Synopsis – Grade 9 Math Term II

Chapter 4: Linear Equations in Two Variables

An equation of the form, ax + by + c = 0, where a, b, and c are real numbers, such that a and

b are both not zero, is called a linear equation in two variables.

For example, 2x + 3y + 10 = 0, 3x + 7y = 0

Equations of the form, ax + b = 0 or cy + d = 0, are also examples of linear equations in

two variables since they can be written as ax +0. y + b = 0 or 0.x + cy + d = 0

respectively.

Solution of a linear equation in two variables

A solution of a linear equation represents a pair of values, one for x and one for y which

satisfy the given equation.

Linear equation in one variable has a unique solution.

Linear equation in two variables has infinitely many solutions.

Geometrical representation of a linear equation in two variables

The geometrical representation of the equation, ax + by + c = 0, is a straight line.

For example: the equation x + 3y = 6 can be represented on a graph paper as follows:

An equation of the form, y = mx, represents a line passing through the origin.

Every point on the graph of a linear equation in two variables is a solution of the linear

equation. Every solution of the linear equation is a point on the graph of the linear

equation.

Equation of lines parallel to the x-axis and y-axis

The graph of x = a is a straight line parallel to the y-axis.

The graph of y = b is a straight line parallel to the x-axis.

Chapter 8: Quadrilaterals

The sum of all the interior angles of a quadrilateral is 360°.

In the following quadrilateral ABCD, A + B + C + D = 360°.

Classification of quadrilaterals

Classification of parallelograms

Diagonals of a parallelogram divide it into two congruent triangles.

If ABCD is a parallelogram, then ABC CDA

In a parallelogram,

opposite sides are parallel and equal

opposite angles are equal

adjacent angles are supplementary

diagonals bisect each other

A quadrilateral is a parallelogram, if

each pair of opposite sides is equal

each pair of opposite angles is equal

diagonals bisect each other

a pair of opposite sides is equal and parallel

Properties of some special parallelograms

Diagonals of a rectangle are equal and bisect each other.

Diagonals of a rhombus bisect each other at right angles.

Diagonals of a square are equal and bisect each other at right angles.

Mid-point theorem and its converse

Mid-point theorem

The line segment joining the mid-point of any two sides of a triangle is parallel to the

third side and is half of it.

In ABC, if D and E are the mid-points of sides AB and AC respectively, then by the

mid-point theorem, DE||BC andBC

DE .2

Converse of the mid-point theorem

A line through the mid-point of one side of a triangle parallel to the other side bisects the

third side.

In the given figure, if AP = PB and PQ||BC, then PQ bisects AB i.e., Q is the mid-point

of AC.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral is a

parallelogram.

Chapter 9: Areas of Parallelograms and Triangles

Two congruent figures have equal areas, but the converse is not true.

If a figure is formed by two non-overlapping regions A and B, then the area of the figure =

Area (A) + Area (B).

Two figures lie on the same base and between the same parallels if they have a common base

and if the opposite vertex (or side) lies on a line parallel to the base.

For example:

In the first figure, parallelograms ABCE and ABDF lie on the same base AB and between the

same parallels AB and CF.

In the second figure, parallelogram ABCE and triangle ABD lie on the same base but they do

not lie between the same parallels.

Parallelograms on the same (or equal) base and between the same parallels are equal in area.

Its converse is also true, i.e., parallelograms on the same base and having equal areas lie

between the same parallels.

The area of a parallelogram is the product of its base and the corresponding height.

Areas of parallelogram and triangle on the same base

If a parallelogram and a triangle lie on the same (or equal) base and between the same

parallels, then the area of the triangle is half the area of the parallelogram.

If a parallelogram and a triangle lie on the same base and the area of the triangle is half

the area of the parallelogram, then the triangle and the parallelogram lie between the

same parallels.

Triangles on the same base and between the same parallels

Triangles on the same base (or equal base) and between the same parallels are equal in

area.

Triangles having the same base and equal areas lie between the same parallels.

The median of a triangle divides it into two congruent triangles.

Thus, a median of a triangle divides it into two triangles of equal area.

Chapter 10: Circles

Two or more circles are said to be congruent if they have the same radii.

Two or more circles are said to be concentric if their centre’s lie at the same point.

Angle subtended by chords at the centre

Chords that are equal in length subtend equal angles at the centre of the circle.

Chords subtending equal angles at the centre of the circle are equal in length.

Perpendicular from the centre to a chord

The perpendicular from the centre of a circle to a chord bisects the chord.

The line joining the centre of the circle to the mid-point of a chord is perpendicular to the

chord.

Perpendicular bisector of a chord always passes through the centre of the circle.

At least three points are required to construct a unique circle.

Equal chords and their distances from the centre

Equal chords of a circle (or congruent circles) are equidistant from the centre.

Chords which are equidistant from the centre of a circle are equal in length.

Angle subtended by an arc of a circle

Two or more chords are equal if and only if the corresponding arcs are congruent.

The angle subtended by an arc at the centre of the circle is double the angle subtended by

the arc at the remaining part of the circle.

For example:

Here, AXB is the angle subtended by arc AB at the remaining part of the circle.

AOB = 2 AXB

1

AXB 30 152

Angles in the same segment of a circle are equal.

PRQ and PSQ lie in the same segment of a circle.

PRQ = PSQ

Angle in a semicircle is a right angle.

Concyclic points

A set of points that lie on a common circle are known as concyclic points.

Here, A, B, D, and E are concyclic points.

If a line segment joining two points subtends equal angles at the two points lying on the

same side of the line segment, then the four points are concyclic.

For example:

Here, if ACB = ADB, then points A, B, C, and D are concyclic.

Cyclic quadrilaterals

A cyclic quadrilateral is a quadrilateral if all four vertices of the quadrilateral lie on a

circle.

Here, ABCD is a cyclic quadrilateral.

The sum of each pair of opposite angles of a cyclic quadrilateral is 180.

If the sum of a pair of opposite angles of a quadrilateral is 180, then the quadrilateral is

cyclic.

The quadrilateral formed by the angle bisectors of interior angles of any quadrilateral is a

cyclic quadrilateral.

Non-parallel sides of a cyclic trapezium are equal in length.

Chapter 11: Constructions

When the base, base angle, and the sum of other two sides of a triangle are given, the

triangle can be constructed as follows.

Let us suppose base BC, B, and (AB + AC) are given.

Steps of construction:

(1) Draw BC and make an angle, B, at point B.

(2) Draw an arc on BX, which cuts it at point P, such that BP (= AB + AC).

(3) Join PC and draw its perpendicular bisector. Let this perpendicular bisector intersect BP

at A.

Thus, ABC is the required triangle.

When the base, base angle, and the difference between the other two sides of the

triangle are given, the triangle can be constructed as follows.

Let us suppose base BC, B, and (AB – AC) are given.

Steps of construction:

(1) Draw BC and make an angle, B, at point B.

(2) Draw an arc on BX, which cuts it at point P, such that BP (= AB – AC).

(3) Join PC and draw its perpendicular bisector. Let this perpendicular bisector intersect BX

at point A. Join AC.

Thus, ABC is the required triangle.

When the base BC, B, and (AC – AB) are given, the triangle can be constructed as

follows:

(1) Draw base BC and B.

(2) Draw an arc, which cuts extended BX on opposite side of BC at point Q, such that

BQ = (AC – AB).

(3) Join QC and draw its perpendicular bisector. Let this perpendicular bisector intersect BX

at point A. Join AC.

Thus, ABC is the required triangle.

When the perimeter and two base angles of the triangle are given, the triangle can be

constructed as follows.

Let us suppose that base angle, B, and C of ABC are given.

Steps of construction:

(1) Draw a line segment PQ of length equal to the perimeter of the triangle and draw the base

angles at points P and Q.

(2) Draw the angle bisectors of P and Q. Let these angle bisectors intersect each other at

point A.

(3) Draw the perpendicular bisectors of AP and AQ. Let these perpendicular bisectors

intersect PQ at points B and C respectively. Join AB and AC.

Thus, ABC is the required triangle.

Chapter 13: Surface Areas and Volumes

Cuboid

Consider a cuboid with dimensions l, b and h.

Lateral surface area = 2h(l + b)

Total surface area = 2(lb + bh + hl)

Volume = l × b × h

Cube

Consider a cube with edge length a.

Lateral surface area = 4a2

Total surface area = 6a2

Volume = a3

Right circular cylinder

Consider a right circular cylinder with height h and radius r.

Curved surface area = 2πrh

Total surface area = 2πr (h + r)

Volume = πr2h

Right circular cone

Consider a right circular cone of height h, slant height l and base radius r.

Slant height, l 2 2h r

Curved surface area = πrl

Total surface area = πr (l + r)

Volume = 21π

3r h

Sphere

Consider a sphere of radius r.

Curved surface area = Total surface area = 4πr2

Volume 34π

3r

Hemisphere

Consider a hemisphere of radius r.

Curved surface area = 2πr2

Total surface area = 3πr2

Volume 32π

3r

Chapter 14: Statistics

The marks of 20 students of a school are as follows.

86 49 52 78 46 54

62 71 92 87 84 45

58 52 50 60 77 85

88 63

The above data can be written in the form of class intervals as follows.

Marks Number of students

40 – 50 3

50 – 60 5

60 –70 3

70 – 80 3

80 –90 5

90 –100 1

This table is called grouped frequency distribution table.

40 – 50, 50 – 60, etc. are class intervals.

40 is the lower limit and 50 is the upper limit of class interval 40 – 50.

The number of students for each class interval is the frequency of that class interval.

Exclusive frequency distribution table

The frequency distribution tables in which the upper limit of any class interval coincides with

the lower limit of the next class interval are known as exclusive frequency distribution tables.

Inclusive frequency distribution table

Consider the following table.

Class interval Frequency

10 – 19 2

20 –29 7

30 – 39 4

40 – 49 1

Here, the upper limit of any class interval does not coincide with the lower class limit of next

class interval. Such frequency distribution table is known as inclusive frequency distribution

table.

It can be converted into exclusive table by subtracting 20 19 1

0.52 2

from the upper limit

and lower limit of each class interval as follows.

Class interval Class interval Frequency

(10 – 0.5) – (19 + 0.5) 9.5 –19.5 2

(20 – 0.5) – (29 + 0.5) 19.5 – 29.5 7

(30 – 0.5) – (39 + 0.5) 29.5 –39.5 4

(40 – 0.5) – (49 + 0.5) 39.5 – 49.5 1

A histogram is a graphical representation of data.

Example: Represent the given data in the form of a histogram.

Height (in cm) Number of students

140 – 150 10

150 – 160 6

160 – 170 15

170 –180 4

Solution:

For this data, histogram can be drawn by taking class intervals along x-axis and frequency

along y-axis and then drawing bars parallel to y-axis.

The histogram for this data is as follows.

If the class intervals are not of uniform width, then in the histogram, the length of bars is

equal to adjusted frequencies. For example, consider the following data.

Class interval Frequency

0 – 10 1

10 – 20 5

20 – 40 8

40 –70 6

Here, minimum class size is 10.

The adjusted frequencies can be calculated by,Frequency

Class width× Minimum class size

Therefore, we obtain the table as:

Class interval Frequency Adjusted frequency

0 – 10 1 1

10 110

10 – 20 5 5

10 510

20 – 40 8 8

10 420

40 – 70 6 6

10 230

The histogram can be drawn by taking class intervals on x-axis and adjusted frequencies on

y-axis.

The frequency polygon for a grouped data is drawn by first drawing its histogram and then

by joining the mid-points of the top of bars. For example, the frequency polygon for the data

given in the previous table can be drawn as follows.

ABCDEF is the required frequency polygon.

Measures of central tendency

Mean, median, and mode are the measures of central tendency.

Mean: Mean is defined by,

Sum of all observations

MeanTotal number of observations

i i

i

f xx

f

Median: To find the median, the observations are arranged in ascending or descending

order.

(i) If the number of observations (n) is odd, then value of

th1

2

n observation is the

median.

(ii) If the number of observations (n) is even, then the mean of the values of

th th

and 12 2

n n

observations is the median.

Mode: The value of the observation that occurs most frequently is called mode. Mode is

the value of observation whose frequency is maximum.

Chapter 15: Probability

Experiment: An experiment is a situation involving chance or probability.

For example, tossing a coin is an experiment.

Outcome: An outcome is the result of an experiment.

For example, getting a head on tossing a coin is an outcome.

Sample space: The set of all possible outcomes of an experiment is called sample space.

The sample space of the experiment of throwing a die is {1, 2, 3, 4, 5, 6}.

Event: An event is the set of one or more outcomes of an experiment.

For example, in the experiment of throwing a die, the event of getting an even number is {2,

4, 6}.

Probability: The empirical (or experimental) probability of an event A is given by

Number of favorable outcomes

Total number of outcomes P A

Example: When a coin is tossed 500 times and on the upper face of the coin tail comes up

280 times, what is the probability of getting head on the upper face of the coin?

Solution: Let A be the event of getting head on the upper face of the coin.

Total number of trials = 500

Number of trials in which tail comes up = 280

Number of trials in which head comes up = 500 – 280 = 220

220 11

P A500 25

The probability of an event always lies between 0 and 1.