math

Multiplying equation (i) by 1 and equation (ii) by 1 and subtracting the resulting equations.
4 3

x - 1 = 1
12 2y 4

x +- 1 =-1
-12 y

-21y - 1 = 1 -1
y 4

or, –1 – 2 = 1 – 4
2y 4

or, – 3 = – 3
2y 4

or, 6y = 12

y=2

Now, substituting the value of y in equation (i), we get

x – 2 =1
3 2

or, x –1=1
3

or, x =2
3

x=6
Hence, x = 6 and y = 2.

Exercise 4.5.1

1. Solve the following equations by substitution method:

(a) 4x + 3y = 5 (b) 4x + 5y = 4 (c) y = 2x – 3
x+y=3
7x – y = 15 5x – 3y = 79
(f) 11x – 7y = 43
(d) 3x + y = 5 (e) 5x + 7y = 1 2x – 3y = 13

2x + 5y = –1 x + 4y = –5

2. Solve the following equations by substitution method:

(a) –3x + 2y = –2 (b) x + y = 23 (c) 8 – 9 =1
2 3 6 x y

5x + 3y = 5 x + y = 17 10 + 6 =7
4 2 4 x y

(d) 14 + x 3 y =5 (e) 3 + 2 =1 (f) 5 – 3 = 1115
x+y – x y x y

21 – x 1 y =2 4 + 3 = 17 2 + 7 = 2115
x+y – x y 6 x y

144 | Mathematics – 9 Algebra

3. Solve the following equations by elimination method:

(a) 2x + y = –1 (b) 14x + 9y = 156 (c) x + y = 6
2x – y = 3
5x – 4y = –9 7x + 2y = 58
(f) 2x + y = 11
(d) x + 2y = 9 (e) 2x – y = –3 x + 3y = 18

x – 1 = 2y 3x + 2y = 20

4. Solve the following equations by elimination method:

(a) 3x + 4y = 2 (b) 15 – 1 = 421 (c) 3 – 2 = 4
x y x y 15

y=x–3 9 + 2 =4 2 + 3 = 9
x y x y 10

(d) 5 + x 2 y =3 (e) x + y = 17 (f) 14 + 40 = 3
x+y – 3 4 12 2x + 3y 2x – 3y 2

20 – x 3 y =1 x – y = – 3 2x 4 3y + 2x 5 3y = 3
x+y – 5 3 5 + – 8

5. Solve the following equations:

(a) 4x – 5y = 7 (b) x + y = 5 (c) 2 + 5 = 11
3x 4y 12

x + 2y = 5 5x – 3y = 17 5 + 1 = 27
(d) 3x + 4y = 4 4x 2y 20

2x – 14 = 3y (e) 4 – x 3 y =1 (f) 3 y – 2 y = 3
x+y – 2x – 2x + 5

12 + x 5 y =4 4 y + 5 y = 2175
x+y – 2x + 2x –

4.5.2 Graphical Method

In this method, we find a few ordered pairs of solutions of each linear equation of two variables in two
separate tables. The pair of solution of each equation from the separate tables are plotted on the same
graph paper and by joining the plotted points two separate straight lines are obtained. The two straight
lines intersect each other at a point, then the co–ordinates of the intersecting point is the required
solution of the given simultaneous equations. Let's learn the process from the following examples.

Worked Out Examples

Example 1: Solve graphically: x + 2y = 8 and 2x – y = 6. Y
Solution: Here,

x + 2y = 8 ... (i) (0,4)

2x – y = 6 ... (ii) (4,2) (8,0)
(3,0)
From equation (i), X’ (0,0) O X

x + 2y = 8 x840 (2,-2) (10,-1)
 x = 8 – 2y y024
From equation (ii), (0,-6)

Y’

Algebra Mathematics – 9 |145

2x – y = 6 x342
y 0 2 –2
or, 2x = y + 6

 x = y + 6
2

We plot the points which are obtained from the above two separate tables and joining
them. We get two separate lines which intersect each other at a point (4, 2) in the
above graph.

Hence, the required solution of the given two simultaneous equations is x = 4 and
y = 2.

Example 2: Solve graphically: x + 3y = 7 and 3x – y = 11.
Solution:
Here,

x + 3y = 7 ––––––––––––– (i) Y

3x – y = 11 –––––––––––– (ii) (6,7)

From equation (i),

x + 3y = 7 Table – 1 (-2,3) (1,2) (4,1)

or, 3y = 7 – x x 1 4 –2 X’ O (0,0) X

 y = 7 – x y 2 1 3
3
(2,-5)

From equation (ii), Table – 2
24
3x – y = 11 x 6 Y’

or, 3x – 11 = y y –5 1 7

 y = 3x – 11

Now, we plot the ordered pairs (1, 2), (4, 1) and (–2, 3) from the table–1 on the graph paper and by
joining them to construct a straight line. Similarly, again we plot the ordered pairs (2, –5), (4, 1) and
(6, 7) from the table– 2 on the same graph and construct another straight line by joining them.

In the above graph, two straight lines which are constructed from the equations (i) and (ii) are
intersected each other at a point (4, 1). It means both straight lines are passing through the point (4, 1).
Therefore, x = 4 and y = 1 are satisfied in the given both equations. Form the above two tables, a pair
numbers (4, 1) is only a common point of the two straight lines.

Hence, the required solution of the given two simultaneous equations is x = 4 and y = 1.

From the solution of the above examples, it is concluded that the following steps are applied to solve
the simultaneous equations of two variables by graphical method.

 Prepare the table and tabulate by putting the values of x and y in the both equations.

 Plot the pair numbers on the graph paper and construct the two straight lines by joining
the pair numbers on the graph.

 Find the point on the graph which is the common point where two straight lines
intersect each other.

 The co–ordinates of the common point is the required solution of the given two
simultaneous equations.

146 | Mathematics – 9 Algebra

Exercise 4.5.2

1. Copy and complete the table of values for x and y. Plot the ordered pairs separately from each
table. Find the common point, which is the solution of the equations from the graphs.

(a) 2x –3 = y x + y = 3

x25 x1 4
y –1
y1

(b) 2x + y = 11 x + 3y = 18

x3 x –3

y9 3 y6 5

(c) 3x – 5y = 10 x – 2y = 4

x0 –5 x4

y1 y –3 1

(d) x + y = 7 x – y = 3

x4 0 x0

y0 y1 3

2. Solve the following simultaneous equations graphically:

(a) 2x + y = 3 (b) x – y + 3 = 0 (c) 2x – 3y = 1

2x – y = 1 3x – y = 1 3x + y = –5

(d) 2x + y = 9 (e) 2x – y = 4 (f) x – 2y = –1

2x – y = 11 x–y=3 2x – y = –4

4.6 Quadratic Equation

Introduction

Let’s consider an equation 4x + 2 = x – 1. This equation has only one variable. i.e x. After solving this
equation, the value of x is –1. It means this equation is satisfied by the value of x = –1. So, the
equation contains only one variable and the highest power of that variable is 1. Such type of the
equation is called a first degree equation of one variable. The first degree equation of one variable is
also called a linear equation in one variable.

But, let’s consider another equation x2 + 5x – 6 = 0. This equation has also only one variable but the
highest power of the variable is 2. When we solve this equation, we get the values of x are 1 and –6. It
means the equation x2 + 5x – 6 = 0 is satisfied by both values of x = 1 and x = –6. So, it is called a
second degree equation of one variable. The second degree equation of one variable is called a
quadratic equation. The solution of any quadratic equation has two values. These two values of
the variable are called the roots of the quadratic equation.

Thus, a quadratic equation is a second degree polynomial equation in one variable. The standard form
of the quadratic equation is ax2 + bx + c = 0, where, a, b, cR and a ≠ 0.

Algebra Mathematics – 9 |147

Types of Quadratic Equation

There are two types of quadratic equation which are (i) Pure quadratic equation (ii) Adfected
quadratic equation.

(i) Pure quadratic equation: A quadratic equation which is in the form of ax2 + c = 0 is called a
pure quadratic equation. It means a quadratic equation which does not contain the term with the
variable containing power 1 is known as a pure quadratic equation. For examples: x2 – 16 = 0,
3y2 – 18 = 0, 5x2 – 4 = 0 etc.

(ii) Adfected quadratic equation: A quadratic equation which is in the form of ax2 + bx + c = 0 is
called an adfected quadratic equation. It means the adfected quadratic equation is a quadratic
equation containing term with the variable having power 1 also. For example: 3x2 + 5x + 2 = 0,
x2 + 2x – 15 = 0 etc.

Solution of a Quadratic Equation

We know that a quadratic equation is a second degree equation. So, we obtain two values of the
variable from the given quadratic equation. It means a quadratic equation has two roots. We can solve
a quadratic equation by the various methods. But in this chapter, we discuss three different methods
for solving the quadratic equation.

4.6.1 Solving a Quadratic Equation by Factorization Method

In this method, the right hand side of the given equation is made zero by transposing the terms to the
left hand side. Then the second degree polynomial in the form of ax2 + c or ax2 + bx + c is factorized
and expressed as the product of two linear factors. After that, each linear factor is separately solved to
get the value of the variable which is in the given equation. The values of the variable are the required
solution of the given equation. Let's learn this process in the following examples.

Worked Out Examples

Example 1: Solve: x2 = 36.
Solution: Here,
x2 = 36
Example 2: or, x2 – 36 = 0
Solution: or, (x)2 – (6)2 = 0
or, (x + 6) (x – 6) = 0

[a2 – b2 = (a + b)(a – b)]
Either, x + 6 = 0 i.e. x = –6
OR, x – 6 = 0 i.e. x = 6

 x = ±6

Solve: 5x2 + 8x = 21.
Here,
5x2 + 8x = 21
or, 5x2 + 8x – 21 = 0

148 | Mathematics – 9 Algebra

or, 5x2 + (15 – 7)x – 21 = 0

or, 5x2 + 1 5x – 7x – 21 = 0

or, 5x(x + 3) – 7(x +3) = 0

or, (x + 3) (5x – 7) = 0

Either, x + 3 = 0 i.e x = –3

OR, 5x – 7 = 0 i.e x = 7
5

 x = –3 and 7 .
5

Example 3: Solve: x 4 3 – x 4 3 = 1 .
Solution: – + 3

Here,

x 4 3 – x 4 3 = 1
– + 3

or, 4(x + 3) – 4(x – 3) = 1
( x – 3)(x + 3) 3

or, 4x + 12 – 4x + 12 = 1
x2 – 9 3

or, 24 = 1
x2 –9 3

or, x2 – 9 = 72
or, x2 – 81 = 0
or, (x)2 – (9)2 = 0

or, (x – 9)(x + 9) = 0

Either, x + 9 = 0 i.e = x = –9

OR, x – 9 = 0 i.e = x = 9

 x = ±9

Example 4: Solve: 4x – 21 = 3x – 11 .
Solution: x – 6 x – 1

Here,

4x – 21 = 3x – 11
x–6 x – 1

or, (4x – 21)(x – 1) = (3x – 11)(x – 6) [ By cross – multiplication ]

or, 4x2 – 4x – 21x + 21 = 3x2 – 18x – 11x + 66

or, 4x2 – 25x + 21 = 3x2 – 29x + 66

or, 4x2 – 3x2 – 25x + 29x + 21 – 66 = 0

or, x2 + 4x – 45 = 0

or, x2 + 9x – 5x – 45 = 0

or, x(x + 9) – 5(x + 9) = 0

Algebra Mathematics – 9 |149

or, (x + 9)(x – 5) = 0

Either, x + 9 = 0 i.e x = –9

OR, x – 5 = 0 i.e x = 5

 x = 5 and –9.

Example 5: Solve: x + 1 + x+2 = 2x + 13 .
Solution: x – 1 x–2 x + 1

Here,

x + 1 + x+2 = 2x + 13
x – 1 x–2 x + 1

or, (x + 1)(x – 2) + (x + 2)(x – 1) = 2x + 13
(x – 1)(x – 2) x + 1

or, x2 – 2x +x – 2 + x2 – x + 2x – 2 = 2x + 13
x2 – 2x – x + 2 x + 1

or, 2x2 – 4 2 = 2x + 13
x2 – 3x + x+1

or, (2x2 – 4) (x + 1) = (x2 – 3x + 2) (2x + 13)

or, 2x3 + 2x2 – 4x – 4 = 2x3 + 13x2 – 6x2 – 39x + 4x + 26

or, 2x3 + 2x2 – 4x – 4 = 2x3 + 7x2 – 35x + 26

or, 2x3 – 2x3 + 2x2 – 7x2 – 4x + 35x – 4 – 26 = 0

or, –5x2 + 31x – 30 = 0

or, –(5x2 – 31x + 30) = 0

or, 5x2 – 25x – 6x + 30 = 0

or, 5x (x – 5) – 6(x – 5) = 0

or, (x – 5) (5x – 6) = 0

Either, x – 5 = 0 i.e x = 5

OR, 5x – 6 = 0 i.e x = 6 = 151
5

 x = 5 and 151 .

Exercise 4.6.1

1. Solve: (b) (y – 7)(2y – 9) = 0 (c) (3x – 1)(x – 5) = 0
(a) (x – 3)(x + 2) = 0 (e) (5 – 6x)(5x + 6) = 0 (f) (9x + 2)(3 – 4x) = 0
(d) (2t – 1)(t – 2) = 0
(b) 4y2 – 9 = 0 (c) 5t2 = 125
2. Solve: (e) y2 = 5
(a) x2 – 25 = 0 (f) x = 9
4 x
(d) x2 – 7 = 29
Algebra
150 | Mathematics – 9

(g) 5x = 20 (h) 3x – 27 = 0 (i) x + 2 = x + 8
4 x 4 x 2 x 8 x

3. Solve:

(a) 2x(3 – x) = 0 (b) y(y – 5) = 0 (c) x2 – 7x = 0

(d) 3x2 – 18x = 0 (e) t2 – 36t = 0 (f) 5m2 + 10m = 0

(g) x2 = x h) t = t2 (i) 3x2 = x
4 12 3 8 2 6

(j) x2 = x (k) 2y2 = y (l) 2 = 7
50 2 3 9 x x2

4. Solve:

(a) (x + 5)(x – 2) = 3(x + 18) (b) (x + 2)(x – 2) = 3x2 – 8

(c) (t – 3)(t + 2) = 10 – t (d) 2(y2 – 8) + y(4 – y) = 4(y + 12)

(e) 3(x + 3)(x – 3) = x2 + 5 (f) 2(p + 3)(p – 2) = 2p + 4

5. Solve: (b) y2 – 15y + 36 = 0 (c) x2 – 5x – 6 = 0
(a) x2 – 11x + 30 = 0 (e) x2 – 9x = 36 (f) 6p2 + 13p = 5
(d) t2 + 2t – 24 = 0 (h) 30x = 8x2 + 27 (i) x2 + 3x – 10 = 0
(g) 3x2 – 5x – 2 = 0

6. Solve: (b) 9t2 + 6t + 1 = 0
(a) 2x2 – 3x + 1 = 0 (d) 2x2 + x – 6 = 0
(c) 3p2 – 10p = 8 (f) (2x – 3)(x – 5) = (x – 3)2

(e) (x – 3)(x + 5) = 48 (h) (x + 4)(2x – 3) = 6
(g) –3y2 + 5y + 8 = 0 (j) 4(x – 2)2 – 5(x – 2) – 6 = 0
(i) x2 – 2ax + a2 – b2 = 0

(k) 11(t + 1)(t + 2) = 38(t + 1) + 9t.

7. Solve:

(a) x+1 = x+7 (b) 8y + 7 = 65y
x+3 2x + 8 y 7

(c) 2x2 + 10 = 7 – 50 + x2 (d) 5 – 5 = 2
15 25 x–2 + 3
x 2

(e) 3 + 3 =8 (f) 4p2 + 5 – 2p2 – 5 = 7p2 – 25
1+x – 10 15 20
1 x

(g) x+2 + x – 2 = 441 (h) x+3 + x – 3 = 221
2– x 2 + x x–3 x + 3

(i) x + 1 – x–1 = 223
x – 1 x+1

8. Solve:

(a) 3x – 7 = x + 1 (b) x+2 = 2x – 3
2x – 5 x – 1 x+3 3x – 7

Algebra Mathematics – 9 |151

(c) 3y – 8 = y – 2 (d) 2 = 3t 1
5y – 2 y + 5 t+1 5t –

(e) t – 1 – t = 221 (f) 2 + 3 = 2
t–1 t x–3 – x
x 1

(g) 1 = 2 3 x – 2 (h) x+2 – x – 2 = 5
x–1 – x x–2 x + 2 6

(i) x + x + 1 = 13
x+1 x 6

9. Solve:

(a) 1 – 1 – 1 1 x =0 (b) x+4 + x – 4 = 10
x+2 x–2 – x–4 x + 4 3

(c) 2t – 3 – 5 – 3t = 5 (d) 2x + 3x – 1 = 5x – 11
3t – 5 2t – 3 2 x–1 x+2 x – 2

(e) x+3 + x – 3 = 2x – 3 (f) y–2 – 11 – 3y = 4y + 13
x+2 x – 2 x–1 y–3 y – 4 y + 1

(g) x+2 + x–2 = 2x + 6 (h) 2x – 1 – 2x + 1 = –232
x–2 x+2 x–3 2x + 1 2x – 1

(i) x–1 + x + 3 = 2(x + 2)
x+1 x – 3 x–2

4.6.2 Solving Quadratic Equation by Completing the Square

We can use this method to solve a quadratic equation when the left hand expression of the given
equation is not easily factorized.

In this method, we transpose the constant term of the equation to the right hand side and the left hand
expression is made a perfect square by adding a suitable term on the both sides. We can make x2 + 2ax
or x2 – 2ax a perfect square by adding a2, i.e. the square of half of the coefficient of x. let’s learn this
method in the following examples.

Worked Out Examples

Example 1: Solve: 4x2 + 20x + 25 = 0.
Solution: Here, 4x2 + 20x + 25 = 0
or, 4x2 + 20x = – 25
or, 4(x2 + 5x) = –25

or, x2 + 5x = – 25
4

or, (x)2 + 2.x.25 + 522 = – 25 + 252
4

or, x + 252 = – 25 + 25
4 4

152 | Mathematics – 9 Algebra

or, x + 52 =0
2

or, x + 5 =0
2

 x = – 5
2

Example 2: Solve: 2x2 + 7x + 4 = 0.
Solution:
Here,

2x2 + 7x + 4 = 0

or, 2x2 + 7x = – 4

or, 2x2 + 7 x = – 4
2

or, (x)2 + 2.x.47 + 47 2 = – 4 + 47 2
2

or, x + 472 = – 4 + 49
2 16

or, x + 742 = –32 + 49
16

or, x + 472 = 17
16

or, x + 7 =± 17
4 4

or, x = – 7 ± 17
4 4

 x = – 7 ± 17
4

Example 3: Solve: 3x2 – 5x – 2 = 0.
Solution:
Here,

3x2 – 5x – 2 = 0

or, 3x2 – 5x = 2

or, 3x2 – 53x = 2

or, (x)2 – 2.x.56 + 652 = 2 + 652
3

or, x – 652 = 2 + 562
3

or, x – 652 = 2 + 25
3 36

or, x – 562 = 24 + 25
36

Algebra Mathematics – 9 |153

or, x – 652 = 49
36

or, x – 652 = ± 672

 x– 5 = ± 7
6 6

Taking + ve, Taking – ve,

x– 5 = 7 x– 5 = – 7
6 6 6 6

Or, x = 7 + 5 or, x = – 7 + 5
6 6 6 6

Or, x = 12 or, x = –2
6 6

x=2  x = – 1
3

x = 2 and – 1
3

Example 4: Solve by completing square: x + 3 – 1 – x = 414 .
Solution: x – 2 x

Here,

x+3 – 1 – x = 414
x–2 x

or, (x + 3).x – (1 – x)(x – 2) = 17
x(x – 2) 4

or, x2 + 3x – x + 2 + x2 – 2x = 17
x2 – 2x 4

or, 2x2 + 3x – 3x + 2 = 17
x2 – 2x 4

or, 4(2x2 + 2) = 17(x2 – 2x)
or, 8x2 + 8 = 17x2 – 34x
or, 17x2 – 34x – 8x2 – 8 = 0
or, 9x2 – 34x = 8

or, (3x)2 – 2 × 3x × 17 + 137 2 = 8 + 137 2
3  

or, 3x – 1372 = 8 + 289
9

or, 3x – 172 = 72 + 289
3 9

or, 3x – 1372 = 361
9

154 | Mathematics – 9 Algebra

or, 3x – 172 = ±1392
3

 3x – 17 = ± 19
3 3

Taking + ve sign, Taking – ve sign,

3x – 17 = 19 3x – 17 = – 19
3 3 3 3

or, 3x = 19 + 17 or, 3x = – 19 + 17
3 3 3 3

or, 3x = 36 or, 3x = – 2
3 3

or, 3x = 12 x = – 2
9

x = 4

x = 4 and – 2
9

Exercise 4.6.2

1. Solve by completing the square:

(a) 9x2 – 6x + 1 = 0 (b) x2+ 8x + 16 = 0 (c) x2 – 7x – 98 = 0
(f) 5y = 12 – 3y2
(d) 3x2 + 2x – 5 = 0 (e) 6t2 + 13t = 5 (i) 2x2 – 5 = x

(g) 25s2 + 16 = 40s (h) 4x – 4x2 = –7

2. Solve the following equations by completing the square:

(a) 3x2 + 5x – 2 = 0 (b) 2y2 + 7y + 4 = 0 (c) 3x2 – 2x – 21 = 0
(f) 5x2 – 8x – 21 = 0
(d) 2t2 – 7t + 5 = 0 (e) 2z2 – 6z + 3 = 0

3. Solve by completing the square:

(a) 3 + 3 =8 (b) x – 2x – 1 =1 (c) x+1 = x+7
1+x 1–x 4 x+1 x+3 2x + 8

(d) 3t – 2 = 3t – 8 (e) 2y – 3 = 3y – 7 (f) z 4 1 – z 5 2 = 3
2t – 3 t+4 y+2 y+3 – + z

(g) x+2 – 1 = 1 (h) x–4 + x 6 1 = 5 (i) 2s – 5 – 25 = s 2s
6 x+2 6 4 + 4 s–3 3 –4

(j) 2x – 1 – 2x + 1 = – 223 (k) x 3 1 + x 2 3 = 2 (l) 2x – 9 – 2x – 7 = 7
2x + 1 2x – 1 – – x 2x – 7 2x – 9 12

Algebra Mathematics – 9 |155

4.6.3 Solving Quadratic Equation by using Formula

Every quadratic equation can be written in the form ax2 + bx + c = 0. The expression ax2 + bx + c has
no simple factors. So, we solve the equation ax2 + bx + c = 0 by completing the square method.

Now,

ax2 + bx + c = 0

or, ax2 + bx = –c [Subtracting ‘c’ both sides]

or, ax2 + b x = – c [Dividing both sides by ‘a’ to make the coefficient of x2 unity]
a a a

or, x2 + b x = – c
a a

or, (x)2 + 2.x. b + 2ba2 = –ac + 2ba2
2a

or, x + 2ba 2 = –4ac + b2
4a2

or, x + 2ba 2 = b2 – 4ac
4a2

or, x + b 2 = ± b2 – 4ac 2
2a 4a2 


or, x + b = ± b2 – 4ac
2a 2a

 x = – b ± b2 – 4ac = –b ± b2 – 4ac
2a 2a 2a

Taking +ve sign, Taking – ve sign,

x = –b + b2 – 4ac x = –b – b2 – 4ac
2a 2a

Thus, –b + b2 – 4ac and –b – b2 – 4ac are two roots of x.
2a 2a

Hence, x = –b  b2 – 4ac is general formula and we can apply it to solve any given quadratic
2a

equation.

Note:

(i) In the formula x = –b  b2 – 4ac , the term b2– 4ac is called discriminant factor.
2a

(ii) If the discriminant factor b2 – 4ac = 0, the equation has single root.

(iii) If b2 – 4ac > 0 i.e. +ve, the equation has two real roots.

(iv) If b2 – 4ac < 0 i.e. –ve, the equation has no real solution i.e. the equation can not be solved.

156 | Mathematics – 9 Algebra

Work Out Examples

Example 1: Solve by using formula: 4x2 – 7 = 0.
Solution:
Here,

4x2 – 7 = 0

or, 4x2 + 0.x – 7 = 0

Comparing this equation with ax2 + bx + c = 0, we get a = 4, b = 0 and c = –7

By using the formula,

x = –b ± b2 – 4ac
2a

= –0 ± 02 – 4 × 4 × 7
2×4

= ±4 8 7

x = ± 7
2

Example 2: Solve by using formula: x2 – 9x + 14 = 0.
Solution:
Here,

x2 – 9x + 14 = 0

Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = –9 and c = 14

 by using formula,

x =–b b2 – 4ac
2a

= –(–9)  (–9)2 – 4×1×14
2×1

=9 81 – 56
2

= 9 2 25 = 9  5
2

Taking +ve sign, Taking –ve sign,

x = 9 + 5 x = 9 – 5
2 2

= 14 =7 = 4 = 2
2 2

 x = 2 and 7.

Algebra Mathematics – 9 |157

Example 3: Solve by using formula: x + 2 – x – 2 = 445 .
Solution: x – 2 x + 2

Here, x + 2 – x – 2 = 454
x – 2 x + 2

or, (x + 2)2 – (x – 2)2 = 24
(x – 2)(x + 2) 5

or, (x2 + 2.x.2 + 22) – (x2 – 2.x.2 + 22) = 24
(x)2 – (2)2 5

or, x2 + 4x +4 – x2 + 4x – 4 = 24
x2 – 4 5

or, 8x 4 = 24
x2 – 5

or, (x2 – 4) × 24 = 8x × 5

or, 24x2 – 96 = 40x

or, 8(3x2 – 12) = 40x

or, 3x2 – 12 = 5x [Dividing both sides by 8]
or, 3x2 – 5x – 12 = 0

Comparing this equation with ax2 + bx + c = 0, we get a = 3, b = –5 and c =–12.

 by using formula

x =–b b2 – 4ac
2a

= (–5)  (–5)2 – 4 × 3 × –12
2×3

=5 25 + 144
6

= 5  6 169

= 5  13
6

Taking +ve sign, Taking –ve sign,

x = 5 + 13 x = 5 – 13
6 6

= 18 = – 8
6 6

=3 = – 4
3

 x = 3 and – 4 .
3

158 | Mathematics – 9 Algebra

Exercise 4.6.3

1. Use the formula to solve the following equations:

(a) 81x2 = 16 (b) 49y2 – 100 = 0 (c) 2t2 – 9 = 0
(f) s2 – 36 = 0
(d) 9y2 – 0.01 = 0 (e) 18x2 – 27x = 0

2. use the formula to solve the following equations:

(a) x2 + 8x + 12 = 0 (b) 4x – 4x2 = –7 (c) 4x2 – 3x = 6

(d) 2y2 + 3 – 6y = 0 (e) t2 – 5t –36 = 0 (f) z2 + 20 – 9z = 0

(g) x2 + 7ax + a2 = 0 (h) 2x2 – 3x = 3 (i) 3s2 + 11s = –10

(j) (x – 3)(2x + 3) = 5 (k) (4x – 3)(8x + 5) = 13 (l) (2x – 3)2 = 4

3. Solve by using the formula:

(a) 6x2 – 5x = 0 (b) x2 + 2x – 24 = 0 (c) x(x – 5) = 24

(d) 5a2 + ax = 41x2 (e) x2 + 4 = x2 + 5 (f) t – 4 + 6 = 5
5 6 4 + 4
t 1

(g) x + x + 1 = 25 (h) x + 2 = 7 (i) 5x + 1 = 3x + 1
+ x 12 3 x 3 7x + 5 7x + 1
x 1

(j) x + 3 + x =1 (k) x+ 3 = 3x – 7 (l) x + 2 – 4–x = 7
x + x+ 2 2x – 3 x – 1 2x 3
x 3

(m) 2t – 1 + t+1 = 4 (n) x + 4 – x – 4 = 2 (o) y + 6 – y + 1 = 1 1
t+1 2t + 1 3 x – 4 x + 4 3 y + 7 y + 2 3y +

4. Solve: (b) lx2 + mx – n = 0
(a) px2 – qx + r = 0 (d) dx2 + ex + f = 0
(c) ax2 – bx – c = 0



Algebra Mathematics – 9 |159

Unit Test

Time: 40 minutes F.M.- 24

Group- A (1 × 1 = 1)

1. If ax + bx + c = 0, what is the value of x?

Group- B (5 × 1 = 10)

2. Factorize: m2 – n2 – mx – nx

3. Simplify: 3x+4 – 3x+3
3x+3

4. If x:y = 3:4, find the value of (x +2y): (4x + y).

5. If a = dc, prove that: a2 + c2 = ac
b b2 + d2 bd

6. Use formula to solve: 2x2 – 3x – 2 = 0

Group- C (2 × 4 = 8)

7. If x = 1 – a–31, prove that x3 + 3x = a – 1a.

a3

8. Solve by completing equation: 2x2 – 7x + 3 = 0

Group- D (1 × 5 = 5)

9. Solve graphically: x – y = 1 and x + y = 5.



160 | Mathematics – 9 Algebra

Answers ____________________________________________________________

Exercise 4.1.1

1. (a) x (3, – 4y) (b) 2 (a – b) (x + 2y) (c) 2a (2x – 3a + 4y)
(f) 3x (x + 4) (x – 4)
(d) 2pq (7p + 3q – 6) (e) (2a + b) (2a – b)

( ) ( )(g)3a + 1 3a – 1 ( ) ( )(h)3x+ 4 3x – 4 (i) (x + y) (x – y) (x2 + y2)
2 2 5y 5y

(j) 3x (3x + 2) (3x – 2) (k) 3 (x + 7) (x – 7) (l) 2 (a + 11) (a – 11)
(m) (a + 2b) (a2 – 2ab + 4b2) (n) (x – 3y) (n2 + 3xy + 9y2)
(o) mn2 (m + n) (m2 – mn + n2)

(p) 2p (2p – 1) (4p2 + 2p + 1) (q) (4m2 + n2) (16m4 – 4m2n2 + n4) ( ) ( )(r)n + 1 n2 –1 + 1
n n2

( ) ( )(s)x2 + 1 x4 –1 + 1 (t) (y + 4) (y – 2) (u) (x – 3) (x – y)
x2 x4

(v) (4 + 2a – b) (4 – 2a + b) (w) 5 (1 – x2)

2. (a) (x + 2) (3x + 2) (b) (x + 3) (x + 4) (c) (a + 4) (a – 1)
(f) (x – 3) (3x + 2)
(d) (a + 6) (a – 1) (e) (2a – 9) (a – 1)

(g) (a + 2) (a – 2) (2a2 + 3) (h) (x + 1) (x2 – x + 1) (2x3 – 3)

( ) ( ) ( )(i)1+y 1 – y 7x2 – 3 ( ) ( )(j)2x–5 x + 1 ( ) ( )(k)2p2 + 1 5 – 2q2
x x y2 y y q2 p2

( ) ( )(l)1 – 2b2 3a2 – 5 (m) (a + b+ 1) (9a + ab – 8) (n) (x + y – 4) (3x + 3y + 2)
a2 b2

(o) (x + y – 2z) (4x + 4y + 3z) ( ) ( )(p)a+1 – 1 7a – 2 – 7 (q) – (a + 3) (a + 9)
a a
(r) (2 – a) (a – 7)
3. (a) (x – y) (x + y – a) (b) (a + b) (a – b – x) (c) (a + b) (a + c)
(e) (x+y) (x–y–z) (f) (a – b)2
(d) (a + 6b) (a – c) (h) (x –1)2 (x + 1) (i) (x + 2) (xy – 3)
(g) (b –1) (ab + c) (k) (2 + x) (4 + x2) (l) (2 – 9x) (2 + x) (2 – x)
(j) (y + 1) (y2 + y + 1) (n) (x – 2y) (x2+ 2xy + 4y2 + 2a) (o) (2a + b)3
(m) (a + b) (a2 – ab + b2 – 1) (q) (2x + y – 1) (2x – y + 1) (r) (2x + 3y + 1) (2x – 3y + 1)
(p) ( x– y) (x2 + xy + y2 – 1) (t) (1 + 2p + 6p2) (1 + 2p – 6p2)
(s) (3a + 2b + 1) (3a – 2b + 1)

Exercise 4.1.2

1. (a) (x2 + 2xy + 2y2) (x2 – 2xy + 2y2) (b) (x2 + 2x + 2) (x2 – 2x + 2) (c) (a2 + 4ab + 8b2) (a2 – 4ab + 8b2)

(d) (18p2 + 6pq + q2) (18p2 – 6pq + q2) (e) (49a2 + 14ab + 2b2) (49a2 – 14ab + 2b2)

( ) ( )(f)x2 + 1 + 1 x2 – 1 + 1 ( ) ( )(g)2x2+ 2x + 1 2x2 – 2x + 1
2x2 2x2 3y 9y2 3y 9y2

( ) ( )(h)2x2 + 2x + 1 2x2 – 2x + 1
5y 25y2 5y 25y2

2. (a) (x2 + xy + y2) (x2– xy + y2) (b) (x2 + x + 1) (x2 – x + 1) (c) (x2 + 3x + 1) (x2 – 3x + 1)

(d) (2p2 + p + 1) (2p2 – p + 1) (e) (m2 + m – 1) (m2 – m – 1) (f) (3a2 + 4a + 2) (3a2 – 4a + 2)

(g) (m2 + 4mn + n2) (m2 – 4mn + n2) (h) (2a2 + 3ab + 11b2) (2a2 – 3ab + 11b2) (i) (x2 + 7x2 + 16) (x2 – 7x2 + 16)

(j) (x2 + 3x – 1) (x2 – 3x – 1) (k) (x2 + 2x – 1) (x2 – 2x – 1) (l) (a2 + 4ab + b2) (a2 – 4ab + b2)

( ) ( )(m) x + 1 + y x – 1 + y ( )( )(n) (5x2 + 7xy + 4y2) (5x2 – 7xy + 4y2) (o) p2 + 3p + 1 p2 – 3p + 1
y x y x q2 q q2 q

( ) ( ) ( ) ( )(p)x2+3 + 2y2 x2 – 3 + 2y2 (q) x2 + 3 + 1 x2 – 3 + 1 ( ) ( )(r)x2+3x – 1 x2 – 3x – 1
2y2 x2 2y2 x2 x2 x2 y2 y y2 y

Algebra Mathematics – 9 |161

( ) ( )(s)2x + 5 – 1 2x – 5 – 1 ( ) ( )(t)x+1– 1 x – 1 – 1 ( ) ( )(u)x2+1 + y2 x2 – 1 + y2
2x 2x x x y2 x2 y2 x2

3. (a) (x + 3y – 6) (x – 3y – 4) (b) (p + q + 4) (p – q – 10) (c) (a + b – 4) (a – b – 2)
(d) (x2 + x – y – 3) (x2 – x + y – 3) (e) (a + b – 14) (a – b + 2) (f) (x – y – 2) x – 3y + 2)
(g) (x – 4y + 55z) (x – 50y – 55z) (h) (a – b + c + d) (a – b – c – d)

Exercise 4.2 (b) 32a (c) (7x)4 (d) (a + b)5
(b) 24 (c) 26
1. (a) 46 (d) a
2. (a) 133 29

(e) (3p + q)5 (f) 1
23a

8

3. (a) x (b) 3a2 (c) x5
y2 b 12

b5

1 (e) 27y3 1
(d) a2 x3
(f) a2

4. (a) 4 (b) 16 (c) 4

(d) 1 (e) 9
8 16

5. (a) 1 (b) 3 (c) 2
4 x

(d) 1 (e) 4
3

6. (a) 125 (b) 1 (c) 16
3 (f) 6

(d) 2 (e) 2
5

3

7. (a) b2 (b) m (c) 9
4x2y2
2

a3

5

(d) 2y2 (e) a3c6 (f) y12
3xz3 x37
5

b3

8. (a) 1 (b) 64 (c) 3 (d) 32
81 5 729

(e) 5 (f) 6
96 7

9. (a) 1 22 (c) 1 (d) xb
(e) x11n
(b) x2(a – c )
10. (a) 1
(e) 1 (f) 1

11. (a) 1 (b) 1 (c) 1 (d) 1
(e) 1
(f) 1

(b) 1 (c) 1 (d) 1

(y – z) (z – x) (y – x)

12. (a) 1 (b) 1 (c) a xyz (d) 1

(e) ( )m + nx (f) ( )m + na (g) 1
y b

13. (a) 2

162 | Mathematics – 9 Algebra

Exercise 4.3

1. (a) 3 (b) 1 (c) –27 (d) 1

(e) –32 (f) 4 (c) 3
(c) 14
2. (a) 1 (b) 1 (d) 1
2 (c) 1 2
(c) –2
(e) 3 (f) 13 (g) 3
(b) 2 (c) –2, –3
3. (a) –1 (f) –12 (d) 5

(e) 2
3

4. (a) 0 (b) 3 (d) 3

(e) –2 (f) –3 (d) 0
(h) 2
5. (a) –1 (b) 2 (d) 1, 2

(e) 3 (f) 3
2 2

6. (a) 1 (b) 1 (e) 2

Exercise 4.4.1 (b) 1:2 (c) 6:1 (d) 14:1
(f) 4:7
1. (a) 3:160 (b) 1:1 (c) 8:105 (d) 196a2:81b2
(e) 50:19 (b) 4x2:49y2 (c) 144:49 (d) 2:3
(b) 7:6 (c) 13a:12b (d) 216x3:1331y3
2. (a) 4:9 (b) 8a3:b3 (c) 125:8 (d) 9x2:8y3
3. (a) 4:25 (b) 5x:1 (c) 6a:4b (d) 7b:18a
4. (a) 3:4 (b) 21y2:16x2 (c) (9y – 5):(2x + 3)
5. (a) 64:343 (b) 3:10, 6:15:20 (c) 3:2, 24:33:16 (d) 0 (e) 0
6. (a) 2:3 (b) 17:37 (c) 4:7 (d) 2:5 (e) 35:9
7. (a) 19:12 (b) 4:3 (c) 25:9 (d) 3:4
8. (a) 16:33, 16:36:33 (b) 5:3 (c) 4:5 (h) 2:3
9. (a) 3:4 (f) 11:8 (g) 1:7
10. (a) 11:7 (b) 5:2 (c) 7:4 (d) –8
11. (a) 3:4 (b) 70, 98 (c) 75, 125 (d) 56, 70
(e) 40°, 60°, 80°
(e) 3:4 (b) Rs. 30, Rs. 42 (c) Rs. 9, Rs. 12 (d) 8 km
12. (a) 2:11 (e) 54
13. (a) 35, 63 (b) 17 (c) 2
(b) 34, 51 (c) 50, 74
(d) 76, 133 (b) 21 years, 28 years (c) 10 years, 35 years
14. (a) Rs. 32, Rs. 36 (b) 54 (c) 20°, 70°

(d) Rs. 30
15. (a) 13
16. (a) 6, 10
17. (a) 24 years, 32 years
18. (a) 35 lit, 20 lit

Exercise 4.4.2 (b) 6 (c) 6 (d) 9 (e) 48
(b) 12 (c) 9 (d) 35 (e) 2
1. (a) 8 (b) 2 (c) 2 (d) 1
2. (a) 21 (b) 37:13 (c) 2 (d) –1:3
3. (a) (i) 2 (ii) 6 (iii) 40
4. (a) 1:3
13. (e) 0

Algebra Mathematics – 9 |163

Exercise 4.5.1

1. (a) x = 2, y = –1 (b) x = 11, y = –8 (c) x = 2, y = 1 (d) x = 2, y = –1
(f) x = 2, y = –3
(e) x = 7 , y = –193 (c) x = 2, y = 3 (d) x = 4, y = 3
9 (b) x = 3, y = 7
(c) x = 3, y = 3 (d) x = 5, y = 2
2. (a) x = 16 , y = 5 (f) x = 3, y = 5 (c) x = 5, y = 6 (d) x = 3, y = 2
19 19 (b) x = 6, y = 8 (c) x = 1, y = 5
(f) x = 3, y = 5 (d) x = 4, y = –2
(e) x = –38 , y = 2 (b) x = 3, y = 2
9 (f) x = 33, y = –18
(b) x = 4, y = 1
3. (a) x = –1, y = 1 (f) x = 2, y = 1
(e) x = 2, y = 7

4. (a) x = 2, y = –1
(e) x = 2, y = 3

5. (a) x = 3, y = 1
(e) x = 81117 , y = –5167

Exercise 4.5.2 (b) x = 3, y = 5 (c) x = 0, y = –2 (d) x = 5, y = 2
(b) x = 2, y = 5 (c) x = –1, y = –2 (d) x = 5, y = –1
1. (a) x = 2, y = 1 (f) x = 3, y = 2
2. (a) x = 1, y = 1

(e) x = 1, y = –2

Exercise 4.6.1

1. (a) 3, –2 (b) 7, 9 (c) 1 , 5 (d) 1 , 2
2 3 2

(e) 5 , –65 (f) –29 , 3
6 4

2. (a) ± 5 (b) ± 3 (c) ± 5 (d) ± 6
2
(g) ± 4
(e) ± 5 (f) ± 6 (c) 0, 7 (h) ±6 (i) ± 4
3. (a) 0, 3 (h) 0, 223 (e) 0, 36
(b) 0, 5 (d) 0, 6
(f) 0, –2 (c) ± 4 (j) 0, 25
(g) 0, 1 (i) 0, 1
3 (c) –1, 6 9
(g) –13 , 2
(k) 0, 1 (l) 0, 312 (c) –32 , 4
6 (g) –1, 232
(k) –181 , 2
4. (a) ± 8 (b) ± 2 (c) ± 5 (d) ± 8
(g) ± 115
(e) ± 4 (f) ± 2 2
5. (a) 5, 6
(b) 3, 12 (d) 4, –6

(e) –3, 12 (f) –221 , 1 (h) 3 , 9 (i) –5, 2
3 2 4

6. (a) 1 , 1 (b) –31 (d) –2, 121
2

(e) –9, 7 (f) 1, 6 (h) –412 ,2

(i) a ± b (j) 114 , 4

7. (a) ± 13 (b) ± 7 (d) ± 34
3 (h) ± 9

(e) ± 1 (f) ± 5 (i) –14 , 1
2

164 | Mathematics – 9 Algebra

8. (a) 3, 4 (b) –1, 5 (c) 4, 512 (d) 1 , 2 (e) –1, 2
(h) –25 , 10 3 (e) 0, 4
(c) 1, 134
(f) –1, 2 (g) 1 , 113 (h) –14 , 1 (i) –3, 2
2 (d) 112 , 4

9. (a) 0, 4 (b) ± 8

(f) 5, 23 (g) 0, 113 (i) 0, 5
7

Exercise 4.6.2

1. (a) 1 (b) –4 (c) –7, 14 (d) –123 , 1
3

(e) –221 , 1 (f) –3, 113 (g) 4 (h) 1 ± 2 2 (i) 1 ± 4 41
3 5 2

2. (a) –2, 1 (b) –7 ± 17 (c) –213 , 3 (d) 1, 212
3 4

(e) 3 ± 3 (f) –125 , 3
2

3. (a) ± 1 (b) 0, 11 (c) ± 13 (d) 1, 1032
2 (f) –12 , 3 (g) –4, 1
(j) –14 , 1 (k) –1, 2
(e) –1, 5 (h) 3, 5

(i) 77 ± 71 (l) 1 , 1441
25 2

Exercise 4.6.3

1. (a) ± 4 (b) ± 137 (c) ± 3 (d) ± 1
9 (f) ± 6 2 30

(e) 0, 112

2. (a) –2, –6 (b) 1 ± 2 2 (c) 3 ± 105 (d) 3 ± 3
2 8 2

(e) –4, 9 (f) 4, 5 (g) (–7 ± 3 5 )a (h) 3 ± 33
(j) –2, 312 2 4

(i) –123 , –2 (k) –78 , –1 (l) 1 , 212
2

3. (a) 0, 5 (b) –6, 4 (c) –3, 8 (d) (1 ± 821)a
6 (g) –4, 3 82
(k) –1, 5
(e) ± 1 (f) 3, 5 (h) 1, 6
(i) –27 , 1
(j) 3 (– 1 ± –3 ) (l) –45 , 3
2

(m) 3 ± 37 n) 4(3 ± 10 ) (o) 3
7

4. (a) q ± q2 – 4pr (b) – m ± m2 + 4ln (c) b ± b2 + 4ac (d) – e ± e2 – 4df
2p 2l 2a 2d



Algebra Mathematics – 9 |165

166 | Mathematics – 9 Algebra

5Chapter

Geometry

Objectives:

At the end of this chapter, the
students will be able to:
 constant the triangles and

quadrilaterals having different
measurement.
 show the properties of
triangle theoretical as well as
experimentally.
 show the properties of
parallelograms theatrical.
 show the similar relation between
triangles and other polygon.
 construct the triangles and
quadrilateral with given
information.
 identify the different parts of
circle and show the properties
of circles theoretical as well as
experimentally.
 solve the logical problems of
circle.

Teaching Materials:

Geometrical instruments, pencils,
compass, setsquare scale, etc.,
board marker, scissors, thread, solid
objects, different programs related in
computer application, chart paper, etc.

5.1 Triangle A
cb
We have already learnt about the triangle in previous classes. Let’s
make simple revision. In the triangle alongside

 Is it a closed figure? Ba C
 How many line segments does it have? What are they?

 How many angles does it have?

 How many intersecting points it has, which are they?

A closed figure formed by three straight lines is called a triangle. A triangle is denoted by a Greek
letter . These straight lines are sides and the points of intersection are called vertices of the triangle.

In the above figure A, B and C are three vertices and AB, BC and CA are the three sides of the
triangle. A, B and C are the three angles of the ABC. The length of the sides of triangle
opposite to each vertex is also represented by small letter of the corresponding vertex, as BC = a,
AC = b and AB = c.

Types of Triangle

According to the measurement of their sides and their angles, the triangles are classified into different
types.

(a) According to sides A

(i) Scalene triangle

If all the sides of a triangle are unequal, then it is called a B C

scalene triangle.

In the adjoining figure, AB  BC  AC, also A  B  C. A
Thus, ABC is a scalene triangle.

(ii) Isosceles triangle

If any two sides of a triangle are equal, then it is called an isosceles triangle.

In the adjoining figure, B C

AB = AC and then the angles opposite to them are also equal i.e. B = C.

Thus, ABC is an isosceles triangle.

(iii) Equilateral triangle A

If all the sides of a triangle are equal, then it is called an
equilateral triangle. In the adjoining figure, AB = BC = AC. And
each angle of an equilateral triangle is always 60o.

i.e. A = B = C.

BC

168 | Mathematics - 9 Geometry

(b) According to angles

(i) Right angled triangle A

If one angle of a triangle is a right angle, then it is called a right
angled triangle. In the adjoining figure, B is a right angle. So,
it is a right angled triangle.

In a right angled triangle, the remaining angles are acute angles B C
and the sum of two remaining angles is 90o. The side opposite to
right angle is called the hypotenuse, which is longest side than A
other two sides of the right angled triangle.

(ii) Acute angled triangle

If all the angles of a triangle are acute angle, then it is called an B C
acute angled triangle. In the adjoining figure, A, B and C A C
are all acute angles. (each angle less than 90o)
B
(iii) Obtuse angled triangle

If one angle of a triangle is more than 90°, then it is called an
obtuse angled triangle.

In the adjoining figure, B is obtuse angle (more than 90o) and
the sum of other two angles of obtuse angled triangle is acute
angles. i.e. (A + C) < 90o

Properties of Triangles

A triangle holds some important properties which have already been discussed in previous classes.
Here, some of the important properties are verified experimentally as well as theoretically.

Theorem - 1

The sum of the angles of any triangle is equal to two right angles (180o). Prove

Given: ABC, ACB and BAC are D A E

To prove: three interior angles of ABC. B C
Construction: ABC + ACB + BAC = 180o
Proof Through the vertex A, draw a line DAE parallel to BC.

S.N Statements S.N Reasons
1. ABC = BAD 1. Being alternate angles, since DE//BC
2. ACB = CAE 2. Being alternate angles, since DE//BC
3. DAE = 180o 3. Being straight angle.
4. Whole part axiom
4. BAD + BAC + CAE = DAE 5. From statements (3) and (4)
5. BAD + BAC + CAE = 180o 6. From statements (1), (2) and (5)
6. ABC + BAC + ACB = 180o
Proved

Geometry Mathematics - 9 |169

Theorem - 2

The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Prove

Given: ABC is a triangle in which the side BC is produced to N where ACN is an exterior
angle of ABC.

To prove: ACN = ABC + BAC

Construction: Through C, draw CD//BA.

Proof

S.N Statements S.N Reasons
1. ABC = DCN 1. Being corresponding angles, since

2. BAC = ACD BA//CD
3. ABC + BAC = DCN + ACD 2. Being alternate angles, since BA//CD
4. ACD + DCN = ACN 3. From statements (1) and (2)
5. ABC + BAC = ACN
4. Whole part axiom
Alternate Method
5. From statements (3) and (4)

Proved

Proof: Statements S.N Reasons
S.N ABC + BAC + ACB = 180 1. Sum of all interior angles of a triangle
1.
ACB + CAN = 180 is 180.
2. ABC + BAC + ACB = ACB + 2. Being supplementary angles.
3. ACN 3. From statements (1) and (2)
ABC + BAC = CAN
4. 4. Cancelling ACB from both sides in
St. (3)

Proved.

Worked Out Examples

Example 1: In the given figure, AB//CD, BCD = 38o and
Solution: CAE = 104o. Find the measure of ACB.
Here, E A B
D
ABC = BCD = 38o [Alternate angles are
equal as AB//CD.]

ACB = CAE - ABC [Exterior angle relation
with its opposite interior angles]

ACB = 104o – 38o = 66o C

170 | Mathematics - 9 Geometry

Example 2: In the given figure PRQ = 2xo, QPR = 4xo P
Solution: and PQR = 3xo. Find the QPR and PQS. 4xo

Example 3: Here,
Solution:
In PQR, PRQ + PQR + QPR = 180o 3xo 2xo
[Sum of interior angles of a triangle] SQ R

or, 2x + 3x + 4x = 180o

or, 9x = 180o

 x = 180o = 20o
9

 QPR = 4 × 20o = 80o

Now, PQS + PQR = 180o [Linear pair]

or, PQS = 180o - 3 × 20o A
= 180° - 60° = 120o 68o

In the figure alongside, OB and OC are angle
bisectors of ABC and ACB respectively. If
BAC = 68o, find the size of BOC.

Here, O
In  ABC,

BAC + ABC + ACB = 180o xx y
B y
[Sum of interior angles of a triangle.]
or, 68o + 2x + 2y = 180o C

or, 2x + 2y = 180o - 68o

or, 2(x + y) = 112o

 x + y = 56o

Again, In  OBC,

OBC + OCB + BOC = 180o [Sum of interior angles of a triangle]

or, x + y + BOC = 180o

or, BOC = 180o - 56o [ x + y = 56o]

 BOC = 124o.

Exercise 5.1

1. Find the value of x, y and z from the given figures.

(a) A (b) P (c) D
5x B 102o
96o

3x 36o 136o x 112o E
B C RS A C

3x
Q

Geometry Mathematics - 9 |171

(d) A (e) A (f) A E
54o x 82o
72o
B 4x D
CD
104o

B xx yx y z 44o
CB CD

(g) A (h) A (i)
36o x 2x
A
yx

5x y 2x z 42o C
B C y BD
D 3x C
B D

2. Find the value of x, y and z in degree.

(a) A P B (b) P B
42o A R

Qx xS

C 36o D C QT D
(c) R Q
(d) T
A B P A
44o
x

D E 128o y z
x RB CS
18o
C A

3. (a) In the adjoining figure, AD and BD are angle bisectors D
of BAC and ABC respectively. If ACB = 74o, find B PC

the ADB.

(b) In the given figure, PS and RS are angle bisectors of Q S 108o
QPR and PRQ respectively. If PSR = 108o, find R

the size of PQR.

172 | Mathematics - 9 Geometry

(c) In the given figure, AB//CD, 2QPR = BPQ and AP B
2PRQ = DRQ. Find the size of PQR.

Q D

CR E
A 108o

(d) In the figure alongside, if CAE = 108o, find the size of Dy 2y C
BCD and ADC. x
2x P
4. (a) In the given figure, AC and BC are angle bisectors of B
C
A

BAP and ABR respectively to meet at C. Prove that:

ACB = 1 (180o - Q). QB R
2

AP B

(b) In the adjoining figure, APQ = RPQ and QR
CRQ = PRQ. Prove that: ABC = 2 (90o - PQR)

C
A

(c) In the given figure, bisectors of BAC and ABC meet

at D. Prove that: ADB = 1 (180o + C) D
2 C

B

(d) In the given figure, ABE = CBE, ACB = ACE, A E
and 2ABF = ACB. Prove that: BEC = BAC - F
ABF.
B CD

Theorem - 3

The sum of any two sides of a triangle is greater than the third sides. Verify experimentally.

Experimental verification

Experiment: Draw three triangles ABC with different shapes and sizes.

A AA

B CB C B C

To Verify: (i) (ii) (iii) Mathematics - 9 |173
Geometry
AB+AC>BC, AB+BC>AC, AC+BC>AB

Verification: Measure the length of sides AB, BC, and CA is each figure and tabulate.

Fig AB BC CA AB + BC AB + CA BC + CA Result

(i) AB+AC>BC, AB+BC>AC, AC+BC>AB

(ii) AB+AC>BC, AB+BC>AC, AC+BC>AB

(iii) AB+AC>BC, AB+BC>AC, AC+BC>AB

Conclusion: The above experiment shows that the sum of any two sides of a triangle is greater
than third side.

Theorem - 4

The angle opposite to the longer side is greater than the angle opposite to the shorter side of any
triangle. Verify experimentally.

Experimental verification

Experiment: Draw three triangles ABC of different shapes and size in which AB is the longest
and AC is the shortest side.

AA C

BC C B B A

(i) (ii) (iii)

To Verify: ACB > ABC
Verification:
Measure the sides AB and AC. Similarly measure the angles opposite to AB and AC
that is ACB and ABC respectively in each figure and tabulate.

Fig AB longest AC shortest ACB angle opposite ABC angle opposite Result
side
side to longest side to shortest side

(i) ACB > ABC

(ii) ACB > ABC

(iii) ACB > ABC

Conclusion: The above experiment shows that the angle opposite to the longer side is greater
than the angle opposite to the shorter side of any triangle.

Converse of theorem - 4

The side opposite to biggest angle is longer than the side opposite to smallest angle of any triangle.
Verify experimentally.

Experimental verification

Experiment: Draw three triangles ABC of different shapes and sizes in which, C is the biggest
and B is the smallest.

174 | Mathematics - 9 Geometry

(i) (ii) (iii)

To Verify: AB > AC

Verification: Measure the biggest C and smallest B. Similarly measure the sides AB and AC
opposite to C and B respectively in each figure and tabulate.

Fig ACB ABC AB (side opposite to AC (side opposite to Result
smallest angle)
(biggest angle) (smallest angle) biggest angle) AB > AC
AB > AC
(i) AB > AC

(ii)

(iii)

Conclusion: The above experiment shows that the side opposite to biggest angle is longer than
the side opposite to smallest angle of any triangle.

Theorem - 5

Of all straight line segments drawn to a given straight line from a given point outside it, the
perpendicular is the least. Verify experimentally.

Experimental verification

Experiment: Draw three different line segments PA, PB and PC to XY from the point P and also

draw PM  XY in each figure. P

PP

X A B M C Y XA M B C Y XA B M CY

To verify: (i) (ii) (iii)
Verification:
PM < AP, PM < BP, PM < CP

Measure the length of the line segment PA, PB, PC and PM in each figure and
tabulate.

Fig. Length of the line segment Result
PA PB PC PM
Perpendicular PM is shorter than other.
(i) Perpendicular PM is shorter than other.
Perpendicular PM is shorter than other.
(ii)

(iii)

Conclusion: The above experiment shows that of all straight line segments drawn to a given
straight line form a given point outside it, the perpendicular is the least.

Geometry Mathematics - 9 |175

Converse of theorem - 5

Of all the straight line segments drawn to a given straight line from a given point outside of it, the
shortest one is perpendicular to the given line. Verify experimentally.

Experimental verification

Experiment: Draw three different line segment PA, PB and PC to XY from the point P and also
draw shortest line segment PM.

PP P

X A B M C Y XA M B C Y XA B M CY

(i) (ii) (iii)

To verify: PM  XY
Verification:
Measure the length of each line segments and angle made by them to XY and
tabulate.

Measurement of length of line segments and angle made by them.

Fig. PA PAY PB PBY PC PCX PM PMX PMY Result

Angle made by
PM to XY is 90o

Angle made by
PM to XY is 90o

Angle made by
PM to XY is 90o

Conclusion: The above experiment shows that of all the straight line segments drawn to a given
straight line from a given point outside of it, the shortest one is perpendicular to the
given line.

Theorem - 6

Base angles of an isosceles triangle are equal. Prove.

Or

If any two sides of a triangle are equal, the angles opposite to them are equal. Prove.

Given:  ABC is an isosceles triangle in which AB = AC.

To prove: ABC = ACB

Construction: From the vertex A, draw AD  BC.

Proof

S.N Statements S.N Reasons
1.
1. In ABD and ACD
(i) Being AD  BC
(i) ADB = ADC (R) (ii) Given (AB = AC)
(iii) Being common side
(ii) AB = AC (H)
By R.H.S
(iii) AD = AD (S)

 ABD  ACD

176 | Mathematics - 9 Geometry

2. ABD = ACD 2. Bing corresponding angles of
3.  ABC = ACB congruent triangles.

3. From (2)

Proved.

Converse of theorem - 6

If two angles of a triangle are equal, the sides opposite to them are also equal. Prove.

Given: In ABC, ABC = ACB
To prove: AB = AC

Construction: From the vertex A, draw AM  BC
Proof

S.N Statements S.N Reasons
1.
1. In ABM and ACM
(i) Given ABC = ACB
(i) ABM = ACM (A) (ii) Being AM  BC (by construction)
(iii) Being common side
(ii) AMB = AMC (A)
By A.A.S
(iii) AM = AM (S) 2. Corresponding sides of the congruent

 ABC  ACB triangles

2.  AB = AC Proved.

Theorem - 7

The bisector of the vertical angle of an isosceles triangle is perpendicular
bisector of the base. Prove.

Given: In ABC, (i) AB = AC (ii) BAD = CAD

To prove: (i) BD = CD (ii) AD  BC
Proof

S.N Statements S.N Reasons
1.
1. In ABD and ACD
(i) Given
(i) AB = AC (S) (ii) Given

(ii) BAD = CAD (A) (iii) Common side of both triangles.

(iii) AD = AD (S) By S.A.S axiom

 ABD  ACD 2.(i) Corresponding sides of the congruent
triangles
2.(i)  BD = CD
(ii) Corresponding angles of the congruent
(ii) ADB = ADC triangles

(iii) AD  BC (iii) Adjacent angles of the linear pair being
equal i.e. ADB = ADC

Proved.

Geometry Mathematics - 9 |177

Theorem - 8

The line segment that joins the vertex and mid-point of the base of an isosceles triangle is
perpendicular to the base and bisects the vertical angle. Prove.

Given: (i) ABC is and isosceles triangle in which AB =AC.

(ii) M is a mid-point of BC or, BM = CM

To prove: (i) BAM = CAM (ii) AM  BC (AMB = AMC)

Proof

S.N Statements S.N Reasons

1. In ABM and ACM 1.

(i) AB = AC (S) (i) Sides of an isosceles triangle.

(ii) AM = AM (S) (ii) Being common side

(iii) BD = CD (S) (iii) M is mid-point of BC i.e. BM = CM

(Given)

 ABM  ACM By S.S.S

2.(i)  BAM = CAM 2.(i) Being corresponding angles of congruent

triangles.

(ii) AMB = AMC (ii) Being corresponding angles of

congruent triangles.

(iii) AM  BC (iii) Adjacent angles of the linear pair being
equal

i.e. AMB = AMC.

A Proved.

Worked Out Examples x

Example 1: From the given figure, find the value of x. 32o
Solution: D
Here, BC

CAD = ADC = 32o [Base angles of the isosceles triangle ADC.]

ACB = CAD + ADC [Exterior angle is equal to the sum of two opposite interior
angles.]

= 32o + 32o

= 64o

ABC = ACB = 64o [ AB = AC in  ABC]

BAC = x = 180o – (ABC + ACB) [ Sum of interior angles of a triangle]

 x = 180o – (64o + 64o) = 180o – 128o = 52o

Example 2: In the given figure, QR = PR = PS. If TPS = 8 4o, find the size of RPS.

Solution: Here, T
Let PQR = xo P 84o

 PQR = QPR = x [ QR = PR]

PRS = PQR + QPR

= x + x = 2x

QR S

178 | Mathematics - 9 Geometry

Again,

PRS = PSR = 2x [ PR = PS]

Now, PQS + PSQ = SPT Exterior angle is equal to the sum of two
or, x + 2x = 84o opposite interior angles.

or, 3x = 84o

 x = 28o

Now, RPS = 180o – (PRS + PSR) [ Sum of interior angles of a triangle]

= 180o – 4x

= 180o – 4 × 28o

= 180o – 112o

= 68o

Example 3: In the adjoining figure, PQR is an isosceles P
Solution: triangle in which PQ = PR and QT = RS. Prove that

PST is also an isosceles triangle.
Here,

Given: In PQR, S TR

To prove: PQ = PR, S and T are two points on QR such that QT = SR. Q
Proof PST is an isosceles triangle.

S.N Statements S.N Reasons
1.(i) QT = SR 1.(i) Given

(ii) QT - ST = SR - ST (ii) Subtracting common ST on both sides
(iii) QS = RT (iii) Remaining facts from statement (ii)

2. In PQS and PRT 2.
(i) Given
(i) PQ = PR (S)

(ii) PQS = PRT (A) (ii) Base angles of an isosceles triangle.
(iii) From statement 1 (iii)
(iii) QS = RT (S)

PQS  PRT By S.A.S axiom
(iv) PS = PT
(iv) Corresponding sides of the congruent
triangles.

3. PST is an isosceles triangle 3. Being PS = PT

A Proved.

Example 4: In the adjoining figure, ABC is an isosceles triangle in
which AB = AC. BO and CO are angle bisectors of

ABC and ACB respectively. Prove that: AO is an O
angle bisector of BAC.

Solution:

Given: (i) ABC is an isosceles triangle in which AB = AC. B C

(ii) BO and CO are angle bisectors of ABC and ACB respectively.
(iii) A and O are joined.

Geometry Mathematics - 9 |179

To prove: OA is an angle bisector of BAC i.e. OAB = OAC.

Proof

S.N Statements S.N Reasons
1.(i) ABC = ACB
1.(i) Base angle of an isosceles triangle.
(ii) OBC = OCB
(ii) OB and OC are angle bisectors of ABC
(iii) OB = OC and ACB respectively.
2. In OAB and OAC
(i) AB = AC (S) (iii) Base angles are equal of OBC.
(ii) OB = OC (S)
(iii) AO = AO (S) 2.

OAB  OAC (i) Given

(iv) OAB = OAC (ii) From statement 1 (iii)

(v) OA is an angle bisector of BAC (iii) Common side.

By S.S.S

(iv) Corresponding angles of congruent
triangles.

(v) From statement 2 (iv)

Proved.

Exercise 5.2

Group 'A'

1. Find the value of x, y and a from the following figures.

(a) A (b) P (c) A
x 44o y x

x 34o
B D CQ S R B DC

(d) A (e) A (f) A
72o 2a a
D 56o E
D BD
2x y a x
x C (h) CB C
B
158o
(g) CH (i)

A 87o A
E x

D 50o G
y

42o x BC D
BF

180 | Mathematics - 9 Geometry

2. Find the values of x, a and y from the following figures. A

(a) E (b) AC (c)

A 102o x a 24o x
xy
D

BC D y 34o 2a y
EB C BD EC

(d) A Y (e) P R (f) A
C y xT xy
45°
Xy 35o 75o O
2xx SQ 112o
B B
C

3. (a) In the given figure, AD = BD and AC = CD. If BAD = E
36o, find the size of CAE. A

BD C

A

(b) In the adjoining figure, ABC is a right angled triangle. If D
AD = CD and BD = BC, find ACB.
BC
(c) In the figure alongside, ABCD is a square in which BD is AD
a diagonal. If AED = 52o, find APB.
PE
BC

(d) In the given figure, ABCD is rhombus where DE  BE. If AD
CAD = 70o, find CDE.

Group 'B' B CE
A
1. In the given figure, ABC is an isosceles triangle in which
AB = AC. If CX  AB and BY  AC, prove that CX = BY and XY
AX = AY. BC

Geometry Mathematics - 9 |181

P

2. In the adjoining figure, RMPQ, QNPR and QN = RM. Prove M N
that PQ = PR and PM = PN. Q
A R
3. In the given figure, AC = BD and AD = BC. Prove that AP = BP D
and PCD = PDC.
P C
4. In the isosceles triangle ABC, AC = BC. If BP and AP are angle BA
bisector of ABC and BAC respectively, prove that CP is also
an angle bisector of ACB. P

BC
A

5. In the given figure, BD = CD = AD. Prove that BAC = 90o.

6. In isosceles ABC, AB = AC. CD and BE are angle bisectors of BDC
ACB and ABC respectively. Prove that CD = BE. A

DE

BC
A

7. In the adjoining figure, M is mid-point of BC. If MD  AB, ME DE
 AC and MD = ME, prove that AB = AC. B MC
A

8. In the given figure, ABC and ADE are equilateral triangles. If

AC = AE, prove that AX = AY. B XY D

9. In the given figure, ABC and ADE are equilateral triangles. CE
A

Prove that BE = CD. B D

CE

182 | Mathematics - 9 Geometry

10. In the adjoining figure, A is the mid-point of DE, BAD = CAE and DA E
ADC = AEB. Prove that BCD = CBE.

11. If the bisector of the vertical angle of a triangle bisects the base, BC
prove that the triangle is an isosceles triangle. A

12. In the adjoining figure, AB = AC, EF//AD and EF = CD. Prove E C
that EG = DG. B FG D
A
5.2 Parallelogram

Quadrilateral

A quadrilateral is a closed figure bounded by four line segments in a B D
plane. In the adjoining figure, ABCD is a quadrilateral where AB, BC, P C
CD and AD are sides of the quadrilateral.
Q S
Trapezium
R
A quadrilateral having a pair of opposite sides parallel to each other is
called trapezium. The pair of parallel sides are called bases of the
trapezium. In the adjoining figure, PQRS is a trapezium in which
PS//QR. PS and QR are bases and PQ and SR are legs of trapezium.

Parallelogram

Parallelogram is a quadrilateral having opposite sides are parallel. In the adjoining figure, DEFG is a
parallelogram in which DE//GF and DG//EF.

Properties of parallelogram E D G
 Opposite sides of the parallelogram are equal. F
 Opposite angles of the parallelogram are equal. A
 Diagonals of the parallelogram bisect each other. B D
 The triangles formed by diagonals are equal in area.
C
Rectangle

Rectangle is a parallelogram having each angle right angle. Or it is a
quadrilateral where all angles are right angles. In the adjoining figure,
ABCD is a rectangle.

Geometry Mathematics - 9 |183

Properties of rectangle AD
 Opposite sides of a rectangle are equal. BC
 All angles of a rectangle are right angles.
 Diagonals of the rectangle are equal. AD
 Diagonals of the rectangle bisect each other. BC
 The triangles formed by diagonals are equal in area.

Rhombus

A rhombus is a parallelogram having all the sides equal. In other words,
a quadrilateral where all sides are equal to each other is called a
rhombus. In the figure alongside, ABCD is a rhombus.

Properties of a rhombus
 All the sides of a rhombus are equal.
 Opposite angles of a rhombus are equal,
 Diagonals of a rhombus bisect each other at right angle.
 Diagonals of a rhombus bisect its vertical angles.
 The triangles formed by diagonals are congruent.

Square

A square is a rectangle having adjacent sides equal. Or square is a
rhombus having each angle right angle. In the adjoining figure, ABCD
is a square.

Properties of a square
 All sides of a square are equal.
 All angles of a square are right angles.
 Diagonals of a square are equal.
 Diagonals of a square bisect each other at right angle.
 Diagonals of a square bisect its vertical angles.
 The triangles formed by diagonals are congruent.

Theorem - 9

The straight lines joining the end points of two equal and parallel straight lines segments towards
the same sides are also equal and parallel. Prove

Given: AB = CD, AB // CD and the ends AB
points A, C and B, D are joined.

To prove: AC = BD and AC // BD.

Construction: Join B and C. CD

184 | Mathematics - 9 Geometry

Proof Statements S.N Reasons
S.N
1. In ABC and BCD
(i)
(ii) AB = CD (S) (i) Given
(iii)
ABC = BCD (A) (ii) Being alternate angles as AB//CD
2.(i)
BC = BC (S) (iii) Common side
(ii)
 ABC  BCD By S.A.S axiom
(iii)
 AC = BD 2.(i) Corresponding sides of congruent
triangles.

ACB = DBC (ii) Corresponding angles of congruent
triangles.

AC//BD (iii) Alternate angles ACB and DBC being
equal.

Proved.

Theorem - 10

The line segments joining the ends of two equal and parallel line segments towards the opposite

sides bisect each other. Prove. AC

Given: (i) AB = CD and AB//CD.

To prove: (ii) The end points A, D and B, C are joined O
which intersect each other at O. BD

AD and BC bisect each other at O i.e. AO
= DO and BO = CO.

Proof

S.N Statements S.N Reasons

1. In AOB and  COD. 1.

(i) ABO = DCO (A) (i) Being alternate angle as AB//CD

(ii) AB = CD (S) (ii) Given

(iii) BAO = CDO (A) (iii) Being alternate angles as AB//CD.

 AOB  COD By A.S.A.

2. AO = DO and BO = CO 2. Corresponding sides of congruent
triangles.

Proved.

Theorem - 11

Opposite sides of a parallelogram are equal. Prove it.

Given: PQRS is a parallelogram i.e. PS//QR, PQ//SR.

To prove: PQ = SR

QR = PS

Construction: Join P and R.

Geometry Mathematics - 9 |185

Proof Statements S.N Reasons
S.N
1. In PQR and PSR 1.
(i)
(ii) QPR = PRS (A) (i) Being alternate angles as PQ//SR
(iii)
PR = PR (S) (ii) Common side.
2.
PRQ = SPR (A) (iii) Being alternate angles as PS//QR

 PQR  PSR By A.S.A.

PQ = SR and QR = PS 2. Corresponding sides of the congruent
triangles.

Proved.

Converse of the theorem - 11

If the opposite sides of the quadrilateral are equal, the quadrilateral is a parallelogram. Prove.

Given: ABCD is a quadrilateral in which, AB = DC, AD = BC. A D

To prove: ABCD is a parallelogram (i.e. AB//DC and BC//AD)

Construction: Join A and C.

Proof BC
S.N
1. Statements S.N Reasons
(i) 1.
(ii) In ABC and ADC (i) Given
(iii) (ii) given
AB = DC (S) (iii) Common side.
2. (i) By S.S.S.
BC = AD (S) 2. (i) Corresponding angles of the congruent
triangles.
AC = AC (S) (ii) Alternate angles being equal as BAC
= ACD.
ABC  ADC (iii) Since, AB = DC
3.
BAC = ACD AB //DC
Opposite sides being parallel.
(ii) AB // DC

(iii) BC = AD
BC // AD

3.  ABCD is a parallelogram

Proved.

Theorem - 12

The opposite angles of a parallelogram are equal. Prove.

Given: PQRS is a parallelogram where PQ // SR, PS // QR

To prove: P = R and Q = S.

186 | Mathematics - 9 Geometry

Proof

S.N Statements S.N Reasons

1.(i) P + Q = 180o 1.(i) Sum of co-interior angles as PS//QR

(ii) P + S = 180o (ii) Sum of co-interior angles as PQ//SR

(iii) P + Q = P + S (iii) From statement 1 (i) and (ii).

(iv) Q = S (iv) From statement 1 (iii).

2. Similarly, P = R 2. Same as above statements and reasons.

Proved.

Converse of theorem - 12

If the opposite angles of the quadrilateral are equal, the quadrilateral is a parallelogram. Prove.

Given: ABCD is a quadrilateral where A = C and B = D. A D

To prove: ABCD is a parallelogram (AB//DC, AD//BC)

Proof BC
S.N
1.(i) Statements S.N Reasons

(ii) A + B + C + D = 360o 1.(i) Sum of the angles of a quadrilateral.

(iii) A + D + A + D = 360o (ii) Being B = D and C = A
(iv) (given).

(v) 2A + 2D = 360o (iii) From statement 1 (ii).
A + D = 180o
2.(i) (iv) Dividing both sides by 2.
AB//DC
(ii) (v) Being A + D = 180o or sum of co-
3. interior angles is 180o.

Similarly, A + B = 180o 2.(i) Same as above statements and reasons.

AD//BC (ii) Being A + B = 180o or, sum of co-
interior angles is 180o

ABCD is a parallelogram 3. Opposite sides being parallel.

Proved.

Theorem - 13

Diagonals of a parallelogram bisect each other. Prove. PS
Theoretical proof

Given: PQRS is a parallelogram in which diagonals PR and O
To prove: QS intersect at O.

PO = RO and QO = SO.

Proof QR

S.N Statements S.N Reasons
1.
1. In POQ and ROS
(i) Opposite sides of the parallelogram.
(i) PQ = RS (S) (ii) Being alternate angles as PQ//SR.

(ii) QPO = SRO (A) (iii) Vertically opposite angles are equal.

(iii) POQ = ROS (A) By S.A.A.
2. Corresponding sides of the congruent
POQ  ROS
triangles.
2. PO = RO and QO = SO
Proved.

Geometry Mathematics - 9 |187

Converse of theorem - 13

If the diagonals of the quadrilateral bisect each other, the quadrilateral is a parallelogram. Prove.
A D
Given: ABCD is a quadrilateral in which diagonals AC and

BD are bisect each other at O. (i.e. AO = CO and BO

To prove: = DO) O
ABCD is a parallelogram i.e. AB//DC, AD//BC.

BC

Proof

S.N Statements S.N Reasons
1.
1. In AOB and COD (i) Given
(ii) Vertically opposite angles.
(i) AO = CO (S) (iii) Given
By S.A.S. axiom
(ii) AOB = COD (A) 2. (i) Corresponding angles of the congruent
triangles.
(iii) BO = DO (S) (ii) Being alternate angles, since
OAB=OCD.
AOB  OCD (iii) Corresponding sides of the congruent
3. triangles.
2. (i) OAB = OCD 4. Being AB = DC and AB//DC.
Opposite sides being equal and
(ii) AB//DC parallel.

(iii) AB = DC Proved.
3. AD = BC, AD//BC
4. ABCD is a parallelogram

Worked Out Examples

Example 1: In the given figure, ABCD is a A D
Solution: parallelogram. DCE is isosceles B CE
triangle in which DC = DE. If CDE =
28o, find the size of BAD and ADE.

Here,

DCE + DEC + CDE = 180o [Sum of angles of a triangle]

or, 2DCE + 28o = 180o [DCE = DEC]

or, DCE = 180o - 28o = 1522o
2

 DCE = 76o

or DCE = DEC = 76o

or DCE = 180° - DEC

or ADE = 180o – 76o [DEC = 76o]
= 104o

188 | Mathematics - 9 Geometry

Now, [Alternate angles]
ADC = DCE = 76o [Co-interior angles]

or, BAD + ADC = 180o A
or, BAD = 180o – 76o

= 104o

D

Example 2: In the adjoining figure, ABCD is a rhombus. O
Solution: Diagonals AC and BD intersect at O. AC is BC
produced to E where CD = CE. If CDE = 26o,

find OAD and ODE.

Here,

CDE = CED = 26o [Base angles of an isosceles triangle] E

sOCD = CDE + CED Exterior angle is equal to the sum of two
= 26o + 26o opposite interior angles.
= 52o

 OAD = OCD = 52o [Being AD = CD]

And DOE = 90o [Diagonals of the rhombus bisect at right angle.]

ODE = 90o – OED [Remaining angle of right angled triangle.]

= 90o – 26o

= 64o

Example 3: In the adjoining figure, AB = CD, EB
Solution: AB//CD, 2BE = AE, 2CF = DF. Prove
that BECF is a parallelogram. A

Given: (i) AB = CD, AB//CD, 2BE = AE and 2CF = DF

To prove: BECF is a parallelogram CF D
Proof

S.N Statements S.N Reasons
1.(i) AB = CD 1.(i) Given
(ii) AE + BE = CF + DF (ii) Whole part axiom
(iii) 2BE + BE = CF + 2CF (iii) Given [AE = 2BE and DF = 2CF]
(iv) 3BE = 3CF (iv) Whole part axiom from statement 1

(v) BE = CF (iii).
(vi) BE//CF (v) Division axiom from statement 1(iv).
(vii) EC = BF, EC//BF (vi) Being AB//CD
2.  BECF is a parallelogram (vii) BE = CF and BE//CF.
2. Opposite sides being equal and

parallel.

Proved.

Geometry Mathematics - 9 |189

Exercise 5.3

Group 'A'

1. Find the value of x & y from the following figures.

(a) A D (b) A D E (c) A D
y yy

B x B 74o x x E
108o C C BC

E

(d) DE (e) D (f) D
A y A 30o A
y CE x
x 81o C
BC x
B y
B

(g) A xD (h) A (i) A D

72o
D 52o G ER y
y P
E xx
124o
y B E 32o C

BC BC F

(j) A E D (k) A B (l) P S T
y x 34o
Q 128o R D
Fx y Q E
A
32o C Px
B DC

2. (a) In the adjoining figure, ABCD is a square. If DAE = 34o, find AEF.

B FC
AED

(b) In the given figure, ABCD is a rectangle. E is any point
on AD such that AB = AE = DE. Find BEC.

B C
A D

(c) In the given figure, ABCD is a parallelogram in which BE C
AF is an angle bisector of BAD. If EFC = 58o, find
ABE and AEC. F
Geometry
190 | Mathematics - 9

(d) In the given figure, ABCD is a rhombus and CDE is AD
an isosceles triangle in which CD = DE. If CDE = 42o,
find the size of BDC. B CE
AD
3. (a) In the adjoining figure, ABCD is a square and AEC is
an equilateral triangle. Find the size of BCE. EB C
A F
(b) In the figure alongside, ABC is an isosceles triangle in
which AB = AC and DBCF is a parallelogram. If ECF DE
= 44o, find AED.
BC
(c) In the adjoining figure, ABCD is a square and PBC is AD
an equilateral triangle. Find the size of APD. P
BC

(d) In the adjoining figure, ABCD and PQRS are two AD
squares. If PBQ = 28o, find the size of BRS. P

Group 'B' BQ C S

1. Prove the following statements. R
(a) If a parallelogram has equal diagonals, then it is a rectangle.

(b) If the diagonals of a parallelogram are at a right angle, the parallelogram is a rhombus.

(c) If the diagonals of a rhombus are equal, the rhombus is a square.

(d) If one angle of a rhombus is a right angle, the rhombus is a square.

(e) If one angle of a parallelogram is a right angle, the parallelogram is a rectangle. A

2. In the given sABC and DEF, AB = DE, AB//DE, BC = EF, D
BC//EF. Prove that AC = DF and AC//DF. BC

P E F
A D
3. In the figure alongside, ABCD is a parallelogram. The diagonal
AC is produced to either side to the points P and Q such that BC
AP = CQ. Prove that BP = DQ and BP // DQ. Q

Geometry Mathematics - 9 |191

P S
A
4. In the adjoining figure, PQRS is a parallelogram. A and B are
two points on the diagonal PR such that PA = RB. Prove that B
AQBS is a parallelogram. QR

5. In the figure alongside, ABCD is a parallelogram. M and N are A MD
midpoints of AD and BC respectively. MN and BD intersect at
O. Prove that OM = ON and BO = DO. O C
BN D
6. In the given figure, ABCD is a parallelogram. BM and DN are
angle bisectors of ABC and ADC respectively. Prove that AM
BM = DN, BM//DN.
B NC
7. In the figure alongside, ABCD is a square. E and F are any
points on BC and DC respectively, such that DE = AF. Prove AB
that DPAF. E

P

D FC

8. In the figure alongside, ABCD is a square in which P, Q, R and S AS D
are any points on AB, BC, CD and AD respectively. If BP = CQ =
DR = AS, prove that PQRS is also a square. R
P
9. In the given figure, ABCD is a parallelogram in which AC is a
diagonal. BP and DQ are perpendiculars to AC. D, P, and B, Q B QC
are joined. Prove that PBQD is a parallelogram. AD

10. In the adjoining figure, AM and CN are perpendiculars to P
diagonal BD of the quadrilateral. If AO = CO and BM = DN,
prove that ABCD is a parallelogram. Q
BC

AD

N
MO

BC

192 | Mathematics - 9 Geometry

5.3 Mid-Point Theorem

Theorem – 14 (A)

A line segment drawn through the mid-point of one side of a triangle and parallel to another side

bisects the third side. A

Given: In ABC,

M is the mid-point of BC i.e. BM = CM FN
BA//MN

To prove: AN = CN

Construction: Through N, Draw NF | | BC. B MC

Proof

S.N Statements S.N Reasons
1.(i) BMNF is a parallelogram
(ii) BM = FN 1.(i) Opposite sides are parallel
(iii) BM = MC
(iv) FN = MC (ii) Opposite sides of the parallelogram
2. In AFN and NMC
(iii) Given
(i) FAN = MNC (A)
(iv) From (ii) and (iii) of 1.

2.

(i) Being corresponding angles as
AB//NM.

(ii) ANF = NCM (A) (ii) Being corresponding angles as FN//BC

(iii) FN = MC (S) (iii) From statement 1. (iv)

AFN  NMC By A.A.S.

(iv) AN = CN (iv) Corresponding sides of the congruent
triangles.

Proved.

Theorem - 14 (B)

A line segments joining the mid-points of any two sides of a triangle is parallel to the third side and
A
it is equal to half of the length of the third side.

Given: In ABC, M and N are mid-points of sides M ND
AB and AC respectively. M and N are joined. BC
To prove:
Construction: MN // BC, MN = 1 BC
2

Draw a line CD parallel to BA from C and
produce MN to meet CD at D.

Geometry Mathematics - 9 |193