larger side has

—————- angle opposite to it.

2. Justify the following statements with reasons:

(a) The sum of three sides of a triangle is more than the sum of its

altitudes.

(b) The sum of any two sides of a triangle is greater than twice the

median drawn to the third side.

(c) Difference of any two sides of triangle is less than the third side.

3. Two triangles ABC and DBC have common base BC. Suppose AB =

DC and ∠ABC = ∠BCD. Prove that AC = BD.

4. Let AB and CD be two line segments such that AD and BC intersect

at O. Suppose AO = OC and BO = OD. Prove that AB = CD.

5. Let ABC be a triangle. Draw a triangle BDC externally on BC such

that AB = BD and AC = CD. Prove that ∆ABC ∼= ∆DBC.

6. Let ABCD be a square and let points P on AB and Q on DC be such

that DP = AQ. Prove that BP = CQ.

7. In a triangle ABC, AB = AC. Suppose P is point on AB and Q is a

point on AC such that AP = AQ. Prove that ∆APC ∼= ∆AQB.

8. In an isosceles triangle, if the vertex angle is twice the sum of the base

angles, calculate the angles of the triangle.

9. If the bisector of the vertical angle of a triangle bisects the base, show

that the triangle is isosceles.

10. Suppose ABC is an isosceles triangle with AB = AC. Side BA is

produced to D such that BA = AD. Prove that ∠BCD is a right angle.

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11. Let AB, CD be two line segments such that AB ‖ CD and AD ‖ BC.

Let E be the midpoint of BC and let DE extended meet AB in F . Prove

that AB = BF .

Glossary

Congruency: two geometrical figures are identical in both shape and size.

Superpose: one figure sitting exactly on the other.

Corresponding sides: while comparing two triangles, we index the sides

of the triangles in an ordered way.

Corresponding angles: indexing of the angles in an ordered way.

SAS postulate: Side-Angle-Side postulate.

ASA postulate: Angle-Side-Angle postulate.

SSS postulate: Side-Side-Side postulate.

RHS theorem: Right angle-Hypotenuse-Side theorem.

Triangle inequality: any side of a triangle is smaller than the sum of the

remaining two.

Points to remember

• Two triangles are congruent if we can superpose one on the other.

• Two triangles are congruent if two sides and the included angle of

one triangle are respectively equal to the corresponding sides and the

corresponding angle of the other.(SAS)

• Two triangles are congruent if two angles and the common side to

these angles of one triangle are respectively equal to the correspond-

ing angles and the corresponding side of the other.(ASA)

• Two triangles are congruent if three sides of one triangle are respec-

tively equal to three corresponding sides of the other. (SSS)

• Two right angled triangles are congruent if they have equal hypotenuse

and, apart from hypotenuse, a side of one triangle is equal to a side

of of the other.(RHS)

• Any side of a triangle is smaller than the sum of the other two.(Triangle

inequality)

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CHAPTER 3 UNIT 4Note: The sections 3.4.5, 3.4.6, 3.4.12 and 3.4.13 are optional in

this unit. This need not be used for examination.

CONSTRUCTION OF TRIANGLES

After studying this unit, you learn:

• to construct a triangle:

– when three sides are given;– when two sides and included angle are given;– when two angles and included side are given;– when two sides and altitude on third side are given;– one side and hypotenuse of a right angled triangle are given;– an isosceles triangle, whose base and height are given;– perimeter and ratio of the sides of a right triangle are given;– whose perimeter and base angles are given;– when the base, sum of the other two sides and one base angle of

a right triangle are given;– when the base, difference of the other two sides and one base

angle of a right triangle are given.

3.4.1 Introduction

In earlier classes, you have learnt that given a triangle, there are six

elements associated with it, namely, three sides and three angles. You may

be wondering whether you need all these to be known for constructing a

triangle. If all are known, it is well and good. But in a variety of practical

situations, you may not know all these. If only two are known, you cannot

hope to construct a triangle. Even if three of these are known, you may not

be able to construct. For example, If two sides and an angle ( not included

angle ) are given, then it is not possible to construct such triangle.

We will take up different situations when we will be able to construct

a triangle. Of course along with these basic six elements, you can also

associate many other elements like medians, angle bisectors altitudes. You

get several other combinations. And constructing triangles with minimum

number of data which includes these additional elements is interesting

and challenging. We will not go beyond the traditional constructions.

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3.4.2 When three sides are given

Example 1 Construct a triangle ABC in which AB = 5 cm, BC= 4.3 cm

and AC= 4 cm.

Solution: We follow several steps in the construction:

1. Draw a line segment which is sufficiently long using ruler.

2. Locate points A and B on it such that AB = 5 cm.

3. With A as centre and radius 4 cm, draw an arc(see figure).

4. With B as centre and radius 4.3 cm, draw another arc cutting the

previous arc at C.

5. Join AC and BC.

Then ABC is the required triangle.

A B

C

5 cm

4 cm4.3 cm

Think it over!

The arc with B as centre and radius 4.3 cm always cuts the arc

with A as center and radius 4 cm, whenever AB=5 cm. Can you

relate this to triangle inequality?

Exercise 3.4.2

1. Construct a triangle ABC in which AB=5 cm and BC= 4.6 cm and

AC= 3.7 cm.

2. Construct an equilateral triangle of side 4.8 cm.

3. Construct a triangle PQR, given that PQ =5.6 cm, PR=7 cm and

QR=4.5 cm.

4. Construct a triangle XY Z in which XY =7.8 cm, Y Z=4.5 cm and

XZ=9.5 cm.

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Constructions 293

5. Construct a triangle whose perimeter is 12 cm and the ratio of their

sides is 3:4:5.

3.4.3 When two sides and their included angle are given

Example 2. Construct a triangle PQR, given that PQ=4 cm, QR=5.2 cm

and ∠Q = 60◦.Solution: Steps of construction:


2. Locate points Q and R on it such that QR = 5.2 cm.

3. At Q, construct a line segmentQM , sufficiently large, such that ∠MQR =

60◦; use protractor to measure 60◦.4. With Q as center and radius 4 cm., draw an arc cutting QM at P ; join

PR.

Then, PQR is required the triangle.

Q R

M

P

60

5.2 cm

4 cm

Think it over!

Without using protractor, can you construct a line segment QM

at Q such that ∠MQR = 60◦?

Exercise 3.4.3

1. Construct a triangle ABC, in which AB=4.5 cm, AC=5.5 cm and

∠BAC = 75◦.2. Construct a triangle PQR in which PQ=5.4 cm, QR=5.5 cm and ∠PQR =

55◦.3. Construct a triangleXY Z in whichXY =5 cm, Y Z=5.5 cm and ∠XY Z =

100◦.

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294 Unit 4

4. Construct a triangle LMN in which LM=7.8 cm, MN=6.3 cm and

∠LMN = 45◦.

3.4.4 When two angles and included side are given

Example 3. Construct a triangleXY Z in whichXY = 4.5 cm and ∠X = 100◦

and ∠Y = 50◦.Solution: Steps of construction


2. Locate points X and Y on it such that XY = 4.5 cm.

3. Construct a line segment XP such that ∠PXY = 100◦; construct a

line segment Y Q such that ∠XYQ = 50◦.

4. Extend XP and Y Q to intersect at Z.

Then XY Z is the required triangle.

X Y

Z

QP

100 50

4.5 cm

Think it over!

Suppose it is given that XY = 4.5 cm, ∠X = 100◦ and ∠Y = 80◦.Can you construct a triangle now? Where does the construction

break down and why?

Exercise 3.4.4

1. Construct a triangle ABC in which AB=6.5 cm, ∠A = 45◦ and ∠B =

60◦.

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Constructions 295

2. Construct a triangle PQR in which QR=4.8 cm, ∠Q = 45◦ and ∠R =

55◦.

3. Construct a triangle ABC in which BC=5.2 cm, ∠B = 35◦ and ∠C =

80◦.

4. Construct a triangle ABC in which BC=6 cm, ∠B = 30◦ and ∠C = 125◦.

3.4.5 To construct an equilateral triangle of given height(Optional)

Example 4. Construct an equilateral triangle of height 3.2 cm.

Solution: First observe that the altitudes from any vertex to the opposite

sides of an equilateral triangle are all of equal length (prove this state-

ment). Hence we can define the height of an equilateral triangle as this

common value of three altitudes.

Steps of construction

1. Draw any line segment XY .

2. Take any point M on XY . Draw ZM ⊥ XY .

3. With M as center and radius 3.2 cm, draw an arc, cutting MZ at A.

4. Construct ∠MAB = 30◦ and ∠MAC = 30◦, with B and C on XY .


Z

A

MB C

3.2

cm

30 30

X Y

Think it over!

Why is that ABC so constructed is equilateral? Can you think

of a proof?

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296 Unit 4

Exercise 3.4.5

1. Construct an equilateral triangle of height 4.5 cm. Measure approxi-

mate length of the its side.

2. Construct an equilateral triangle of height 5.2 cm. Measure approxi-


3. Construct an equilateral triangle of height 6. cm. Measure approxi-


3.4.6 When two sides and altitude on the third side aregiven (Optional)

Example 5. Construct a triangle ABC in which AB =4.5 cm, AC= 5 cm

and length of perpendicular from A on BC is 3.6 cm.

Solution: Steps of construction

1. Draw a line segment XY .

2. Take a point M on XY .

3. Draw ZM ⊥ XY , with MZ sufficiently large.

4. With M as center and radius 3.6 cm, draw an arc, cutting MZ at A.

5. With A as center and

radii 4.5 cm and 5 cm,

draw arcs cutting XY

at B and C respectively;

join AB and AC.

Then ABC is the re-

quired triangle.

Z

M

A

3.6

cm

4.5 cm

5 c

m

X YBC

Think it over!

(1) Why is that the arcs with centre A and radii 4.4 cm and 5 cm

cut the line segment XY ? Which part of the data ensure it?

(2) There are two more triangles possible other than the one

given. Construct them.

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Constructions 297

Exercise 3.4.6

1. Construct a triangle PQR in which PQ =5.5 cm, PR=6.2 cm and

length of the perpendicular from P on QR is 4 cm.

2. Construct a triangle MNP in which MN =4.5 cm, MP= 5.2 cm and

length of perpendicular from M on NP is 3.8 cm.

3.4.7 To construct a right triangle whose one side andhypotenuse are given

Example 6. Construct a right triangle LMN in which ∠M = 90◦, MN = 4

cm and LN=6.2 cm.

4 cmM N

L

6.2 cmP

X Y

Solution:


1. Draw a line segment

XY .

2. Locate M,N on XY such

that MN =4 cm.

3. Construct a line

segment MP , suffi-

ciently large, such that

∠NMP = 90◦.

4. With N as centre and ra-

dius 6.2 cm draw an arc,

cutting MP at L; join

NL.Then, LMN is the required triangle.

Think it over!

Why is that the arc with centre N and radius 6.2 cm cuts the

line segment MP? Which part of the data ensures it?

Exercise 3.4.7

1. Construct a right angle triangle ABC in which ∠B = 90◦, AB = 5 cm

and AC=7 cm.

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298 Unit 4

2. Construct a right angle triangle PQR in which ∠R = 90◦, PQ = 4 cm

and QR=3 cm.

3. Construct a right angle triangle ABC in which ∠B = 90◦, BC = 4 cm

and AC=5 cm.

3.4.8 To construct an isosceles triangle whose base andcorresponding altitude are given

Example 7. Construct an isosceles triangle ABC in which base BC=5.8

cm and altitude from A on BC is 4.8 cm.

P

B CM

A

5.8 cm

4.8

cm

X Y

Solution:


1. Draw a line segment BC

whose length is 5.8 cm.

2. Draw the perpendicular

bisector of BC; call it MP ,

with M on BC.

3. With M as centre and ra-

dius 4.8 cm, draw an arc

cutting MP at A; join AB

and AC.

Then ABC is the required tri-

angle.

Think it over!

Which results on triangle is used to conclude that ABC is the

required triangle?

Exercise 3.4.8

1. Construct an isosceles triangle ABC in which base BC=6.5 cm and

altitude from A on BC is 4 cm.

2. Construct an isosceles triangle XY Z in which base Y Z=5.8 cm and

altitude from X on Y Z is 3.8 cm.

3. Construct an isosceles triangle PQR in which base PQ=7.2 cm and

altitude from R on PQ is 5 cm.

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Constructions 299

3.4.9 To construct an isosceles triangle when its altitudeand vertex angle are given

Example 8. Construct an isosceles triangle whose altitude is 4 cm and

vertex angle is 80◦.

P

40 40

4

cm

B M C

A

X Y

Solution:


1. Draw a line segment XY .

2. Take a point M on XY and

draw a line MP ⊥ XY .

3. With M as centre and ra-

dius 4 cm, draw an arc cutting

MP at A.4. Construct B and C on XY such that ∠MAB = 80◦/2 = 40◦ and ∠MAC =

80◦/2 = 40◦.Then ABC is the required triangle.

Think it over!

Why is triangle ABC isosceles? Which results on triangles are

used to conclude that ABC is the required triangle?

Exercise 3.4.9

1. Construct an isosceles triangle whose altitude is 4.5 cm and vertex

angle is 70◦.2. Construct an isosceles triangle whose altitude is 6.6 cm and vertex

angle is 60◦.3. Construct an isosceles triangle whose altitude is 5 cm and vertex an-

gle is 90◦.

3.4.10 To construct a triangle whose perimeter and ratioof the sides are given

Example 9. Construct a triangle ABC, whose perimeter is 12 cm and

whose sides are in the ratio 2:3:4.


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300 Unit 4

1. Draw a line segment and locate points X, Y such that XY = 12 cm.

2. Draw a ray XZ, making an acute angle with XY and drawn in the

down-ward direction.

3. From X, locate (2+3+4)= 9 points at equal distances along XZ.

4. Mark point L,M,N on XZ such that XL=2 parts, LM=3 parts and

MN= 4 parts.

5. Join NY . Through L and M , draw LB ‖ NY and MC ‖ NY , intersect-

ing XY in B and C respectively.

6. With B as center and BX as radius, draw an arc; with C as a centre

and CY as radius, draw an arc cutting the previous arc at A.

7. Join AB and AC.

A

XC

Y

N

L

12 cmB

M

Z


Think it over!

Why is that the sides of ABC are in the required ratio? Is it pos-

sible to construct a triangle if the sides are in the ratio 2:3:5?

Exercise 3.4.10

1. Construct a triangle ABC, whose perimeter is 13 cm and whose sides

are in the ratio 3:4:5.

2. Construct a triangle PQR, whose perimeter is 14 cm and whose sides


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Constructions 301

3. Construct a triangle MNP , whose perimeter is 15 cm and whose sides


3.4.11 To construct a triangle whose perimeter and baseangles are given

Example 10. Construct a triangle ABC whose perimeter is 12.5 cm and

whose base angles are 60◦ and 75◦.

B C

RS

QP

60 75

12.5 cm

M

LA


1. Draw a line segment and locate points P,Q such that PQ=12.5 cm.

2. Construct rays−→PR such that ∠QPR = 60◦ and

−→QS such that ∠PQS =

75◦.3. Draw the bisectors PL and QM of ∠QPR and ∠PQS respectively. Let

these intersect at A.

4. Draw the perpendicular bisector of AP and AQ and let these intersect

PQ at B and C respectively.

5. Join AB and AC.


Think it over!

What properties of triangles ensure that we get required base

angles? And the perimeter is also as required?

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302 Unit 4

Exercise 3.4.11

1. Construct a triangle ABC whose perimeter 12 cm and whose base

angles are 50◦ and 80◦.

2. Construct a triangle XY Z whose perimeter 15 cm and whose base


3. Construct a triangle ABC whose perimeter 12 cm and whose base


3.4.12 To construct triangle when its base, sum of theother two sides and one base angle are given (Optional)

Example 11. Construct a triangle ABC in which AB=5.8 cm, BC+CA=8.4

cm and ∠B = 60◦.

B A

D

C

5.8 cm

60

8.4

cm

X Solution:


1. Draw a line segment AB of

length 5.8 cm.

2. Draw a line segment BX,

sufficiently large such that

∠ABX = 60◦.

3. From the segment BX, cut

off line segment BD of length

8.4 cm.4. Join AD.

5. Draw the perpendicular bisector of AD and let it meet BD at C.

6. Join AC.


Think it over!

How is that the sum of CA and CB is equal to the given sum?

Exercise 3.4.12

1. Construct a triangle ABC in which BC=3.6 cm, AB + AC=4.8 cm and

∠B = 60◦.

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Constructions 303

2. Construct a triangle ABC in which AB + AC=5.6 cm, BC=4.5 cm and

∠B = 45◦.

3. Construct a triangle PQR in which PQ+ PR=6.5 cm, QR=5.4 cm and

∠Q = 40◦.

3.4.13 To construct a triangle when its base, difference ofthe other two sides and one base angle are given (Optional)

Example 12. Construct a triangle ABC in which base AB= 5 cm, ∠A = 30◦

and AC − BC=2.5 cm.

B

C

D

X

2.5 cm

A5 cm

Solution:

Steps of construction1. Draw a line segment AB of

length 5 cm.2. Draw another line segment

AX, sufficiently large, such

that ∠BAX = 30◦.3. From the segment AX, cut

off line segment AD=2.5 cm,

which is equal to (AC−BC).

4. Join BD.

5. Draw the perpendicular bisector of BD and let it cut AX at C.

6. Join BC.


Think it over!

Can you see why AD is equal to the difference AC − BC? Can

you take AC −BC > AB and still construct a triangle?

Exercise 3.4.13

1. Construct triangle ABC in which BC=3.4 cm, AB − AC=1.5 cm and

∠B = 45◦.

2. Construct triangle ABC in which BC=5 cm, AB − AC=2.8 cm and

∠B = 40◦.

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304 Unit 4

3. Construct triangle ABC in which BC=6 cm, AB − AC=3.1 cm and

∠B = 30◦.

Additional problems on “Constructions of triangles”

1. Construct a triangle ABC in which AB=5cm, BC=4.7cm and AC=4.3cm.

2. Construct a triangle ABC in which AB=5cm, BC=5cm and AC=4.3cm.

3. Construct a triangle PQR in which PQ=4cm, QR=4.5cm and ∠Q = 60◦.

4. Construct a triangle PQR in which PQ=4cm, ∠P = 60◦ and ∠Q=60◦.

5. Construct a triangle ABC in which AB=3.5cm, AC=4cm and length of

the perpendicular from A to BC is 3cm.

6. Construct an isosceles triangle ABC in which base BC=4.5cm and

altitude from A on BC is 3.8cm.

7. Construct an isosceles triangle whose altitude is 5 cm and whose


8. Construct an isosceles triangle whose altitude is 5 cm and whose


9. Construct an equilateral triangle of height 3.5 cm. (Optional)

10. Construct an equilateral triangle of height 4.3 cm. (Optional)

11. Construct right angle triangle LMN in which ∠M = 90◦, MN=4.5 cm

and LN=5.6 cm.

12. Construct right angle triangle PQR in which ∠Q = 90◦, QR=4.5 cm

and ∠R = 50◦.





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Constructions 305

15. Construct a triangle ABC, whose perimeter is 13.5 cm and whose

base angles are 60◦ and 75◦.

16. Construct a triangle ABC, whose perimeter is 12.5 cm and whose

base angles are 50◦ and 80◦.

17. Construct a triangle XY Z in which Y Z=4.5 cm, ∠Y = 60◦ and sum of

other two sides is 7.5 cm. (Optional)

18. Construct a triangle ABC whose perimeter is 9 cm and the angles are

in the ratio 3:4:5.

19. Construct a triangle ABC whose perimeter is 12 cm and the angles


20. Construct a triangle ABC in which BC=4.5 cm, ∠B = 35◦ and differ-

ence between the other two sides is 2.8 cm. (Optional)

Glossary

Perpendicular bisector: the line which is perpendicular to the given line

segment and also bisects the line segment.

Angle bisector: the line which bisects the given angle.

Perimeter: the length of the boundary of any plane figure.

Altitude: the perpendicular from a point to a line; this is also used for the

length of such a perpendicular.

Arc: part of the circumference of a circle.

Base angle: any of the angles formed by the base of a triangle, with the

other sides.

Vertex angle: the angle at the top of an isosceles triangle.

Points to remember

• At least three parameters are needed to construct a triangle.

• Not all combination of three parameters will enable one to construct

a triangle.

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CHAPTER 3 UNIT 5

QUADRILATERALS


• recognise quadrilaterals from a given list of figures;

• list out common properties of a quadrilateral;

• solve the sums related to quadrilaterals;

• classify different types of quadrilaterals and recognise their distinct

properties;

• transform the problems which occur in daily life related to quadrilat-

erals to numerical sums and solve them.

3.5.1 Introduction

You have learnt earlier that a triangle is a plane figure bounded by three

sides. Triangles are classified based on the measures of sides and angles.

Activity 1: Name the triangles based on their sides. Name the triangles

based on their angles.

Now let us take four points on the plane and see what we obtain on

joining them in pairs in some order.

Fig 2 Fig 3 Fig 4Fig 1

A

B

B

C

C

DD D

CBA D

AA B

C

If all the points are collinear, that is they all lie on the same straight line,

we obtain a line segment (Fig 1). If three points out of the four are collinear,

we get a triangle (Fig 2). If no three points out of four points are collinear,

we get a closed figure with four sides (Fig 3 and Fig 4). Such a closed

figure having four sides formed by joining four points, no three of which

are collinear, in an order, is called quadrilateral.

Look at the following figures:-

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Quadrilaterals 307

C

D

A

BQ

R

S

P

OPZ Y

XWNM

They are closed figures. Their boundary consists of four line segments.

The common name to all the above plane figures is quadrilateral.

A quadrilateral is named by re-

ferring to its vertices in a partic-

ular order. In the adjoining fig-

ure, we can read it as ABCD or

ADCB, we cannot read it ADBC

and such names. You may keep

in mind that the vertices should

be read in such a way that if

you join adjacent letters in the

name, then there should not be

any crossing of line segments.

For

B

C

D

A

example in the name ADBC, we see that AC and BD cross each other.

Observe the following figures.

CD

BA

B

C

D

A Are these two figures quadri-

laterals? No they are not

quadrilaterals; in the first,

there is crossing of sides; in

the second, the sides are not

all line segments.

So we may redefine the quadrilateral as: A quadrilateral is the union of

four line segments that join four coplanar points, no three of which

are collinear and each segment meet exactly two other lines, each at

their end point.

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308 Unit 5

Here again we do not distinguish between the closed curve which

is the union of four line segments and the plane figure which is

bounded by these four line segments. The context makes it clear

which form is taken.

Observe that a quadrilateral consists of four sides and four internal

angles. Based on these internal angles, we can classify quadrilaterals in

to two types: convex and concave quadrilaterals.

A quadrilateral is convex if each of the internal angle of the quadri-

lateral is less than 180◦. Otherwise it is called a concave quadrilat-

eral.

Convex quadrilateral

A

D

C

B

Concave quadrilateral

M

N

K

L

Think it over:

Suppose you are given four sticks of different lengths. Can you

put them together and make a quadrilateral?

Like triangles, quadrilaterals also enjoy nice properties. We will study

some of them in the coming sections.

3.5.2 Properties of quadrilaterals

Let ABCD be a quadrilateral.

1. The points A,B,C and D are called the vertices of the quadrilateral.

2. The segments AB, BC, CD and DA are the four sides of the quadri-

lateral.

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Quadrilaterals 309

3. The angles ∠DAB , ∠ABC, ∠BCD

and ∠CDA are the four angles of

the quadrilateral.

4. The segments AC and BD are

called as the diagonals of the

quadrilateral.B

C

D

A

Note: A quadrilateral has four sides, four angles and two diagonals.

In all it has ten elements.

Adjacent sides and opposite sides

1. Two sides of a quadrilateral are said

to be adjacent sides or consecu-

tive sides if they have a common

end point. In the adjoining figure, AB

and AD are adjacent or consecutive

sides. Identify the other pair of adja-

cent sides. B

C

D

A

2. Two sides are said to be opposite sides, if they do not have a common

end point. In the above figure AB and DC are opposite sides. Identify

the other pair of opposite sides.

Adjacent angles and opposite angles.

3. Two angles of a quadrilateral are adjacent angles or consecutive

angles, if they have a side common to them. Thus ∠DAB and ∠ABC

are adjacent or consecutive angles. Identify the other pair of adjacent

angles.

4. Two angles of a quadrilateral are said to be opposite angles, if they

do not contain a common side. Here ∠DAB and ∠BCD are opposite

angles. Identify other pair of opposite angles.

Diagonal property

The diagonal AC divides the quadrilateral in to two triangles, namely, tri-

angle ABC and triangle ADC. Name the two triangles formed when the

diagonal BD is drawn.

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310 Unit 5

Angle sum property

Activity 2: Take a cut-out of a quadrilateral drawn on a card board. Cut

it along the arms making each angle of the quadrilateral as in the figure

below. The four pieces so obtained are numbered as 1,2,3 and 4. Arrange

the cut outs as shown in next figure. Are they meeting at a point? What

can you say about the sum of these four angles? The sum of the measures

of the four angles is 360◦.

3

2

4

1 1 32

4

Note: Recall the sum of the angles at any point is 360◦.

Theorem 1. The sum of the angles of quadrilateral is 360◦.

Given: ABCD is a quadrilateral.

To prove: ∠A + ∠B + ∠C + ∠D = 360◦.Construction: Draw the diagonal AC.

2

31

5

64A

D

C

B

Proof: In triangle ADC,

∠1 + ∠2 + ∠3 = 180◦;(angle sum property.) In triangle

ABC,

∠4 + ∠5 + ∠6 = 180◦.( again angle sum property.)

Adding these,

∠1 + ∠4 + ∠2 + ∠5 + ∠3 + ∠6 = 360◦.

But ∠1 +∠4 = ∠A and ∠3 +∠6 = ∠C. Therefore ∠A+∠D+∠B +∠C = 360◦.

Thus the sum of the angles of the quadrilateral is 360◦.

Example 1. The four angles of a quadrilateral are in the ratio 2:3:4:6.

Find the measures of each angle.

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Quadrilaterals 311

Solution:

Given: The ratio of the angles as 2:3:4:6.

To find: The measure of each angle.

Observe that 2+3+4+6 = 15 (sum of the terms of the ratio). Thus 15 parts

accounts for 360◦. Hence

15 parts −→ 360◦;

2 parts −→ 360◦

15× 2 = 48◦;

3 parts −→ 360◦

15× 3 = 72◦;

4 parts −→ 360◦

15× 4 = 96◦;

6 parts −→ 360◦

15× 6 = 144◦.

Thus the angles are 48◦, 72◦, 96◦ and 144◦. Can you see that their sum is

360◦ ?

Example 2. In a quadrilateral ABCD, ∠A and ∠C are of of equal measure;

∠B is supplementary to ∠D. Find the measure of ∠A and ∠C.

Solution: We are given ∠B + ∠D = 180◦. Using angle-sum property of a

quadrilateral, we get

∠A+ ∠C = 360◦ − 180◦ = 180◦.

Since ∠A and ∠C are of equal measure, we obtain ∠A = ∠C =180◦

2= 90◦.

Can you draw a quadrilateral in which ∠A = ∠C = 90◦ and ∠B,∠D are

complementary ?

Example 3. Find all the angles in the given quadrilateral below.

Solution: We know that ∠P+∠Q+∠R+

∠S = 360◦ (angle sum property). Hence

x+ 2x+ 3 + x+ 3x− 7 = 360◦.

This gives 7x − 4 = 360◦ or 7x = 364◦.Therefore x = 364/7 = 52◦.

S

P

Q

R

x

2x+3

3 −7

xx

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312 Unit 5

We obtain, ∠P = 52◦; ∠R = 52◦; ∠Q = 2x + 3 = 2 × 52 + 3 = (104 + 3) = 107◦;∠S = 3× 52◦ − 7◦ = 156◦ − 7◦ = 149◦. Check that ∠P +∠Q+∠R +∠S = 360◦.

Exercise 3.5.2

1. Two angles of a quadrilateral are 70◦ and 130◦ and the other two angles

are equal. Find the measure of these two angles.

2. In the fig, suppose ∠P and ∠Q

are supplementary angles and

∠R = 125◦. Find the measures

of ∠S.125

P Q

R

S

3. Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth

angle is 90◦. Find the measures of the other three angles.4. In the adjoining figure, ABCD

is a quadrilateral such that

∠D + ∠C = 100◦. The bisectors

of ∠A and ∠B meet at ∠P . De-

termine ∠APB. A B

CD

P

3.5.3 Trapezium

Based on the nature of the sides or angles, a quadrilateral gets special

name.

Set I Set II

KRS

P Q

L

MN

BA

CD

E

F

G

H

XY

Z

W

Observe the above figures given in two sets. Discuss with your friends

what is the difference you observe in the first set and the second set of

quadrilaterals. [Note: the arrow mark indicates parallel lines].

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Quadrilaterals 313

The first set of quadrilaterals have a pair of opposite sides which are

parallel. Such a quadrilateral is known as trapezium. In the first set

each quadrilateral is a trapezium. In the second set no quadrilateral is a

trapezium.

5

4

3

3

5

5

4

4

3

Activity 3:

Take identical cut-outs of congru-

ent triangles of sides 3 cm, 4 cm

and 5 cm. Arrange them as shown

in the figure. Which figure do you

get? It is a trapezium. Which are

the parallel sides?

Can you get a trapezium in which non parallel sides are equal?

3

3

3

33

3

3

3

3

BA

D CActivity 4:

Cut three identical equilateral

triangles and place them as

shown in the figure. Measure AD

and BC. Are they equal? Mea-

sure angles A and B.

Are they equal? Measure angles D and C. Are they equal? Measure AC

and BD. Are they equal?

The special name given to the above trapezium is isosceles trapezium.

You see that in an isosceles trapezium;

1. the non parallel sides are equal;

2. the base angles are equal;

3. the adjacent angles corresponding to parallel sides are supplemen-

tary;

4. the diagonals are equal.

Think it over!

Is there a trapezium whose all angles are

equal?

Example 4. In the figure ABCD, suppose AB ‖ CD; ∠A = 65◦ and∠B = 75◦. What is the measure of ∠D and ∠C?

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314 Unit 5

BA

CD

65 75

Solution: Observe that, ∠A +

∠D = 180◦ (a pair of adjacent an-

gles of a trapezium is supplemen-

tary). Thus 65◦ + ∠D = 180◦. This

gives ∠D = 180◦ − 65◦ = 115◦. Sim-

ilarly ∠B + ∠C = 180◦, which gives

75◦ + ∠C = 180◦. Hence

∠C = 180◦ − 75◦ = 105◦.Example 5. In an isosceles trapezium PQRS, ∠P and ∠S are in the ratio

1:2. Find the measure of all the angles.

P Q

RS

Solution: In an isosceles trapezium

base angles are equal; ∠P = ∠Q.

Let ∠P = x◦ and ∠S = 2x◦. Since

∠P + ∠S = 180◦ (one pair of adjacent

angles of a trapezium is supplemen-

tary), we get

x◦ + 2x◦ = 180◦.

Thus 3x◦ = 180◦ or x◦ = 180◦/3 = 60◦. Hence ∠P = 60◦ and∠S = 2× 60◦ = 120◦. Since, ∠P = ∠Q, we get ∠Q = 60◦. But, ∠Q+∠R = 180◦

(one pair of adjacent angles are supplementary). Hence we also get

∠R = 180◦ − 60◦ = 120◦.

Exercise 3.5.3

1. In a trapezium PQRS, PQ ‖ RS; and ∠P = 70◦ and ∠Q = 80◦. Calculatethe measure of ∠S and ∠R.

2. In a trapezium ABCD with AB ‖ CD, it is given that AD is not parallel

to BC. Is ∆ABC ∼= ∆ADC? Give reasons.

3. In the figure, PQRS is an isosce-

les trapezium; ∠SRP = 30◦, and

∠PQS = 40◦. Calculate the angles

∠RPQ and ∠RSQ.

P Q

RS

30

40

4. Prove that the base angles of an isosceles trapezium are equal.

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Quadrilaterals 315

5. Suppose in a quadrilateral ABCD, AC = BD and AD = BC. Prove

that ABCD is a trapezium.

3.5.4 Parallelograms

Look at the following sets of quadrilaterals.

Set I Set II

How many pairs of parallel sides do you see in each of the quadrilater-

als in set I ? How many pairs of parallel sides do you see in each of the

quadrilaterals in set II ? What is your conclusion?

A quadrilateral in which both the pairs of opposite sides are parallel is

called a parallelogram.

The quadrilaterals in set II are not parallelogram, while all the quadri-

laterals in set I are parallelograms. Can you see that a parallelogram is

a particular type of trapezium? Hence, whatever properties are true for

trapezium also hold good for parallelograms. You will see that additional

properties are also true because the parallelness of one more pair of sides.

E A E B

CD

Activity 5:

Take a rectangular card board,

ABCD and mark a point E on AB

(as shown in the fig). Join CE

by dotted line and cut the card

board along CE. Place the trian-

gular part EBC to the left of the

rectangle such that BC coincides

with AD to get a quadrilateral.

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316 Unit 5

Which type of quadrilateral is this? It is a parallelogram.

Trace the above cut-out card board in your book and measure the op-

posite sides, opposite angles, Repeat the same activity with two more par-

allelograms and mark them as PQRS and KLMN and tabulate the results

as follows.

Parallelo- (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

gram

ABCD ∠A = ∠B = ∠C = ∠D = AB = BC = CD = DA =

PQRS ∠P = ∠Q = ∠R = ∠S = PQ = QR = RS = SP =

KLMN ∠K = ∠L = ∠M = ∠N = KL = LM = MN = NK =

What relations are there among angles? What relations are there among

sides? Do you observe that opposite angles are equal and opposite sides

are also equal? These observations can be proved logically.

Proposition 1. In a parallelogram, opposite sides are equal and opposite

angles are equal.

13

A B

CD

4 2Proof: Let ABCD is a parallelogram.

Join BD. Then ∠1 = ∠2, and

∠3 = ∠4. (Why ? See figure.) In tri-

angles ABD and CBD, we observe

∠1 = ∠2, ∠4 = ∠3, BD (common).

Hence ∆ABD ∼= ∆CDB (ASA postulate). It follows that

AB = DC, AD = BC and ∠A = ∠C.

Similarly join AC, and you can prove that ∆ADC ∼= ∆CBA. Hence ∠D =

∠B.

Activity 6:

Trace a parallelogram as in the previous case on a card board; draw the di-

agonals and mark the point of intersection. Measure the intercepts formed

by the point of intersection. What is the inference drawn from this activ-

ity?

The diagonals bisect each other. Thus you may describe the properties of

parallelogram as:

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Quadrilaterals 317

1. The opposite sides are equal and parallel.

2. The opposite angles are equal.

3. The adjacent angles are supplementary

4. The diagonals bisect each other.

5. Each diagonal bisects the parallelogram into two congruent triangles.

Example 6. The ratio of two sides of parallelogram is 3:4 and its perimeter

is 42 cm. Find the measures of all sides of the parallelogram.

Solution: Let the sides be 3x and 4x. Then the perimeter of the parallelo-

gram is 2(3x + 4x) = 2 × 7x = 14x. The given data implies that 42 = 14x, so

that x = 42/14 = 3. Hence the sides of the parallelogram are 3 × 3 = 9 cm

and 3× 4 = 12 cm.

Example 7. In the adjoining figure, PQRS is a parallelogram. Find x and

y in cm.Solution: In a parallelogram, we know

that the diagonals bisect each other.

Therefore SO = OQ. This gives 16 =

x + y. Similarly, PO = OR, so that

20 = y + 7. We obtain y = 20− 7 = 13cm.

Substituting the value of y in the first

relation, we get 16 = x + 13. Hence

x = 3cm.

16

x+y20

y +7

P Q

R S

O

Exercise 3.5.4

1. The adjacent angles of a parallelogram are in the ratio 2:1. Find the

measures of all the angles.

2. A field is in the form of a parallelogram, whose perimeter is 450 m

and one of its sides is larger than the other by 75 m. Find the lengths

of all sides.

3. In the figure, ABCD is a parallelo-

gram. The diagonals AC and BD in-

tersect at O; and ∠DAC = 40◦, ∠CAB =

35◦; and ∠DOC = 110◦. Calculate the

∠ABO,∠ADC,∠ACB, and ∠CBD.

4035

110

B

CD

A

O

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318 Unit 5

4. In a parallelogram ABCD, the side DC is produced to E and ∠BCE =

105◦. Calculate ∠A,∠B,∠C, and ∠D.

5. Prove logically the diagonals of a parallelogram bisect each other.

Show conversely that a quadrilateral in which diagonals bisect each

other is a parallelogram.

6. In a parallelogram KLMN , ∠K = 60◦. Find the measures of all the

angles.

7. Let ABCD be a quadrilateral in which ∠A = ∠C, and ∠B = ∠D. Prove

that ABCD is a parallelogram.

8. In a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that

ABCD is a parallelogram.

3.5.5 Special kinds of parallelograms

There are special kind of parallelograms which enjoy different types of

properties. We study them here.

Rectangle

Activity 7:

Take a sheet from your note book and paste it on a card board; cut the

cardboard along the boundary, measure all the sides, all the angles. Write

down the observations in the following chart. Repeat the activity with

sheets of different sizes.

Parallelogram (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

ABCD ∠A = ∠B = ∠C = ∠D = AB = BC = CD = DA =

PQRS ∠P = ∠Q = ∠R = ∠S = PQ = QR = RS = SP =

From the above activity, you can infer that:

• all angles are equal to 90◦;

• opposite sides are equal;

• opposite sides are parallel;

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Quadrilaterals 319

The special name given to such a parallelogram is rectangle. A rectan-

gle is a parallelogram whose all angles are right angles.

Activity 8:

Take a rectangular sheet of paper and name it as ABCD. Fold it along

the diagonals. Mark the point of intersection as O. Answer the following

questions:

When you fold along any of the diagonals, are the two triangles formed

congruent? Measure the length of the diagonals. Are they equal? Measure

the line segments OA, OC, OB and OD. Do you find any relation among

the length of these segments?

Diagonal properties of a rectangle

(i) The diagonals of a rectangle are equal.

(ii) The diagonals of a rectangle bisect each other.

Example 8. In a rectangle XYWZ, suppose O is the point of intersection

of its diagonals. If ∠ZOW = 110◦, calculate the measure of ∠OYW .

O

X Y

110

WZSolution:- We know ∠ZOW = 110◦.Hence, ∠WOY = 180◦ − 110◦ = 70◦ (sup-plementary angles). Now OYW is an

isosceles triangle, as OY = OW . Hence

∠OYW = ∠OWY = (180◦ − 70◦)/2 =

110◦/2 = 55◦. (Can you suggest an al-

ternate method?)

Example 9. In a rectangle RENT , the diagonals meet at O. If OR = 2x+ 4

and OT = 3x+ 1, find x.

O

R E

NT

2x +4

3x+1

Solution: Observe that OR = OT (di-

agonal bisect each other and they are

equal in a rectangle). Hence

2x+ 4 = 3x+ 1.

This implies that 4−1 = 3x−2x = 3 = x.

Hence x = 3.

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320 Unit 5

Rhombus

1 234

A

B

C

D

5cm 5cm

5cm5cm

Activity 9:

Construct four identical right an-

gle triangles with measures 3 cm,

4 cm and 5 cm using card board.

Arrange them as shown in the fig-

ure, on a plane sheet. Draw the

boundary of the figure. Measure

the sides and angles of the figure.

Tabulate them. Repeat this with

right triangles of different dimen-

sions. What do you observe?Can you conclude that ∠A = ∠C, ∠B = ∠D; AB = BC = CD = DA?

Such a figure is called a rhombus. A rhombus is a parallelogram in

which all the four sides are equal. Being a parallelogram, a rhombus

has all the properties of a parallelogram and more:(i) all the sides of a rhombus are equal;

(ii) opposite sides are parallel;

(iii) the diagonals bisect each other at right angles;

(iv) the two diagonals divide the rhombus into four congruent right angled

triangles;

(v) angles are also bisected by the diagonals.

Example 10. The diagonals of a rhombus are 24 cm and 10 cm. Calculate

the area of the rhombus.

A

D B

C

12

cm

5 cm

Solution: We are given that AC = 24

cm; BD = 10 cm. We know that the di-

agonals of a rhombus bisect each other

at right angles. Let O be the point of in-

tersection of these diagonals. Then we

have AO = CO = 12 cm and BO = DO =

5 cm. We also know that AOD is a right

angled triangle. Hence the area of AOD

is

1

2×OA×OD =

1

2× 12× 5 = 30cm2.

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Quadrilaterals 321

Since a rhombus has four congruent right triangles, is area is 4× 30 = 120

cm2.

Example 11. In a rhombus ABCD, ∠BAC = 38◦. Find (i) ∠ACD, (ii) ∠DAC

and (iii) ∠ADC.

Solution: We know that ∠BAC = 38◦. But in a rhombus ABCD, since

ABC is an isosceles triangle, we see that ∠BAC = ∠ACB = 38◦. Moreover

∠DAC = 38◦, since the diagonal AC bisect ∠A. Since ADC is also an

isosceles triangle, we get ∠ACD = ∠DAC = 38◦. Finally,

∠ADC = 180◦ − (∠DAC + ∠DCA)

= 180◦ − (38◦ + 38◦)

= 180◦ − 76◦

= 104◦. A

D B

C

Square

There is a type of parallelogram which is simultaneously a rectangle and

a rhombus. All its angles are equal and all its sides are equal. Recall what

you have for triangles: triangle which is at the same time have all angles

equal and all sides equal. These are equilateral triangles. In the case

of triangles, you have seen that whenever all the angles are equal, all the

sides are also equal. Conversely, if all the sides of a triangle are equal, then

all the angles are also equal. But when you move to quadrilaterals, you

do not have such a nice property. A rectangle is a quadrilateral in which

all the angles are equal, but the sides need not be equal. On the other

hand a rhombus is a quadrilateral in which all the sides are equal, but all

the angles need not be equal. A quadrilateral in which all the angles are

equal and all the sides are equal is given a special name; a square. Thus a

square is an a quadrilateral in which all the nice properties come together.

A square is a parallelogram in which

A B

CD ((i) all the sides are equal;

(ii) each angle is a right angle;

(iii) diagonals are equal;

(iv) diagonals bisect at right an-

gles.

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322 Unit 5

Think it over!

(iii) and (iv) are consequences of (i) and (ii). Can you prove them?

A square can also be defined as:

(a) a rectangle in which adjacent sides are equal;

(b) a rhombus in which each angle is 90◦.

Think it over!

Suppose the perimeters of a square and a rhombus are equal. Do

they have equal area?

Example 12. A field is in the shape of a square with side 20 m. A pathway

of 2 m width is surrounding it. Find the outer perimeter of the pathway.

Solution: Width of the pathway

is 2 m. Length of the side of the

outer square= (20+2+2)= 24 m.

Hence perimeter = 4× 24 = 96 m..

A B

CD

2 m

20 m

Example 13. The square field has area 196 m2. Find the length of the

wire required to fence it around 3 times.

Solution: Suppose s is the side-length of a square. Then its area is s2 sq.

units. We are given that s2 = 196m2. Therefore S = 14m. Thus perimeter =

4× s = 4× 14 = 56 m.

The length of the wire required to fence around it 3 times is 56× 3 = 168 m.

Kite

You have seen that in a rhombus, a diagonal divides the rhombus in to

two congruent isosceles triangles. Suppose we take two isosceles triangles

whose bases are of equal length and glue them together to get a quadrilat-

eral. You get a special type of quadrilateral. Such a quadrilateral is called

a kite.

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Quadrilaterals 323

B

A

O

C

D

Kite is a quadrilateral in which two isosce-

les triangles are joined along the common

base. In the adjoining figure AB = AD,

BC = CD, and BD is the common base. It is

important to observe that triangles ABC and

ADC are congruent, but the triangles ABD

and CBD need not be congruent. Can you see

that if ABD and CBD are also congruent, the

quadrilateral ABCD reduces to a rombhus?

Properties of kite

Like rhombus, kite has some properties which we record here:

1. There are two pairs of equal sides; AB = AD and CB = CD in the

previous diagram.

2. One of the diagonals bisect the other diagonal perpendicularly; the

diagonal BD is bisected perpendicularly by AC in the previous figure.

3. One of the diagonals bisect the apex angles; the diagonal AC bisects

the apex angles ∠A and ∠C.

Example 14. In the figure PQRS is a kite; PQ = 3 cm and QR = 6 cm.

Find the perimeter of PQRS.

Solution: We have PQ = PS = 3

cm, QR = SR = 6 cm. Hence the

perimeter = PQ+QR+RS+PS =

3 + 6 + 6 + 3 = 18 cm.

O

Q

P R

S

Exercise 3.5.5

1. The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm.

Calculate the measure of all the sides.

2. In the adjacent rectangle ABCD,

∠OCD = 30◦. Calculate ∠BOC. What

type of triangle is BOC?A

CD

B

O

30

3. All rectangles are parallelograms, but all parallelograms are not rect-

angles. Justify this statements.

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324 Unit 5

4. Prove logically that the diagonals of a rectangle are equal.

5. The sides of a rectangular park are in the ratio 4:3. If the area is 1728

m2, find the cost of fencing it at the rate of Rs. 2.50/m.

6. A rectangular yard contains two

flower beds in the shape of con-

gruent isosceles right triangle.

The remaining portion is a yard

of trapezoidalA B

CEFD

shape (see fig). whose parallel sides have lengths 15 m and 25 m.

What fraction of the yard is occupied by the flower bed?

7. In a rhombus ABCD, ∠C = 70◦. Find the other angles of the rhombus.

8. In a rhombus PQRS, ∠SQR = 40◦ and PQ = 3 cm. Find ∠SPQ, ∠QSR

and the perimeter of the rhombus.

9. In a rhombus PQRS, if PQ = 3x− 7 and QR = x+ 3, find PS.

10. Let ABCD be a rhombus and ∠ABC = 124◦. Calculate ∠A, ∠D and

∠C.

11. Rhombus is a parallelogram: justify.

12. In a given square ABCD, if the area of triangle ABD is 36 cm2, find

(i) the area of triangle BCD; (ii) the area of the square ABCD.

13. The side of a square ABCD is 5 cm and another square PQRS has

perimeter equal to 40 cm. Find the ratio of the perimeter of ABCD to

perimeter of PQRS. Find the ratio of the area ABCD to the area of

PQRS.

14. A square field has side 20 m. Find the length of the wire required to

fence it four times.

15. List out the differences between square and rhombus.

16. Four congruent rectangles are

placed as shown in the figure.

Area of the outer square is 4

times that of the inner square.

What is the ratio of length to

breadth of the congruent rect-

angles?A B

CD

RS

QP

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Quadrilaterals 325

Additional problems on “Quadrilaterals”

1. Complete the following:

(a) A quadrilateral has ————————— sides.

(b) A quadrilateral has ————————— diagonals.

(c) A quadrilateral in which one pair of sides are parallel to each

other is —————————.

(d) In an isosceles trapezium, the base angles are —————————.

(e) In a rhombus, the diagonals bisect each other in ———————

—— angles.

(f) In a square, all the sides are —————————.

2. Let ABCD be a parallelogram. What special name will you give it:

(a) if AB = BC?

(b) if ∠BAD = 90◦?

(c) if AB = AD and ∠BAD = 90◦?

3. A quadrilateral has three acute angles each measuring 70◦. What is

the measure of the fourth angle?

4. The difference between the two adjacent angles of a parallelogram is

20◦. Find measures of all the angles of the parallelogram.

5. The angles of a quadrilateral are in the ratio 1:2:3:4. Find all the

angles of the quadrilateral.

6. Let PQRS be a parallelogram with PQ= 10 cm and QR= 6 cm. Calcu-

late the measures of the other two sides and the perimeter of PQRS.

7. The perimeter of a square is 60 cm. Find its side-length.

8. Let ABCD be a square and let AC = BD = 10 cm. Let AC and BD

intersect in O. Find OC and OD.

9. Let PQRS be a rhombus, with PR = 15 cm and QS = 8 cm. Find the

area of the rhombus.

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326 Unit 5

10. Let ABCD be a parallelogram and suppose the bisectors of ∠A and

∠B meet at P . Prove that ∠APB = 90◦.

11. Let ABCD be a square. Locate points P , Q, R, S on the sides AB,

BC, CD, DA respectively such that AP = BQ = CR = DS. Prove that

PQRS is a square.

12. Let ABCD be a rectangle and let P,Q,R, S be the mid-points of AB,

BC, CD, DA respectively. Prove that PQRS is a rhombus.

13. Let ABCD be a quadrilateral in which the diagonals intersect at O

perpendicularly. Prove that AB +BC + CD +DA > AC +BD.

14. Let ABCD be a quadrilateral with diagonals AC and BD. Prove the

following statements ((Compare these with the previous problem);

(a) AB +BC + CD > AD; (b) AB +BC + CD +DA > 2AC;

(c) AB +BC +CD+DA > 2BD; (d) AB +BC +CD+DA > AC +BD.

15. Let PQRS be a kite such that PQ > PS. Prove that ∠PQR > ∠PSR.

(Hint: Join QS.)

16. Let ABCD be a quadrilateral in which AB is the smallest side and CD

is the largest side. Prove that ∠A > ∠C and ∠B > ∠D.(Hint: Join AC

and BD.)

17. In a triangle ABC, let D be the mid-point of BC. Prove that AB+AC >

2AD.(What property of quadrilateral is needed here?)

18. Let ABCD be a quadrilateral and let P,Q,R, S be the mid-points of AB,

BC, CD, DA respectively. Prove that PQRS is a parallelogram.(What

extra result you need to prove this ?)

Glossary

Quadrilateral: a linear figure on a plane consisting of four line segments,

which are placed in an ordered way such that the adjacent segments meet

only at their end points.

Convex quadrilaterals: a quadrilateral in which each interior angle is less

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Answers 327

than 180◦.Concave quadrilateral: a quadrilateral in which some angle exceeds 180◦.Diagonal: the line segment joining opposite vertices of a quadrilateral.

Trapezium: a quadrilateral in which a pair of sides are parallel.

Parallelogram: a quadrilateral in which two pairs of sides are parallel.

Rhombus: a quadrilateral in which all the sides are equal.

Square: a quadrilateral in which all the sides are equal and all the angles

are equal to 90◦.Rectangle: a quadrilateral in which all the angles are equal to 90◦.Kite: a quadrilateral formed by a pair of isosceles triangle glued along a

common side.

Points to remember• The sum of all the four angles of a quadrilateral is 360◦.• For a quadrilateral, equiangularity is not the same as equilaterality,

unlike for triangles.

• The diagonals intersect perpendicularly in a rhombus and a kite.

• Any quadrilateral in which diagonals bisect each other is a parallelo-

gram.


Exercise 3.1.3

3. (i) 60◦; (ii) 18◦; (iii) 135◦; (iv) 90◦; (v) 30◦; (vi) 65◦.

Exercise 3.1.4

1. ∠DML = 45◦; ∠BLQ = 45◦; ∠MLB = 135◦; ∠CMP = 45◦; ∠CML = 135◦;MLA = 45◦; ∠QLA = 135◦. 2. 40◦.

Additional problems on “Axioms, postulates and theorems”

1. (i) A. (ii) D. (iii) B. (iv) B. (e) D.

3. 3. 8. 40◦. 9. (i) 190◦; (ii) 22◦30′. 10. 144◦.

Exercise 3.2.1

1. (i)→ (C); (ii) → (D); (iii)→ (A); (iv) → (B). 2. (i) scalene; (ii) scalene;

(iii) scalene; (iv) isosceles; (v) scalene; (vi) scalene; (vii) scalene; (viii)

equilateral; (ix) isosceles; (x) isosceles.

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328 Answers

Exercise 3.2.2

1. 85◦. 2. 55◦. 3. 65◦ each. 4 30◦, 60◦ and 90◦. 5 x = 50◦, ∠A = 65◦,∠B = 35◦, ∠C = 80◦. 6. 50◦, 60◦ and 70◦.

Exercise 3.2.3

1. If Ext∠B = 136◦ and Ext∠C = 104◦, then ∠A = 60◦, ∠B = 44◦ and

∠C = 76◦. 3. (i) 130◦; (ii) 56◦; (iii) 35◦; (iv) 52◦; (v) 40◦. 4. ∠TRS = 50◦,∠PSQ = 80◦. 5. The other interior opposite angle is 90◦ and the third

angle is 60◦. 6. 180◦.

Additional problems on “Theorems on triangles”

1. (a) 180◦; (b) the interior; (c) larger; (d) one; (e) one.

2. (a) A. (b) B. (c) A. (d) C. (e) D.

3. 110◦. 4. 35◦ each. 5. 36◦, 54◦, 90◦. 6. 45◦, 60◦, 75◦. 7. 90◦.8. ∠C = 30◦, ∠B = 60◦, ∠A = 90◦. 9. x = 90◦. 10. ∠A = 80◦, ∠B = 65◦,∠C = 35◦. 11. 30◦, 50◦, 100◦. 12. 110◦. 13. 45◦, 45◦, 90◦. 14. 36◦, 72◦,72◦. 15. 110◦. 16. 30◦, 30◦, 120◦.Exercise 3.3.3

1. ∠B = ∠C = 65◦. 2. 58◦. 3. (i) 110◦; (ii) 55◦; (iii) 20◦; (iv) 40◦.4. ∠ACD = 120◦, ∠ADC = 30◦.Exercise 3.3.7

1. BC is the largest and CA is the smallest. 2. ∠C < ∠A < ∠B.

Additional problems on “Congruency of triangles”

1. (a) largest (b) less (c) larger (d) smaller (e) less (f) larger

9. 30◦, 30◦, 120◦.Exercise 3.5.2

1. 80◦ each. 2. 55◦. 3. 54◦, 81◦ and 108◦. 4. 50◦.

Exercise 3.5.3

1. ∠S = 110◦ and ∠R = 100◦. 3. ∠RPQ = 30◦ and ∠RSQ = 40◦.

Exercise 3.5.4

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Answers 329

1. 120◦, 60◦, 120◦, 60◦. 2. 150 m and 75 m. 3. ∠ABO = 35◦, ∠ADC = 105◦,∠ACB = 40◦, ∠CBD = 70◦. 4. ∠A = ∠C = 75◦, ∠B = ∠D = 105◦.6. 60◦, 120◦, 60◦, 120◦.

Exercise 3.5.5

1. 10 cm, 5 cm, 10 cm, 5 cm. 2. ∠BOC = 60◦; ∆BOC is equilateral.

5. . 420. 6.1

5. 7. 110◦, 70◦, 110◦. 8. ∠SPQ = 100◦, ∠QSR = 40◦ and

perimeter= 12 cm. 9. 8. 10. 56◦, 124◦, 56◦. 12. (i) 36 cm2; (ii) 72 cm2.

13. (i) 1:2 (ii) 1:4. 14 320 m. 16. 3:1.

Additional problems on “Quadrilaterals”

1. (a) four; (b) two; (c) a trapezium; (d) equal; (e) right; (f) equal.

2. (a) rhombus; (b) rectangle; (c) square.

3. 150◦. 4. 80◦, 100◦, 80◦, 100◦. 5. 36◦, 72◦; 108◦, 144◦. 6. RS = 10 cm,

SP = 6 cm; perimeter is 32 cm. 7. 15 cm. 8. OC = OD = 5 cm. 9. 120

cm2.

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CHAPTER 4 UNIT 1

MENSURATION

After studying this unit you learn to:

• recognise the cubes and cuboidal objects used in our day to day life;

• list out the properties of cubes and cuboids;

• relate formulae to given problems;

• substitute the data in the given formula and solve the problems.

4.1.1 Introduction

When we look at an empty box, an empty bowl and an empty container,

it has some space and we can keep things in that empty space. A class

room has space for the students to sit in.

A solid occupies fixed amount of space. Solids occur in different shapes.

Observe the following diagrams. These shapes (cuboid, cube, cylinder,

sphere, cone, triangular prism etc) are known as three dimensional ob-

jects.

Cuboid Cube Cylinder Cone

Triangular

Prism

Do you see that each solid occupies some space. Each solid also has

some surface and hence has associated surface area. Since each occupies

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Mensuration 331

space, it has some volume. We shall study these aspects corresponding to

some simple figures.

Observe the following figures

wooden box match box book case Almirah

These are in the shape of cuboids.

4.1.2 Surface area of a cuboid

Let us understand the cuboids by doing an activity.

Activity 1:

Take a box in the shape of cuboid and cut it open along one edge, open

up the lids and spread it over a sheet of paper and fasten it with pins.(See

the figure)

A

H G

E F

D C

B

6

5

1 2 3 4

How many faces has a

cuboid? Find the num-

ber of edges and vertices.

A cuboid has 6 faces, 12

edges and 8 vertices. Do

you see that a cuboid has

rectangular faces?.

Any face of a cuboid may be called its base (can you give the reason?).

The four faces which meet the base are called the lateral faces of a cuboid.

In the given figure above, the cuboid has 6 faces. They are ABCD,

EFGH, EFBA, HGCD, EHDA and FGCB. Any two adjacent faces of a

cuboid meet in a line segment, called an edge of the cuboid. The 12 edges

are AB, BC, CD, DA, EF , FG, GH, EH, AE, DH, GC and BF . The point of

intersection of three edges of a cuboid is a vertex of the cuboid. The eight

vertices are A, B, C, D, E, F , G and H.

Activity 2:

Take any cuboidal box. Fix a base (observe any face can be taken as

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332 Unit 1

base). Place it vertically and wrap a thick sheet of paper such that it just

fits around the surface. Remove the paper and measure the area of the

paper. It is the lateral surface area (L.S.A) of the cuboid. Do this with

different cuboids.

Cut open the cuboidal box

and lay it flat on a sheet of

paper. Let the length of the

base be l; breadth of the

base be b; and the height

of the cuboid correspond-

ing to this base be h units. VI

IIIIII IV

V

b

l

hh

b

b

l

l

b bll

Compute 2(lh + bh) and compare this with the area of the lateral surface

you have measured. Do they match? What is 2(lh+ hb+ bl)?

If you change the base, you may observe that the units l, b, h do

not change; only their representation as length, breadth and height may

change.

There are 6 faces in a cuboid. All the faces are rectangular in shape.

The sum of the areas of all the six surfaces is called the total surface

area (T.S.A) of the cuboid. Let us find a formula for total surface area and

lateral surface area of a cuboid. The total surface area of cuboid is

Area of I + area of II + area of III + area of IV + area of V + area of VI (see

fig)

Hence

A = (l × h) + (l × b) + (b× h) + (l × h) + (b× h) + (l × b) sq. units.

Thus

A = 2lh+ 2lb+ 2bh = 2(lh+ lb+ bh) sq. units.

The lateral surface area of a cuboid is

area I + area II + area III + area IV.

Hence

L.S.A = (l × h) + (b× h) + (l × h) + b× h = 2h(l + b)

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Mensuration 333

sq.units.

Here you may observe that the lateral surface area of a cuboid

depends on the base you choose for the cuboid. If you change the

base the lateral surface area changes. However, the total surface area

of the cuboid remains the same.

Cube

A cuboid whose length, breadth and height are all equal is called a cube.

Ice cubes, sugar cubes, dice are some examples of cubes.

R

R

G

GGG

4 cm

4 cm

4 cm

Activity 3:

Construct a cube of side 4 cm by

using card board. Paint any two

opposite sides with red paint and

remaining with green. Place the

cube on table with one of the red

faces as base. Identify the num-

ber of green faces bounding the

red faces.

The green faces are called lateral faces of the cube. What is the area of

the lateral faces? Do you see that: L.S.A of this cube is 4 × 16 = 64 cm2;

T.S.A of this cube is 6× 16 = 96 cm2.

l

l

l

Observe the adjoining figure of a

cube of side l units. It has six

square faces. Area of each face

of the cube is l2 units. Area of 6

faces is therefore 6l2 units. And

the area of 4 lateral faces is 4l2

units.

Think it over!

What should be the maximum length of the ladder which

can be placed from the bottom of the floor to reach the

opposite corner of the roof of a room which is in the shape

of a cube?

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334 Unit 1

Example 1. Find the lateral surface area and total surface area of a cuboid

which is 8 m long, 5 m broad and 3.5 m high.

Solution: We are given l = 8 m, b = 6 m, h = 3.5 m. We know that

L.S.A = 2h(l + b) = 2× 3.5(8 + 6) = 7× 14 = 98 m2.

Similarly,

T.S.A = 2(lb+bh+lh) = 2(8×6+6×3.5+8×3.5) = 2(48+21+28) = 2×(97) = 194 m2.

Example 2. How many tiles each of 30 cm × 20 cm are required to cover

the floor of hall of dimension 15 m by 12 m?

Solution: Since the tiles are in cm2, we have to convert the dimensions of

the hall to cm first. It will be 1500 cm by 1200 cm. Hence the area of the

floor is

1500× 1200 = 1800000 cm2.

Area of each tile is 30 × 20 = 600 cm2. Hence the number of tiles required

to cover the floor is:1800000

600= 3000.

Example 3. Find the length of each side of a cube having the total surface

area is 294 cm2.

Solution: Given T.S.A of a cube as 294 cm2, we have to find its length l.

We know T.S.A of a cube is equal to 6l2. Thus 6l2 = 294 or l2 = 294/6 = 49.

Hence l = 7 cm.

Example 4. The total surface area of a cube is 600 cm2. Find the lateral

surface area of the cube.

Solution: Given T.S.A of a cube as 600 cm2, we have to find its L.S.A. But

we know T.S.A = 6l2, where l is the length of the cube. Hence 600 = 6l2 or

l2 = 100 units. Taking square-root, we get l = 10 cm. But we also know

that L.S.A. = 4l2. Hence

L.S.A. = 4× 10× 10 = 400 cm2.

Example 5. Find the area of a metal sheet required to make a cube of

length 2 m. Find the cost of metal sheet required at the rate of 8/ m2 to

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Mensuration 335

make the cube.

Solution: We know that l = 2 m. We have to find T.S.A. of the cube. But

T.S.A. = 6l2 = 6× 2× 2 = 24 m2.

The cost of the metal sheet required is therefore 24× 8 = 192 rupees.

Exercise 4.1.2

1. Find the total surface area of the cuboid with l = 4 m, b = 3 m and

h = 1.5 m.

2. Find the area of four walls of a room whose length 3.5 m, breadth 2.5

m and height 3 m.

3. The dimensions of a room are l = 8 m, b = 5 m, h = 4 m. Find the cost

of distempering its four walls at the rate of 40/ m2.

4. A room is 4.8 m long, 3.6 m broad and 2 m high. Find the cost of

laying tiles on its floor and its four walls at the rate of 100/ m2.

5. A closed box is 40 cm long, 50 cm wide and 60 cm deep. Find the

area of the foil needed for covering it.

6. The total surface area of a cube is 384 cm2. Calculate the side of the

cube.

7. The L.S.A of a cube is 64 cm2. Calculate the side of the cube.

8. Find the cost of white washing the four walls of a cubical room of side

4 m at the rate of 20/ m2.

9. A cubical box has edge 10 cm and another cuboidal box is 12.5 cm

long, 10 cm wide and 8 cm high.

(i) Which box has smaller total surface area?(ii) If each edge of the cube is doubled, how many times will its T.S.A

increase?

4.1.3 Volume of cubes and cuboidsAmount of space occupied by a three dimensional object is called its

volume. Volume of a room is bigger than the volume of a brick or shoe

box. Remember we use square units to measure the area of a region or a

surface. Similarly, we use cubic units to measure the volume of a solid,

as solids are three dimensional objects.

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336 Unit 1

l

l

l

l

b

h

The figures above are cube and cuboid which occupy space in three

dimension and hence posses volume. We are interested in finding their

volume.

Activity 4:

Take four cubes each of

length 1 unit and arrange

as shown in figure. Take

another 4 cubes of same

size and arrange them on

the top of the cubes ar-

ranged earlier, as shown

in the next figure.

Do you get another cube? You have used 8 cubes, i.e, the volume of the

new cube is 8 cubic units. Here l = b = h = 2 units.

Volume of this cube = 2 units× 2 units× 2 units = 8 cubic units.

In general volume of a cube = side × side × side. Thus

V = l × l × l = l3

cubic units. Cubic units are used to measure volume:

1 cm3 = 1 cm × 1 cm × 1 cm;

1 m3 =1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 106 cm3.

Volume of a cuboid

Activity 5:

Take 24 cubes of equal size. Arrange them to form a cuboid. You can

arrange them in many ways to get a cuboid. Observe the following table:

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Mensuration 337

l b h l × b× h

12 2 1 24 cubic units



1 212

12

6

4 3

4

Since we have used 24 cubes in making these cuboids, volume of each

cuboid is 24 cubic units. Thus we may arrive at the conclusion that the

volume of each cuboid is equal to the product of its length, breadth and

height: volume of the cuboid = l × b × h. Since l × b = area of the base, we

can also write that

volume of the cuboid = area of the base× height.

Think it over!

There are 36 cubes having of side-length 1cm. How many

cuboids of different dimensions can be prepared by using

all of them.

Example 6 A match box measure 4 cm, 2.5 cm and 1.5 cm. What will be

the volume of the packet containing 12 such boxes.

Solution: The volume of each box is 4 × 2.5 × 1.4 = 15 cm3. Volume of a

packet containing 12 such boxes is 15× 12 = 180 cm3.

Example 7 How many 3 meter cubes can be cut from a cuboid measuring

18 m × 12 m × 9 m.

Solution: Volume of the cuboid = 18 × 12 × 9 m3. Volume of each cube to

be cut is = 3 × 3 × 3 m3. Hence number of cubes that can be cut out from

the cuboid is18× 12× 9

3× 3× 3= 72.

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338 Unit 1

Exercise 4.1.3

1. Three metal cubes whose edges measure 3 cm, 4 cm and 5 cm re-

spectively are melted to form a single cube. Find (i) side-length (ii)

total surface area of the new cube. What is the diifernece between the

total surface area of the new cube and the sum of total surface areas

of the original three cubes?

2. Two cubes, each of volume 512 cm3 are joined end to end. Find the

lateral and total surface areas of the resulting cuboid.

3. The length, breadth and height of a cuboid are in the ratio 6:5:3. If

the total surface area is 504 cm2, find its dimension. Also find the

volume of the cuboid.

4. How many m3 of soil has to be excavated from a rectangular well 28

m deep and whose base dimensions are 10 m and 8 m. Also find the

cost of plastering its vertical walls at the rate of 15/m2.

5. A solid cubical box of fine wood costs 256 at the rate 500/m3.

Find its volume and length of each side.

Additional problems on “Mensuration”

1. Find the total surface area and volume of a cube whose length is 12

cm.

2. Find the volume of a cube whose surface area is 486 cm2.

3. A tank, which is cuboidal in shape, has volume 6.4 m3. The length

and breadth of the base are 2 m and 1.6 m respectively. Find the

depth of the tank.

4. Find the area of four walls of a room having length, breadth and

height as 8 m, 5 m and 3 m respectively. Find the cost of white-

washing the walls at the rate of Rs. 15/m2.

5. A room is 6 m long, 4 m broad and 3 m high. Find the cost of laying

tiles on its floor and four walls at the cost of Rs. 80/m2.

6. The length, breadth and height of a cuboid are in the ratio 5:3:2. If its

volume is 35.937 m3, find its dimension. Also find the total surface

area of the cuboid.

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Mensuration 339

7. Suppose the perimeter of one face of a cube is 24 cm. What is its

volume?

8. A wooden box has inner dimensions l = 6 m, b = 8 m and h = 9 m and

it has uniform thickness of 10 cm. The lateral surface of the outer

side has to be painted at the rate of Rs. 50/ m2. What is the cost of

painting?

9. Each edge of a cube is increased by 20%. What is the percentage

increase in the volume of the cube?

10. Suppose the length of a cube is increased by 10% and its breadth is

decreased by 10%. Will the volume of the new cuboid be the same as

that of the cube? What about the total surface areas? If they change,

what would be the percentage change in both the cases?

Glossary

Mensuration: finding the area of a planar objects or volume of a three

dimensional objects.

Solids: objects which occupy space in three dimensions.

cube: a cuboid having equal length, breadth and height.

Lateral surface: the surface of a cuboid which is neither a base nor the

top surface.

Edges: the line along which surfaces meet.

Surface area: the area of the faces of a cuboid.

Volume: measure of the space occupied by a solid.

Points to remember

• A solid is three dimensional figure; it occupies space(in three dimen-

sions).

• A solid has two quantities associated with it; its surface area and it

volume.

• Area is measured in square units whereas volume is measured in

cubic units.

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340 Answers


Exercise 4.1.2

1. 45 m2. 2. 36 m2. 3. 4160. 4. 5088. 5. 14,800 cm2. 6. 8

cm. 7. 4 cm. 8. 1280. 9. (i) cube. (ii) 4 times.

Exercise 4.1.3

1. (i) 6 cm; (ii) 216 cm2; (iii) 84 cm2. 2. L.S.A= 384 cm2 and T.S.A=

640 cm2. 3. dimension: length=12 cm, breadth= 10 cm, height=6 cm;

volume= 720 cm3. 4. 2240 m3 and 15,120. 5. volume = 0.512 m3;

side-length= 80 cm.

Additional problems on “Mensuration”1. 864 cm2, 1728 cm3. 2. 729 cm3. 3. 2 m. 4. 78 m2 and 1,170.

5. 6,720. 6. 1.65 m, 0.99 m, 0.66 m; T.S.A= 6.7918 m2. 7. 216

cm3. 8. 8,931.5 9. 1.728 % 10. Volume decreases by 1% and T.S.A

decreases by 2%.

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OPTIONAL PROBLEMS

These problems are included to pose challenge to those stu-

dents who are looking for it. These are neither for class

room discussion nor for examination.

1. Find a proper fraction greater than 1/3, given that the fraction does

not change if the numerator is increased by a positive integer and the

denominator is multiplied by the same positive integer.

2. Find all rational numbers p/q such that

p

q=

p2 + 30

q2 + 30.

3. Show that the number of distinct remainders which occur when a

perfect square is divided by an odd prime p is (p + 1)/2.

4. Find the number of positive divisors of 22, 32, 42, 52, 102. Do you

see that the number of divisors is odd? Prove the proposition that

the number of positive divisors of a perfect square is always an odd

number.

5. Find all odd natural numbers n for which there is a unique perfect

square strictly between n2 and 2n2.

6. A person was born in 19-th century. His age was x years in the year

x2. If he passed away in 1975, what was his age at the time of his

demise?

7. There is a unique 4-digit number n = abcd such that n2 also ends in

abcd. Find this number.

8. In the adjoining figure, you are given

a skeleton of a 4 × 4 magic square

which uses numbers from 1 to 16.

Find A and B. Hard: Complete the

magic square.

14 11 5 A

8

12 3

B

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342 Optional Problems

9. (Hard) Prove that the magic sum of a 3×3 magic square is three times

the central number.

10. The sum of the squares of three natural numbers is divisible by 9.

Prove that one can select two among these three such that their dif-

ference is divisible by 9.

11. Suppose p and p2 + 2 are prime numbers. Prove that p3 + 2 is also a

prime number.

12. Find all perfect squares which when divided by 11 give a prime as

quotient and 4 as remainder.

13. The number 60 is written on a board. Two players take turn and

play the following game: they can subtract any positive divisor of

the number on the board and replace the number by the result of

this subtraction. The player who writes 0 on the board wins. Which

player wins, the first or the second?

14. There are three pile of stones: one with 10 stones, another with 15

stones and the third with 20. Two players play the following game: at

each turn, a player can choose one of the pile and divide it in to two

smaller pile. The player who cannot do this is the loser. Who will win,

the first or the second?

15. The numbers 1 to 20 are written on a black board in a row. Two

players take turns and put either + sign or − sign between these

numbers according to their choice, one at a time. After all the signs

are put, the sum is evaluated. The first player wins if the number

obtained is even and the second wins if the number is odd. Who

wins?

16. Suppose a natural number n is such that m2 < n < (m + 1)2 for some

natural number m. If n− l = m2 and n+ k = (m+ 1)2, prove that n− kl

is a perfect square.

17. If x, y, z are integers such that x2 + y2 = z2, prove that one of x, y is

divisible by 3.

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Optional Problems 343

18. (Hard) Suppose x, y, z are three natural numbers such that they do not

have any common factor and x2 + y2 = z2. Prove that xyz is divisible

by 60.

19. For any x, suppose [x] denotes the largest integer not exceeding x; for

example [2.5] = 2 and [−1.6] = −2. Find all positive real numbers a

such that a[a] = 8.

20. How many positive integers less than 1000 are 6 times the sum of

their digits?

21. How many palindromes between 1000 and 10000 are there which are

divisible by 7?

22. (Hard) Suppose there are two mirrors inclined at an angle 8◦. A ray of

light starts at the point A gets reflected in B and then gets reflected

in C and so on n times in successive mirrors until it hits a mirror at

right angle and then traces back the same path. What is the largest

possible value of n?(The following figure is an example of 7 reflections.)

O A

8

B

C

23. Show that the shortest distance from a point to a line is the perpen-

dicular distance.

24. (Hard) Let P be any point inside a triangle ABC. Prove that

BC + CA+ AB

2< PA+ PB + PC < BC + CA+ AB.

25. Let D be the mid-point of the side BC in a triangle ABC. Prove that

AB + AC > 2AD.

26. Suppose AD and BE are respectively the medians drawn from A and

B on to the opposite sides in a triangle ABC. If BC > CA, prove that

BE > AD.

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344 Optional Problems

27. Let ABCD be a quadrilateral in which AB is the smallest side and CD

is the largest side. Prove that ∠A > ∠C and ∠B > ∠D.

28. In the adjoining figure, a corner of

a shaded star is at the mid-point

of each side of the large square.

What fraction of the large square is

shaded? A B

CD

P

Q

R

S

29. In the adjoining figure, the outer

equilateral triangle ABC has area 1

and the points D,E, F are such that

DB = EC = FA and each equal to

one-fourth the side of the triangle

ABC. What is the area of DEF . A D B

E

C

F

30. In the quadrilateral ABCD, AB = 5, BC = 17, CD = 5 and DA = 9. It

is known that BD is an integer. What is BD?

31. In a triangle ABC, AB = 2AC. Let D,E be points respectively on the

segments AB,BC such that ∠BAE = ∠ACD. Let F be the point of

intersection of AE and CD. Suppose CFE is an equilateral triangle.

What is ∠ACB?

32. Let ABC be a triangle and let AD be the bisector of ∠A with D on BC.

Prove that AB/AC = BD/DC.

33. Prove that the sum of the interior angles of pentagon is 540◦. What

do you expect for a hexagon? What about an octogon? Can you

generalise this to a general n-gon? Can you prove your guess? What

tools you need?

34. Suppose ABCD is a parallelogram. Equilateral triangles CBX and

DCY are constructed externally respectvely on the sides DC and CB.

Prove that AXY is an equilateral triangle.

35. One dimension of a cube is increased by 1, another is decreased by 1

and the third remains as it is. The volume of the resulting cuboid is 5

less than that of the original cube. What is the volume of the original

cube?

36. A solid cube has side length 3 units. A 2×2 square hole is cut into the

centre of each face. The edge of each square is parallel to the sides of

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Optional Problems 345

the cube and each cut goes all the way through the cube. What is the

volume of the resulting solid?

37. Let ABCD be a trapezium in which AB ‖ CD and AD ⊥ DC. Suppose

AB > BC and draw CM ⊥ AB. Suppose BC = 5 cm, MB = 3 cm and

DC = 8 cm. Find the perimeter of ABCD. What is the area of ABCD?

38. The diagonals of a rhombus are 24 cm and 10 cm respectively. Find

its sides.

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