yearly plan of mathematics 2015 2016 second semester grade 7 · pdf fileyearly plan of...

Yearly plan of mathematics 2015 – 2016 Second semester grade 7

Teacher's Name & Signature: Senior Teacher's Signature: Supervisor's Signature: Principal's Signature:

Only minimum level of objectives are given in the yearly plan and more can be added further.

The teacher is strongly advised to give more examples and exercises (from the approved list of Math books) than the ones provided in the yearly plan.

The teacher may solve the problems in any scientific way (rather than the suggested methods in the yearly plan).

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Identify the algebraic term and algebraic expression.

3 apples n apples (in a basket)

↓ ↓ constant variable

A constant is a number of its own that is used to represent a known

quantity.

A variable is a letter or a symbol that is used to represent an unknown

quantity.

Algebraic term :

is either a number or a number multiplied by one or more variables.

Examples of algebraic term: 3 , m , 2n

3 m 2n

A (numerical) coffectiont of an algebriac term :

is a number that is multiplied by a variable or variables.

The degree of an algebriac term (in one variable) :

is the highest power of the variable.

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Example of Algebraic Term Numerical Coefficient Degree

3𝑥5 3 5

𝑥14 → rewrite as 1𝑥14 1 14

7 → rewrite as 7𝑥0 7 0 1

2𝑝 → rewrite as

1

2𝑝1

1

2 1

e.g.(1): Choose the correct answer :

1) The numerical coefficient of the algebraic term 2𝑥3𝑦4𝑧5 is :

a) 2 b) 3 c) 4 d) 5

2) The numerical coefficient of the algebraic term (−5𝑎)2 is :

a) −25 b) −10 c) 10 d) 25

3) The degree of the algebraic term (−5𝑎3)2 is :

a) 2 b) 3 c) 5 d) 6

An algebraic expressions is an expression formed from any

combination of numbers and variables by using the operations of

addition, subtraction, multiplication, division, exponentiation (raising

to powers), or extraction of roots.

contain any number of algebraic terms.

do not contain an equality sign (=).

Example : 3 + m + 2n ( + + 2× )

The degree of algebraic expression (in one variable):

Is the highest degree of its algebraic terms.

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1) One of the following is an algebraic term :

a) 5𝑥2 b) 5𝑥2 + 1 = 2 c) 5𝑥2 − 𝑥 d) 5𝑥2 + √𝑥

2) One of the following is not an algebraic expression :

a) 5𝑥2 b) 5𝑥2 + 1 = 2 c) 5√𝑥 𝑦3

d) 5 𝑧 𝑦

𝑥2

3) The number of algebraic terms in the expression

2𝑎 + 3 + 𝑎𝑏2 is :

a) 2 b) 3 c) 4 d) 5

4) The degree of the algebraic expression 4𝑧 + 3 𝑧2 is :

a) 1 b) 2 c) 3 d) 4

5) The degree of the algebraic expression 5𝑥3 − 3𝑥 + 2 equals

the degree of :

a) 2𝑥 − 5𝑥2 + 3 b) 3𝑥 + 4 − 2𝑥4 + 3𝑥3

c) 2𝑦3 + 3𝑦2 d) 5𝑦3 − 3𝑦4 + 2

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Evaluate algebraic expressions.

e.g.(1): Evaluate (3𝑛 + 5) when:

a) 𝑛 = 2

b) 𝑛 = 1

3

c) 𝑛 = −4

e.g.(2): Evaluate (2𝑥2 + 3𝑥𝑦 − 2) when 𝑥 = 1 and 𝑦 = 2

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Use and define the terms monomial, binomial, trinomial and polynomial.

A polynomial is an algebraic sum, in which no variables appear in

denominators or under radical signs, and all variables that do appear

are raised only to positive-integer powers.

The degree of a polynomial (in one variable) :

Is the highest degree of its algebraic terms.

Polynomials are usually written in descending order of terms. the term

with the highest exponent in the polynomial is usually written first.

A monomial is a polynomial that has one term.

A binomial is a polynomial that has two terms.

A trinomial is a polynomial that has three terms.

Types of

polynomial Example

Descending order of

terms

Degree of

polynomial

Monomial 2𝑥9 2𝑥9 9

Binomial 10𝑦 − 7𝑦2 −7𝑦2 + 10𝑦 2

Trinomial 25𝑎4 − 𝑎8 + 1.3𝑎3 −𝑎8 + 25𝑎4 + 1.3𝑎3 8


1) One of the following algebraic expressions is a polynomial:

a) 1

√2𝑥2 − √8 + 3.7𝑥 b) 2𝑥 +

1

2𝑥− 4

c) (𝑥2 − 2𝑥) ÷ (𝑥2 + 𝑥) d) 6 + √𝑥 − 𝑥 − 15𝑥2

2) One of the following expressions is a binomial:

a) 2𝑥2 b) 𝑥2 + 𝑥 − 4

c) 𝑥2 − 2𝑥4 + 𝑥3 − 1 d) 1 − 15𝑥2

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Add and subtract polynomials (in one variable).

e.g.(1):Write each polynomial in descending order of terms :

a) 𝑥2 − 2𝑥 + 𝑥3 − 1 b) 𝑦2 + 2𝑦5 + 2 − 𝑦3

To add or subtract two polynomials, we combine like terms.

Two terms are like terms if they each have the same variables

and the corresponding variables are raised to the same powers.

Polynomials can be added/subtracted horizontally or vertically.


1) One of the following pairs of terms is a pair of like terms:

a) 2𝑎 , 2𝑏 b) 2𝑎 , 𝑎2

c) 2 , 2𝑎 d) 𝑎2 , 2𝑎2

e.g.(3): Find :

a) (𝑥2 − 6𝑥 + 5) + (−3𝑥2 + 5𝑥 − 9)

= 𝑥2 − 3𝑥2⏟ −6𝑥 + 5𝑥⏟ +5 − 9⏟ Group like terms.

= −2𝑥2 − 𝑥 − 4 Add like terms.

b) (−5𝑎3 + 3𝑎 − 7) + (4𝑎2 − 3𝑎 + 7)

When adding vertically, we line up the like terms :

−5𝑎3 + 3𝑎 − 7 4𝑎2 − 3𝑎 + 7−5𝑎3 + 4𝑎2

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e.g.(2): Find (3𝑥2 + 2𝑥 − 5) − (4𝑥2 − 7𝑥 + 2)

Change the sign of each term of the second polynomial.

= (3𝑥2 + 2𝑥 − 5) + (−4𝑥2 + 7𝑥 − 2)

= 3𝑥2 + (−4𝑥2)⏟ +2𝑥 + 7𝑥⏟ +(−5) + (−2)⏟ Group like terms.

= −𝑥2 + 9𝑥 − 7 Add like terms.

e.g.(3): Subtract (5𝑦2 − 7𝑦 − 6) from (4𝑦3 − 3𝑦 + 2) When subtracting vertically, we line up the like terms :

4𝑦3 − 3𝑦 + 2

− (5𝑦2 − 7𝑦 − 6)

Now change the sign of each term of the second polynomial and

add the like terms :

4𝑦3 − 3𝑦 + 2

−5𝑦2 + 7𝑦 + 64𝑦3−5𝑦2 + 4𝑦 + 8

e.g.(4): Jamila traveled 2𝑥 + 50 miles in the morning and 3𝑥 − 10

miles in the afternoon. Write a polynomial that represents the total

distance that she traveled.

e.g.(5): Write an expression for the perimeter of the given figure :

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Multiply a monomial by a monomial.

⇒ Involving revision of exponents rules and sign rules

e.g.(1): Simplify : a) (−4𝑥2). (3𝑥) b) −2𝑎2. −3𝑎3

Multiply the numbers and the variables separately

a) (−4𝑥2). (3𝑥) = [(−4). (3)](𝑥2. 𝑥)

= −12𝑥3

b) −2𝑎2. −3𝑎3 = [(−2). (−3)](𝑎2. 𝑎3)

= 6𝑎5

e.g.(2): What is the area of a square with a side length of (5𝑦) 𝑐𝑚 ?

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Solving equations of one variable of first degree using concrete materials

and manipulatives.

An equation is a mathematical sentence in which

two expressions are joined by an equal sign (=).

To keep the balance level, what you do to one side of the "=" you

should also do to the other side.

e.g.(1): Given that represents 1 and represents 𝑥 , write the

given problem as an equation then solve it to find the value of 𝑥.

2𝑥 + 7 = 𝑥 + 10 𝑥 = 3

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e.g.(2): Solve 3𝑥 + 3 = 2𝑥 + 5 using algebra tiles.

Represent the equation by

algebra tiles.

Eliminate common algebra

tiles from both sides.

Write the solution:

𝑥 = 2

e.g.(3): Solve 2𝑥 + 3 = 11 using algebra tiles.

Represent the equation by algebra tiles. Then eliminating

common algebra tiles from both sides.

We want to get 𝑥 alone for a solution. First, we can make two

groups of equal numbers of tiles on each side of the bar. Then

we can remove one set of the tiles from each side of the bar.

x = 4

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Solving equations of one variable of first degree using guess and check

method.

e.g.(1): solve 2𝑥 + 1 = 3 by guess and check method.

Step1: Guess a number to replace the missing variable with.

we will guess 𝑥 = 2.

Step2: Insert the guessed number into the equation.

2(2) + 1 = 3 Step3: Solve the equation.

2(2) + 1 = 3

4 + 1 = 3

5 = 3

𝑥 is not equal to 2 since 5 is not equal to 3.

Step4: Guess a new variable if you are incorrect as in the above

example.

we will guess 𝑥 = 1

Step5: Insert the new number into the equation.

2(1) + 1 = 3 Step6: Solve the equation.

2(1) + 1 = 3

2 + 1 = 3

3 = 3

The guess of 𝑥 = 1 is correct since the equation solves to 3 = 3.

Step7: Continue to guess new variables if you have not gotten a

correct answer.

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Solving equations of one variable of first degree algebraically Of the

following forms:

𝑥 ± 𝑎 = 𝑏

𝑎𝑥 = 𝑏

𝑥

𝑎= 𝑏

𝑎𝑥 ± 𝑏 = 𝑐 where a, b and c are integers

Revision: Complete the following:

e.g.(1): Solve 𝑥 + 5 = 12 algebraically.

What you are aiming for is an answer like "𝑥 = ...", and the plus

5 is in the way of that.

So, let us have a go at adding

additive inverse of 5 to both sides: 𝑥 + 5 + (−𝟓) = 12 + (−𝟓)

A little arithmetic

(5-5 = 0 and 12-5 = 7) becomes: 𝑥 + 0 = 7

Which is : 𝒙 = 𝟕

(Quick Check: 7 + 5 = 12)

Number Additive Inverse Multiplicative Inverse

3 -3 1

3

…….. 1

4 ………

…….. ……… -7

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e.g.(2): Solve 5𝑥 – 3 = 12 algebraically.

5𝑥 − 3 = 12 5𝑥 − 3 + 3 = 12 + 3

5𝑥 = 15

1

5 × 5𝑥 = 15 ×

1

5

𝑥 = 3

Adding additive inverse of

(−3) to both sides

Multiplying by multiplicative

inverse of (15)

(Quick Check: 5 × 3 – 3 = 15 – 3 = 12)

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Translate between a word expression and its variable representation to

apply and solve problems.

e.g.(3): Find the solution for each of the followings:

1. Five added to some unknown number is equal to eight.

2. Fifty-four decreased by twice a number gives ten.

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Define, illustrate and identify a sequence.

Introduction:

Find the next term and describe the pattern:

a) 2, 4, 6, 8, ….

b) 2, 4, 8, 16, ….

a) The next term is 14. We can get the next term by adding 2.

b) The next term is 32. We can get the next term by multiplying

by 2.

The above patterns are called sequences.

A Sequence is a set of numbers that are in order.

The first sequence is called an arithmetic sequence and the second

pattern is called a geometric sequence.

An Arithmetic Sequence is made by adding some value each time.

Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, ...

This sequence has a difference of 3 between each number.

The sequence is continued by adding 3 to the last number each time.

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The value added each time is called the "common difference" and it

could be positive or negative value.

In General you could write an arithmetic sequence as follow:

𝑎 , 𝑎 + 𝑑 , 𝑎 + 𝑑 + 𝑑 , … Where :

𝒂 is the first term

𝒅 is the common difference (between one term and the next)

e.g.(1): Find the next three terms in the following arithmetic sequences:

a) 3, 8, 13, 18, …

b) 25, 23, 21, …

a) The first term in this sequence is 3 and the common difference

is 5 between each number. The sequence is continued by

adding +5 to the last number each time.

3, 8, 13, 18, 18+5 , 18+5+5 , 18+5+5+5 , …

3, 8, 13, 18, 23, 28, 33 , …

b) The first term in this sequence is 25 and the common

difference is -2 between each number. The sequence is

continued by adding -2 to the last number each time.

25, 23, 21, 21+ (-2) , 17+ (-2)+ (-2) , 15+ (-2)+ (-2)+ (-2) , ...

25, 23, 21, 19, 17, 15, ...

e.g.(2): Write the first three terms of the arithmetic sequences:

1) 𝑎 = 6 and 𝑑 = 11

2) 𝑎 = 6 and 𝑑 = −5

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A Geometric Sequence is made by multiplying by some value each

time.

Example: 2, 4, 8, 16, 32, 64, 128, 256, ...

This sequence has a factor of 2 between each number.

The sequence is continued by multiplying by 2 each time.

The value multiplied each time is called the "common ratio".

In General you could write a geometric sequence as follow:

𝑎 , 𝑎 × 𝑟 , 𝑎 × 𝑟 × 𝑟 , … Where :

𝒂 is the first term

𝒓 is the common ratio (the factor between the terms)

e.g.(1): Choose the correct answer:

One of the following sequences is a geometric sequence:

a) 5 , 9, 13, … b) 1, 4, 9, …

c) 2 , 6, 18, … d) 9 , 7 , 5, …

e.g.(2): Write the next three terms in the geometric sequences 2, 8, 24,…

e.g.(2): Write the first three terms of the geometric sequences:

1) 𝑎 = 9 and 𝑟 = 1

3

2) 𝑎 = 1 and 𝑟 = −2

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Using Schedules and Time lines (24-hour time).

How many minuets in 1 hour?

In the following digital clock , what does 06 and 30

represent?

In 12-hours clock, we use the words A.M. and P.M. to refer to time that

is before noon and after noon.

In 24-hour clock, the words A.M. and P.M.

are not used. We used 4-digit number.

24 Hour Clock: the time is shown as how

many hours and minutes since midnight.

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Converting A.M./P.M. to 24 Hour Clock

From 1:00 A.M. to 12:59 P.M. From 1:00 P.M. to 11:59 P.M.

no change add 12 Hours

11:20 A.M. = 11:20

12:30 P.M. = 12:30

4:45 P.M. = 16:45

11:50 P.M. = 23:50

e.g.(1): Convert the following times from the 12-hour clock to 24-hour:

12-hour clock 24-hour clock

7 : 27 P.M.

8 : 45 A.M.

12 : 55 P.M.

5 : 28 P.M.

Converting 24 Hour Clock to A.M./P.M. :

From 1:00 to 11:59 From 13:00 to 23:59

just make it "A.M." subtract 12 Hours, then make it

"P.M."

1:15 = 1:15 A.M.

11:25 = 11:25 A.M.

14:55 = 2:55 P.M.

23:30 = 11:30 P.M.

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e.g.(2): Convert the following times from the 24-hour clock to 12-hour:

24-hour clock 12-hour clock

11 : 00

21 : 39

06 : 19

17 : 30

Adding and subtracting time in 24-hour clock :

Adding Times : Add the hours and add the minutes separately.

e.g.(1): A boy starts eating his breakfast at 15:30. it takes 15 minutes.

What time does he finish?

15 30 min+ 15 min 15 45 min

e.g.(2): A cartoon movie started at 16:30. It took 1:45 minutes to

finish. What time does the movie end?

Method 1 : 1 6 30 min+ 1 45 min 17 75 min = 18: 15 min

17 75 min = 17 + 60 min + 15 min

= 18 + 15 min

Method 2 :

1 hour 30 min 15 min

16:30 17:30 18:0 18:15

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Subtracting Times :

Subtract the hours and subtract the minutes separately.

e.g.(1): A plane left an airport at 14:30. It arrived at 17:15. How long

was the flight? Method 1 :

17 15 min = 16 + 60 min + 15 min

= 16 + 75 min

17 15 min−1 4 30 min

==

16 75 min− 14 30 min 2 h 45 min

Method 2 :

30 min 1 hour 1 hour 15 min

14:30 15:30 6:00 17:00 17:15

e.g.(2): Said went to his school at 7:15 and got home at 14:05 . How

long was he at the school?

e.g.(3): Find the missing start time or end time ( in 24-hour clock):

Start time End time Time Duration

09 : 17 -------------- 3h 27 min

-------- 07 : 08 44 min

23 : 45 --------- 25 min

-------- 02 : 22 6h 10 min

T

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Solve problems involving time zones.

A time zone is one of the areas into which the world is divided where

the time is calculated as being a particular number of hours behind or

ahead at Greenwich (GMT), in UK .

There are 24 time zones in the world, demarcated approximately by

meridians at 15° intervals, an hour apart.

e.g.(1): India is five hour ahead of the UK time. If the time in GMT is

13:05, what is the time in India?

For each movement to the left For each movement to the right

Subtract 1 hour Add 1 hour

India is located in the fifth area east of GMT line. Therefore, the

time in India= 13:05+5 = 18:05

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General Rule :

If you are travelling East to a country which is ahead of your own,

you will add hours to the time.

If you are travelling West to a country which is behind your own,

you will subtract hours from the time.

e.g.(2): Muscat is 4 hours ahead of the UK time and New York is 5

hours behind the UK time. If the time in Muscat is 20:00, what is

the time in New York?

The time in New York is (20:00 – 9 = 11:00).

e.g.(3): Bermuda is 4 hours behind the UK time. If it is 1641 in

Bermuda, what is time in UK?

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Extend the coordinate plane to 4 quadrants.

Coordinate Plane :

The plane determined by a horizontal number line, called the x-axis,

and a vertical number line, called the y-axis, intersecting at a point

called the origin which is written as (0,0).

x-axis : is a horizontal scale. As you go to the right from zero, the

values are positive. As you go to the left from zero, the values are

positive.

y-axis : is a vertical scale. As you go up from zero the numbers are

positive. As you go down from zero the values are negative.

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The two axes divide the plane into four areas called quadrants :

1. The first quadrant contains

all the points with positive x

and positive y coordinates.

2. The second quadrant contains

all the points with negative x

and positive y coordinates.

3. The third quadrant contains

all the points with negative x

and negative y coordinates.

4. The fourth quadrant contains

all the points with positive x and negative y coordinates.

Quadrant Sing of coordinates

first (x , y )

second (- x , y )

third (- x , - y )

forth (x , - y )

e.g.(1): In which quadrant is located the point (x, y), such that x and y

both are negative numbers?

e.g.(2): In which quadrant are located the following points:

point Quadrant

(3 , 2)

(-5 , 7)

(- 4, -1)

(9, - 2)

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An ordered pair : is a pair of numbers that name a point on the

coordinate plane. They are written in parentheses as (1 , 2).

To write the order pair of a point:

Step 1: Locate first number by looking down in the horizontal axis.

Step 2: Locate second number by looking in the vertical axis.

Step 3: Write the ordered pair (first number, second number)

e.g.(3): Write the order pair for each point in the given coordinate plane:

e.g.(4): plot the ordered pair A(−4, 3)

Step 1: Start at the origin.

Step 2: The x-coordinate of the given

ordered pair is - 4. So, move 4 units to the

left of the y-axis.

Step 3: The y-coordinate of the given

ordered pair is 3. So, move 3 units up from

the x-axis.

Step 4: Draw a dot and label it A.

e.g.(5): Plot the given points :

A(2,2) , B(-2,1) , C(4,1) , D(1,-3) , E(-3,0)

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Translations of figures on the coordinate plane .

Introduction:

Ask a student to stand in front of the class.

Ask him to move one step. He most probably will move forward

although you didn't determine the direction. Discuss that with them.

Ask him to move to the right. He should ask how many steps to move.

Discuss this as well.

Now determine the distance and the direction and ask the volunteer to

move. Ask the students what has been changed in their friend? His

shape, age, length, weight…etc

So, what we have done now is called translation.

Translation is a geometrical transformation which moves every point

of a shape by same distance in the same direction.

(𝑥, 𝑦) → (𝑥 + 𝑎 , 𝑦 + 𝑏)

by counting how many units the shape has moved up or down and left or right.

If we want to say that the shape gets moved 30 units to the right and

40 units up, we can write:

(x, y) → (x + 30 , y + 40) This says "all the x and y coordinates will become x+30 and y+40"

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(x, y) → (x + a , y + b)

To describe the translation, we use

arrow notation rule:

Where a and b represents the distance

moved on x-axis and y-axis

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e.g.(1): On the given coordinate

plane, translate the

triangle ABC 2 units to

the left and 4 units down.

Method 1 : Move each vertex 2

units to the left and 4 units down.

Method 2:

Find the translation rule:

(x, y) → (x + a , y + b) 2 units to the left a = - 2

4 units down b = - 4

Therefore, the translation rule is:

(x, y) → (x − 2 , y − 4)

A(-1,3) A'(-1-2,3-4)

A'(-3,-1)

B(2,3) B'(2-2,3-4)

B'(0,-1)

C(1,1) C'(1-2,1-4)

C'(-1,-3)

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e.g.(2):

Describe the transformation of the

following rectangle ABCD:

The rectangle ABCD has been

translated ……… units up/down and

…… units left/right.

e.g.(3): The vertices of a triangle ABC are given by the coordinates

A(-3,2), B(1,1), C(-3,-1). Draw the triangle then translate it

according to the rule (x, y) (x, y -2).

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Enlarge and reduce figures on the coordinate plane.

An enlargement is a geometrical transformation which enlarges or

reduces a shape by a given scale factor around a given center point

called center of enlargement.

Center of enlargement: is a point from which we draw lines joining

the corresponding vertices on the image and original shape.

Scale Factor is an amount by which the image grows or shrinks.

Scale factor can be calculated by one of the following ratios:

= 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎 𝑠𝑖𝑑𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑚𝑎𝑔𝑒

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑠ℎ𝑎𝑝𝑒

=𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑣𝑒𝑟𝑡𝑖𝑥 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑚𝑎𝑔𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑒𝑛𝑙𝑎𝑟𝑔𝑒𝑚𝑒𝑛𝑡

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑣𝑒𝑟𝑡𝑖𝑥 𝑜𝑛 𝑡ℎ𝑒 𝑠ℎ𝑎𝑝𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑒𝑛𝑙𝑎𝑟𝑔𝑒𝑚𝑒𝑛𝑡

If the scale factor is greater than 1, the image is an enlargement.

If the scale factor is between 0 and 1, the image is a reduction.

□

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e.g.(1): Enlarge triangle PQR

with O as the center of

enlargement and a scale

factor of 2.

Step 1: Measure line OP̅̅ ̅̅ .

Step 2: Extend the line OP̅̅ ̅̅ to the

point P′ such that OP′̅̅ ̅̅ ̅ = 2OP̅̅ ̅̅ .

Step 3: Repeat the steps for all

the vertices; point Q to

get Q' and point R to get

R'.

Step 4: Join the points P' Q' R' to

form the image of the

triangle.

e.g.(2): The opposite figure

shows an enlargement

transformation. Find the

center of enlargement.

The center of enlargement is

found by drawing straight lines to

join corresponding vertices on the

shape and its image. These lines

are then extended until they meet.

The point which they meet is the

center of enlargement O.

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e.g.(3): Find the scale factor in the given figures:

Scale factor =

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎 𝑠𝑖𝑑𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑚𝑎𝑔𝑒

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑖𝑑𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑠ℎ𝑎𝑝𝑒

=𝐴′𝐵′̅̅ ̅̅ ̅̅ ̅

𝐴𝐵̅̅ ̅̅ ̅=4

2= 2

Given that 𝑅′𝑂̅̅ ̅̅ ̅ = 4 and 𝑅𝑂̅̅ ̅̅ = 8.

Scale factor =

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑣𝑒𝑟𝑡𝑖𝑥 𝑅′ 𝑓𝑟𝑜𝑚 𝑂

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑣𝑒𝑟𝑡𝑖𝑥 𝑅 𝑓𝑟𝑜𝑚 𝑂

=𝑅′𝑂̅̅ ̅̅ ̅̅

𝑅𝑂=4

8=1

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Identify, name and list the properties of the following polygons:

convex

concave

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A Convex Polygon :

A polygon that has all interior

angles less than 180ο.

A Concave Polygon:

A polygon that has one or more

interior angles greater than 180°

Properties of convex polygons : Properties of concave polygons:

1) All the diagonals of a convex

polygon lay entirely inside

the polygon.

2) A straight line drawn through

a convex polygon crosses at

most two sides.

1) Some of the diagonals of a

concave polygon will lay

outside the polygon.

2) At least one straight line can

be drawn through a concave

polygon that crosses more

than two sides.

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http://www.mathopenref.com/polygoninteriorangles.html



http://www.mathopenref.com/polygondiagonal.html

http://www.mathopenref.com/polygondiagonal.html






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e.g.(1): Classify the followings into concave and convex polygons:

(a) (b) (c) (d)

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Recognize, draw, name and describe:

Complementary angles (add-up to 90̊)

Supplementary angles (add-up to 180̊)

Vertically opposite angles

Alternate interior and exterior angles

Corresponding angles

Same side interior angles

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Complementary angles (add-up to 90̊)

Two angles are complementary if the sum of their measure is 90°

∠ ABC and ∠ 𝐸𝐹𝐺 are complementary angles because 30° + 60° = 90°

e.g.(1): In the given figure, 𝑥°and 50° are

complementary angles. Find the size of ∠ x.

𝑥 = 90° − 50° = 40°

e.g.(2): Find the complementary angle for each of the given angles:

Angle Complementary angle

27° -------------------------------------

76.5° --------------------------------------

32.8° -------------------------------------

e.g.(3): In the given figure, find the value of x.

54° + (4 𝑥)° = 90° (4 𝑥)° = 90° − 54° (4 𝑥)° = 36°

𝑥 = 36

4= 9

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Supplementary angles (add-up to 180̊)

Two angles are supplementary if the sum of their measure is 180°

∠ABC and ∠𝐸𝐹𝐺 are supplementary angles because 120° + 60° = 180°

e.g.(1): In the given figure, 𝑥°and 50° are supplementary angles. Find the

size of ∠ x.

180° − 50° = 130°

e.g.(2) : find the supplementary angle for each of the given angles:

Angle Supplementary angle

39° -------------------------------------

145.3° --------------------------------------

170° -------------------------------------

e.g.(3): From the given figure, find the value of x.

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e.g.(4) : from the given figure

a) Name the complementary angles.

b) Name the supplementary angles.

c) find the size of angles x and y.

a) ∠𝐴𝐵𝐷 𝑎𝑛𝑑 ∠𝐴𝐵𝐸are complementary angles.

b) ∠𝐴𝐵𝐶 𝑎𝑛𝑑 ∠𝐴𝐵𝐷 are supplementary angles.

∠𝐷𝐵𝐸 𝑎𝑛𝑑 ∠𝐸𝐵𝐶 are supplementary angles.

c) 𝑥° = 90° − 37° = 53° and 𝑦° = 180° − 37° = 143°

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Vertically opposite angles

Vertical Angles are the angles opposite each other when two lines

cross.

a° and b° are vertical angles, and they are equal.

𝑎° = 𝑏°

c° and d° are vertical angles, and they are equal.

𝑐° = 𝑑°

e.g.(1): From the given figure

1) Name the vertically opposite angles.

2) Find the size of angle x.

1) Vertically opposite angles are :

( ∠𝐴𝐵𝐸, ∠𝐶𝐵𝐷)

( ∠𝐴𝐵𝐶, ∠𝐸𝐵𝐷)

2) 𝑥 = 45° because ∠𝐴𝐵𝐸 = ∠𝐶𝐵𝐷 ( Vertically opposite angles)

e.g.(2): Find the value of x :

(a) (b)

(2𝑥 + 15)° = 40° (8𝑥 − 17)° = 143° 2𝑥 = 40 − 15 8𝑥 = 143 + 17

2𝑥 = 35 8𝑥 = 160

𝑥 = 17.5 𝑥 = 2

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Alternate interior and exterior angles

A Transversal is a line which crosses two lines.

Line C is the transversal.

When two lines are crossed by a transversal, the pairs of angles

on opposite sides of the transversal but outside the two lines are

called Alternate Exterior Angles.

When two lines are crossed by a transversal, the pairs of angles

on opposite sides of the transversal but inside the two lines are

called Alternate Interior Angles.

Alternate Exterior Angles.

∠B = ∠G

∠A = ∠H

Alternate Interior Angles.

∠C = ∠F

∠D = ∠E

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e.g.(1): In the given figures, find the missing values (write the reason).

Y = 128° (Vertically opposite angles) (2𝑥 + 10)° = (120)° X = 52° (supplementary angles) (Alternate Exterior Angles)

𝑍 = 128° (Alternate Exterior Angles) 𝑥 = 55°

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Corresponding angles

When two lines are crossed by a transversal, the angles in matching

corners are called corresponding angles.

corresponding angles :

𝐴° = 𝐸° 𝐵° = 𝐹° 𝐶° = 𝐺° 𝐷° = 𝐻°

e.g.(1): From the given figure, name the corresponding angles?

∠𝐴𝐵𝐶 , ∠𝐻𝐹𝐸

∠𝐴𝐵𝐷, ∠𝐺𝐹𝐻

∠𝐷𝐵𝐹, ∠𝐺𝐹𝐻

∠𝐶𝐵𝐹, ∠𝐸𝐹𝐻

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Two angles that lie on the same side of a transversal and between the

lines cut by the transversal, in corresponding positions with respect

to the two lines that the transversal intersects.


𝐿°𝑎𝑛𝑑 𝑁° 𝑍°𝑎𝑛𝑑 𝑀°


add up to 180°.

𝐿° + 𝑁° = 180° 𝑍° + 𝑀° = 180°

e.g.(1): In the given figure,

if ∡𝐶𝐵𝐹 = 100° find the size of

∡𝐺𝐹𝐵.

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Use a compass and a straight edge to construct the bisector of an angle.

An angle bisector is a line that divides an angle into two equal angles.

∠ABC = ∠CBD

e.g.(1): From the opposite figure, complete :

a) If ∠MON = ∠NOP then NO̅̅ ̅̅ is called an

…………………..

b) If NO̅̅ ̅̅ is an angle bisector of ∠MOP and

∠MON = 20° , then ∠NOP = ……….

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e.g.(1): Use a compass and a straight edge to draw a bisector of angle

PQR.

Step1: Start

with angle

PQR that we

will bisect.

Step2: Place the compass' point

on the angle's vertex Q.

Step3: Draw an arc across each

leg of the angle.

Step4: Place the compass on the

point where one arc crosses a leg

and draw an arc in the interior of

the angle.

Step5: Without changing the

compasses setting repeat for the

other leg so that the two arcs

cross.

Step6: Using a straightedge or

ruler, draw a line from the vertex

to the point where the arcs cross.

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Use a compass and a straight edge to construct an angle congruent to a

given angle.

e.g.(1): The opposite angles are congruent,

find x.

e.g.(2): Use a compass and straight edge to draw an angle congruent to

the angle BAC.

Step1: Start with an angle BAC that

we will copy, make a point P that will

be the vertex of the new angle.

Step2: From P, draw a ray PQ. This

will become one side of the new angle.

Step3: Place the compasses on

point A, set to any convenient width. Step4: Draw an arc across both

sides of the angle creating the points J and K.

Step5: Without changing the

compasses' width, place the compasses' point on P and draw a similar arc there, creating point M as shown.

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Step6: Set the compasses on K and adjust its width to point J.

Step7: Without changing the compasses' width, move the compasses to M

and draw an arc across the first one, creating point L where they cross.

Step8: Draw a ray PR from P through L.

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Use a compass and a straight edge to construct a line parallel to a given

line.

Two line segments are parallel if they never meet/ intersect.

AB̅̅ ̅̅ 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 CD̅̅ ̅̅ (AB̅̅ ̅̅ ∥ CD̅̅ ̅̅ )

e.g.(1): Which of the following lines are parallel lines ?

e.g.(2): Use a compass and a straight edge to draw

a line parallel to the line AB̅̅ ̅̅ and passes

through the point Q.

e.g.(3): Use a compass and a straight edge to draw a line parallel to the

line PQ̅̅ ̅̅ .

Step1: Start with a point R above

PQ̅̅ ̅̅ . Draw a transverse line through R and across the line PQ at an angle, forming the point J where it intersects the line PQ.

Step2: With the compasses' width

set to about half the distance between R and J, place the point on J, and draw an arc across both lines.

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Step3: Without adjusting the

compass' width, move the compass to R and draw a similar arc to the

one in step 2.

Step4: Set compass' width to the

distance where the lower arc crosses the two lines.

Step5: Move the compasses to

where the upper arc crosses the transverse line and draw an arc across the upper arc, forming point

S.

Step6: Draw a straight line through

points R and S.

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Construction of a Parallelogram

A Parallelogram is a quadrilateral with both pairs of opposite sides

parallel.

To construct a parallelogram using a compass and a straight edge:

Step 1: Follow steps of drawing

two parallel lines to get the

following figure:

Step 2: With the compasses' width

set to any distance, place the point on J, and draw an arc across JQ forming point F.


compass' width, move the compass to R and draw a similar arc to the one in step 2 forming point L.


points F and L.

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Construction of a Rhombus A rhombus is a parallelogram with all sides equal in length

To construct a rhombus using a compass and a straight edge:



following figure:


set to the same distance between R and J, place the point on J, and draw an arc across JQ forming point F.


compass' width, move the compass to R and draw a similar arc to the one in step 2 forming point L.


points F and L.

i.e 1) Using a compass and a ruler, draw a rhombus with 4 cm side length.

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Construction of a trapezoid

A trapezoid is a quadrilateral which has only one pair of opposite

parallel sides.

Draw a trapezoid with bases' lengths 3 cm and 5 cm, using a compass

and a ruler:



following figure:


set to distance of 5 cm, place the point on J, and draw an arc across JQ forming point F.


set to distance of 3 cm, place the point on R, and draw an arc across RS forming point L


points F and L.

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Calculate the area (using a formula) for a trapezoid.

Activity: Finding the Area of a Trapezoid Required materials: Cards, pencils, scissors, ruler and tape.

Take a card and create a trapezoid by cutting off two triangles.

Label the bases and the height on the card.

Trace the trapezoid onto another card to create a congruent

trapezoid. Tape the two trapezoids together to create a

parallelogram.

The area of the parallelogram is (𝑏1 + 𝑏2) × ℎ .

Since the parallelogram is formed from 2 congruent trapezoids,

the area of the trapezoid is half that of the parallelogram.

Area of a Trapezoid = 1

2 × (𝑏1 + 𝑏2) × ℎ

e.g.(1): Calculate the area for each of the following trapezoids:

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Calculate the area (using a formula) for a composite figure.

Introduction: Write the formula of the area for each of the following shapes:

……………

……………

……………

……………

……………

……………

……………

e.g.(1): Calculate the area of the following figures:

a) b)

e.g.(2): In the opposite figure, calculate the area of

shaded part.

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Measure and calculate the surface area of a cube.

A face is one of the flat sides of the 3D object.

How many faces in a cube?

What is the shape of the face in a cube?

A net of a solid is a figure which can be folded to form the solid.

Surface area is the sum of the areas of the faces of a solid figure.

The area of a square 𝐴 = 𝑎 × 𝑎 = 𝑎2

So, the Surface area of a cube is: 𝑆𝐴 = 6 𝑎2 (square units)

Lateral surface area of a solid is the sum of the surface areas of all

its faces excluding the bases of the solid.

lateral surface area of a cube will be the area of four faces:

𝑎 × 𝑎 × 4 = 4𝑎2

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e.g.(1): For the given cube, find:

a) Its lateral surface area b) Its surface area

a) lateral surface of a cube = 4𝑎2

LA = 4(2𝑐𝑚)2

LA = 4 × 4 𝑐𝑚2

L 𝐴 = 16 𝑐𝑚2

b) Surface area 𝑜𝑓 𝑐𝑢𝑏𝑒 = 6𝑎2

SA = 6 × 2

SA = 6 × 4 = 24 𝑐𝑚2

e.g.(2): What is the surface area of a cube with an edge of 5 𝑚?

Surface area 𝑜𝑓 𝑐𝑢𝑏𝑒 = 6𝑎2

SA = 6 × (5)2

SA = 150 𝑚2

e.g.(3): The surface area of a cube is 96 𝑚2. What is the length of an

edge of the cube?

Surface area 𝑜𝑓 𝑐𝑢𝑏𝑒 𝑆𝐴 = 6𝑎2

96 𝑚2 = 6 𝑎2 ( ÷ 6 )

16 𝑚2 = 𝑎2 ( taking the square root)

𝑎 = 4 𝑚

e.g.(4): How much gift wrap is needed to cover a box

shown in the picture?

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e.g.(5): If the surface area of a Rubik's Cube is

216 𝑐𝑚2.

a) what is the area of one face?

b) what is the length of one edge of the cube?

c) what is the lateral surface of a cube?

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Measure and calculate the surface area of a rectangular prism.

Surface Area of a rectangular prism :

SA = 2𝑙ℎ + 2𝑤ℎ + 2𝑙𝑤

(lateral surface are + area of the two bases)

lateral surface Area of a rectangular prism :

LA = (2 𝑙 + 2 𝑤 ) × ℎ

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e.g.(1): Find the surface area and lateral surface

area for the opposite rectangular prism?

SA = 2lw + 2lh + 2wh

SA = 2 × 5 × 4 + 2 × 5 ×7 + 2 × 4 × 7

SA = 40 + 70 + 56

SA = 166 𝑐𝑚2

LA = (2 l + 2 w ) x h

LA= (2× 7 + 2 × 4) × 4

LA= 88 𝑐𝑚2

e.g.(2): How many square centimeters of

cardboard does Ahmed need to make a

rectangular prism with length of 6 cm ,

width of 5 cm, and height of 4cm?

SA = 2lw + 2lh + 2wh

SA = 2 × 6 × 5 + 2 × 6 ×4 + 2 × 5 × 5

SA =158 cm2

e.g.(3): Salim needs to paint the sides of a rectangular prism. The prism

has a length of 25 mm, a width of 15 mm, and a height of 9 mm.

How much paint does he need to cover the sides?

LA = (2 l + 2 w ) x h

LA= (2× 25 + 2 × 15) × 9

LA= 720 mm2

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Estimate, then calculate (using a formula) the volume of a cube.

Determine the edge of cubes when the volume is known.

Volume of a cube = 𝑎 × 𝑎 × 𝑎 = 𝑎³ where 𝑎 is the length of each edge of the cube

e.g.(1): Find the volume of the following solids:

(a) (b)

Volume of a cube = 𝑎³ Volume of a cube = 𝑎³ = 4 × 4 × 4 = 8 × 8 × 8

= 64 cm3 = 512 cm3

e.g.(2): How much water is needed to fill one cube ice in the container

that is 5 𝑚𝑚 deep?

Volume of a cube = (5)3 = 125 𝑚𝑚3

e.g.(3): Find the edge of a cube with a volume of 729 𝑐𝑚3.

(Use prime factorization)

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Estimate, then calculate (using a formula) the volume of a rectangular

prism.

Determine one dimension of rectangular prisms when the volume and

two dimensions are known.

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 rectangular prism = 𝐿𝑒𝑛𝑔𝑡ℎ × 𝐵𝑟𝑒𝑎𝑑𝑡ℎ × 𝐻𝑒𝑖𝑔ℎ𝑡

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 rectangular prism = ℎ × 𝑙 × 𝑤

e.g.(1): Find the volume of the following solids:

(a)

(b)

𝑉𝑜𝑙𝑢𝑚𝑒 = 5 × 4 × 6 = 120 𝑐𝑚3 𝑉𝑜𝑙𝑢𝑚𝑒 = 4 × 3 × 8 = 96 𝑐𝑚3

e.g.(2): A box has a length of 11.2 cm and width of 11.2 cm and a height

of 13 cm what is the volume of the tissue box?

𝑉𝑜𝑙𝑢𝑚𝑒 = 11.2 × 11.2 × 13

= 1630.72 𝑐𝑚3

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e.g.(3): The dimensions of 𝑎 rectangular prism are 3 , 5𝑥 ,2𝑥2 find:

1) The volume of the rectangular prism (in terms of 𝑥).

2) The volume of the rectangular prism when 𝑥 = 2.

1) 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎 rectangular prism = (3). ( 5𝑥). ( 2𝑥2) = (3 × 5 × 2)( 𝑥 × 𝑥2 ) = 30𝑥3

2) Volume = 30 × (2)3 = 240 cubic unit

e.g.(4): The following rectangular prism has a volume of 64 𝑐𝑚3. Find

its height.

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Convert Metric units of volume.

When you change from Larger

units to Smaller units⟹ Multiply.

e.g.(1): Complete:

25.6 𝑐𝑚3 = ⋯𝑚𝑚3

25.6 𝑐𝑚3 = 25.6 × 𝟏𝟎𝟎𝟎 = 256 00 𝑚𝑚3

When you change from Smaller

units to Larger units⟹ Divide.

e.g.(2): Convert

9000 𝑐𝑚3 = ⋯𝑚3 = 9000 ÷ 𝟏𝟎𝟎𝟎 𝟎𝟎𝟎

= 0.009 𝑚3

e.g.(1): Complete:

a) 3 𝑚3 + 125 000 𝑚𝑚3 = ……𝑚𝑚3

b) 35000 𝑚𝑚3 + 6 𝑐𝑚3 = ……𝑐𝑚3

e.g.(2): A box has a volume of 15 𝑐𝑚3. Express this volume in 𝑚𝑚3.

e.g.(3): A cubic tank has an edge of length 2 𝑚. Calculate it's volume in

cubic millimeters.

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Convert a variety of units:

1 mL of water at 4℃ has a volume of 1 cm3 and a mass of 1 g

1 L of water at 4℃ has a volume of 1000 cm3 and a mass of 1 kg

1000 L = 1 kL = 1 m3

1000 L of water at 4℃ has a mass of 1000 kg

A cube of 1 𝑐𝑚 on each side holds 1 𝑚𝑙 of water, and the water would

weigh 1 g.

1 𝑚𝑙 of water weighs exactly 1 g when the water is at a temperature of

4℃. At other temperatures, 1 𝑚𝑙 of water weighs approximately 1 g.

Volume Capacity Mass

1 𝑐𝑚3 = 1 𝑚𝑙 = 1 g

A cube of 10 𝑐𝑚 on each side holds 1 𝑙 of water, and the water would

weigh 1 kg.

Volume Capacity Mass

1000 𝑐𝑚3 = 1 𝑙 = 1 kg

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60 | P a g e

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e.g.(1): a) What is the weight in kilograms of 3.6 𝑚𝑙 of water at 4℃ ?

b) Find the weight of water in grams having a volume of 3 𝑙 at

4℃.

a) 3.6 𝑚𝑙 = 0.0036 𝑙 = 0.0036 kg

b) 3 𝑙 = 3000 𝑚𝑙 = 3000 g

e.g.(1) : A rectangular prism container filled with water. It's dimensions

are: 3m, 2m and 1m. Calculate:

a) Volume of water in the container (in 𝑐𝑚3)

b) Capacity of the container (in Liter)

c) Weight of water in the container (in kg)

a) Volume = 3 × 2 × 1 = 6 𝑚3 = 6 000 000 𝑐𝑚3

b) Capacity = 6 000 000

1000= 6000 𝐿𝑖𝑡𝑒𝑟

c) Weight = 6000 kg

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