9 Green C Math Class

9.1 Equation

Two variable linear equations can be solved by using following method:
a. Elimination method b. Substitution method c. Graphical method

a. Elimination method:

In this method, one variable is eliminated to find value of other variable. When we solve
a pair of simultineous equation, we should use the following steps:

Step I : Making coefficient of any one variable of equation by multiplying or dividing
the equation by constant value.

Step II : Add or subtract the new equations and eliminate any one variable.

Step III : Substitute the value of obtained variable in any one of the given equations,
then we will get the values of both variables.

Worked Out EXAMPLES

EXAMPLE 1 Solve the following equations by elimination method
Solution :
4x + 5y = 5

x + 3y = 3 Here, we should make coefficient of y equal.
Here, So, L.C.M. of 5 and 3 is 15 then,
3 × 5 = 15
4x + 5y = 5 ................ (i) 5 × 3 = 15

x + 3y = 3 .................. (ii)

Multiplying equation (i) by 3 and equation (ii) by 5 then subtracting eqn
(ii) from equation (i)

12x + 15y = 15

– 5x –+ 15y =–15

7x = 0
0
or, x = 7

∴ x = 0

Putting the value of x in equation (i)

4x + 5y = 5

or, 4 × 0 + 5y = 5

or, 0 + 5y = 5

or, 5y = 5
5
y = 5 =1

∴ x = 0
y = 1

∴ The values of x and y are 0 and 1 respectively.

GREEN Mathematics Book-9 145

EXAMPLE 2 Solve by elimination method:

3 + 4 =2
x y

6 – 1 = 7
x y 4

Solution : The given equations are:

3 + 4 = 2 ..................... equation (i)
x y

6 – 1 = 7 ..................... equation (ii)
x y 4

Multiplying equation (ii) by 4 and adding with equation (i) we get,

3 + 4 = 2
x y

24 – 4 = 7
x y

3 + 24 = 9
x 1

or, 27 = 9
x 1

x = 27 =3
9

Substituting the value of x in equation (i)

We get,

3 + 4 =2
x y

or, 3 + 4 =2
3 y

or, 4 =1
y

or, y = 4
∴ The values of x and y are 3 and 4 respectively.

146 GREEN Mathematics Book-9

Substitution method:
To solve simultaneous equations by this method, the value of one variable should be
calculated from any one of equations and then substitute the value. So obtained on these
equations. We use the following steps to solve the equations by subtitution.

Step I : Express any one variable in terms of other variables.

Step II : Substitute this value in another equation, we will get the value of one variable.

Step III : Put the value of one variable which is obtained in step II.

EXAMPLE 1 Solve the following equations:

Solution : 4x + 5y = 5 ............ (i)

x + 3y = 3 ............ (ii)

From equation (ii)

x + 3y = 3

x = (3 – 3y)

Substituting the value of x in equation (i)

4x + 5y = 5

or, 4(3 – 3y) + 5y = 5

or, 12 – 12y + 5y = 5

or, 12 – 7y = 5

or, – 7y = 5 – 12

or, – 7y = – 7

or, y = –7
–7

∴ y = 1

Putting the value of y in equation (i)

4x + 5y = 5

or, 4x + 5 × 1 = 5

or, 4x + 5 = 5

or, 4x = 5 – 5

or, 4x = 0

x = 40
x = 0
∴ x = 0
y = 1

GREEN Mathematics Book-9 147

EXAMPLE 2 Solve the following equations by substitution method.

2 + 3 = 7, 5 – 6 =8
x y x y

Solution : Here,

2 + 3 =7 ..................... equation (i)
x y

5 – 6 =8 ..................... equation (ii)
x y

From equation (i)

2 + 3 =7
x y

or, 2 = 7y–3
x y

or, x = 2y ..................... equation (iii)
7y–3

Substitute the value of x in equation (ii)

5 – 6 =8
x y

or, 5 – 6 =8
2y y

7y–3

or, 35y – 15 – 6 =8
2y y

or, 35y – 15 – 12 = 8
2y

or, 35y – 27 = 16y

or, 19y = 27

y = 27
19

Again, putting the value of y in equation (iii)

y = 2y 3
7y –

2 . 27 54
19
= = 19 = 54 = 9
27 189 – 57 132 22
7. 19 –3
19

∴ The value of x and y are 9 and 27 respectively.
22 19

148 GREEN Mathematics Book-9

EXERCISE: 9.1

A. Very Short Questions

1. Solve the following equations by substitution method.

a. x + y = 2 b. x + 2y = 6 c. y = 3x
x + y = 20
x – y = 0 y = 2

2. a. If equation x + 3y = 8 and x = 2, what is the value of y?
b. Find the value of y in the equation 2x – y = 5 if x = 5

B. Short Questions b. 7x + 4y = 4 , 2x + 3y = 16
3. Solve by using elimination method. d. x + 6y = 11, 2x – 3y = 7
f. 2x – y = 5, 3x + 2y = 11
a. 5x – y = 5, x – 2y = 1 h. 2x + y = 9, 2x – y = 11
c. 2x – y = – 3, 3x – y = – 1 j. 7x + y + 9 = 0, 11x + 3 – 4y = 0
e. x + 3y = 8 , 3x – 2y = 2 l. 6x + 12 = 7y, 4y + 11 = 7x
g. 3x – y = 17, 5x + y = 15
i. 7x – y + 9 = 0, 11x + 4y + 3 = 0 n. 5m + n = 10, 14m + 3n = 18

k. 9x – 3y = 27, 4x + 5y = 31 b. 2x – 3y + 7 = 0, 5x + 3y +7 = 0
m. 2x – y = 7, 3x – y = 3
d. x + y = 0, 3x – 2y = 10
42
4. Solve by using substitutions method. f. 10x – y = 17, 9y + 4x = 35

a. x + 2y = 6, x – y = 3 h. 3x – 21 = 5y, 1 – 5x = 3y
c. 3x – 2y = 4, 5x – 2y = 0
e. 4x + 3y = 7, 2x + 2y = 3 j. 3x – 4y = – 5, 4x + 5y – 14 = 0
g. 5x – 7y + 11 = 0, x + y – 5 = 0
i. 2x – 7y = 3, 2x + y = 11 l. 2x – 3y + 1 = 0, – x + 2y – 1 = 0
k. 7x – 8y + 14 = 0, 5x – 9 = 3y
n. 2x – 3y = 5, x – y = 1
m. 5x – 3y = 5, – 3x, + 2y = – 2 2

C. Long Questions

5. Solve the following equations by using substitution method or elimination method

a. 2x + y = 13, 2y – x = 1 b. 2 + 3 = 5, x – y = 0
3 5 x y

c. x – y = 0, x – 1 (y – 1) = 1 d. 1x2 + y6 = 5, 30 – 8 = 1
4 3 6 x y

GREEN Mathematics Book-9 149

e. 6– 9 = 3 , 4 – 3 = 7
5x 10y 10 3x 4y 12

f. 3 + 4 = 25 , 40 + 14 = 42
2x – 3y 2x + 3y 7 2x – 3y 2x + 3y

g. 2 + 5 = 3, 3 + 20 = 1
x–y x+y y–x x+y

h. 20ax – by = ab, – 5ax + 3by = 52ab

i. 2(a – b) x – (a + b) y = a² – b², 3(x – y) = 3b

j. a + b = m, b + a = n
xy xy

9.2 Solving equation by graphical method

To solve two simulteneous equations by graphical method, we should follow the following
steps:
Step I : Draw the graph of the given equations in the same graph paper.
Step II : Identify the intersecting point of two lines.
Step III : Find the values of x and y from the intersecting point of lines.

Worked Out EXAMPLES

EXAMPLE 1 Solve :

Solution : x + y = 6 .............. (i)
x – y = 4 .............. (ii)
From equation (i)
x + y = 6
y = 6 – x

Table for value of x and y

x6025 The points (6, 0), (0, 6), (2, 4) and (5, 1)
y0641 are plotted on graph.

150 GREEN Mathematics Book-9

For equation (ii)
x – y = 4
y = x – 4
Table for equation (ii)

x04 5 2 The points (6, –4), (4, 0), (5, 1) and (2, –2)
y –4 0 1 –2
Y

Putting above points on the same graph, two straight (0, 6)
lines intersect at the point (5, 1).
Therefore the solution of the above equations are given x+y = 0
by x = 5 and y = 1.
x + (2, 4)
X' =6
y

(5, 1) x–y = 4

(0, 0) (4, 0) (6, 0)
X
O x–y=4 (4, 0)

(2, – 2)

(0, – 4)

Y'

EXAMPLE 2 Solve graphically : 3x – 5 = 2x – 3 .
Solution : 43

3x – 5 = 2x – 3 = y (suppose)
43

y = 3x – 5 ................... equation (i)
4

y = 2x – 3 .................... equation (ii)
3

From equation (i)

y = 3x – 5
4

Table for value of x and y

x –1 3 –5
y –2 1 –5

Ploting the points (–1, –2), (3, 1) and (–5, –3) on graph and drawing a line
through the point.

From equation (ii)
y = 2x – 3

3

GREEN Mathematics Book-9 151

Table for value of x and y 10 small square boxes = 1 unit

x 0 3 –3 Y
y –1 1 –3
X' O (3, 1)
Ploting the points (0, –1), (3, 1)
and (–3, –3) in the same graph X
paper.
y = 2x–3 (0, –1)
Two lines are intersecting at 3
point (3, 1).
(–3, –3) (–1, –2)
So, x = 3 and y = 1
(–5, –5) y = 3x–5
∴ The required value of x is 3. 4
Y'

EXERCISE: 9.2

A. Very Short Questions

1. Solve the following equations graphically.

a. 2x + 3 = y, 3x – y + 1 = 0 b. 6x + 12y = 18, 15x + 4y = 19

c. 4x – 3 = 6x –1
5 7

B. Long Questions

2. Solve the following simultaneous equations graphically.

a. x + 3y = 8, 3x – 2y = 2 b. 3x + 2y = 12, 3x – y = 3

c. 3x + 5y = 12, 3x – 5y + 18 = 0 d. 3x – y = – 1, 2x + y = 6

e. x+6 = 38 – 3x f. y – 1 = 2x, y = x + 2
3 5
x y
g. 2x + y = 8, x – y = 1 h. 2 + 4 = 5, 2x – y = 4

i. 3x – y = 17, 5x + y = 15 j. 2x + y = 9, 2x – y = 11

k. 2x + y = 13, 2y – x = 1 l. 2 + 3 = 5, x – y = 0
3 5 x y

152 GREEN Mathematics Book-9

10

Quadratic Equation

Estimated Teaching Periods : 6

In the context where algebra is identified with the theory of equations,
the Greek mathematician Diophantus has traditionally been known as
the "father of algebra" but in more recent times there is much debate over
whether al-Khwarizmi, who founded the discipline of al-jabr, deserves
that title instead.

Contents

10.1 Introduction of quadratic equation
10.2 Factorization method
10.3 Completing square method
10.4 Solving quadratic equation by using formula

Objectives

At the end of this unit, students will be able to:
know about the quadratic equations and its types
solve the quadratic equations by factorization method, completing square method and
by using formula

Materials

Pencil, scale, graph papers, graph chart.
Derivation chart of ax2 + bx + c = 0

GREEN Mathematics Book-9 153

10.1 Introduction of Quadratic Equation

If the variable has the highest exponent 2 in an algebraic equation then it is called second
degree equation. The second degree equation has only one unknown variable, it is known
as quadratic equation. The standard form of quadratic equation is ax² + bx + c = 0.

In the quadratic form of equation, if middle term is missed, it is called pure quadratic
equation. But the quadratic equation in standard form is called adfected quadratic
equation.

Solution of quadratic equation.

1. Factorisation method 2. Completing square method

3. Using quadratic formula.

10.2 Factorization method

To solve quadratic equation by the factorize method, we should use the following steps:
Step I : Remove the brackets or fraction.
Step II : Reduce the equation in the form ax2 + bx + c = 0
Step III : Factorize left side.
Step IV : Write each factor is equal to zero.
Step V : Solve these equation and get two values of variable.
Following worked out example will clear in solving quadratic equation by factorisation
method.

Worked Out EXAMPLES

EXAMPLE 1 Solve the quadratic equation, x² + 7x + 10 = 0 by factorisation method.

Solution : H ere, x² + 7x + 10 = 0

or, x² + x(5 + 2) + 10 = 0

or, x² + 5x + 2x + 10 = 0

or, x(x + 5) + 2 (x + 5) = 0

or, (x + 5) (x + 2) = 0

either, x + 5 = 0 or, (x + 2) = 0

when, x + 5 = 0

or, x = – 5,

when, x + 2 = 0 or, x = – 2

∴ x = – 2 or – 5

154 GREEN Mathematics Book-9

EXAMPLE 2 If 2 and –1 are the roots of a quadratic equation, find the equation.
Solution : Let x be the variable of a quadratic equation
then x = 2
EXAMPLE 3 or x = – 1
Solution : then (x – 2) (x + 1) = 0
or, x2 + x – 2x – 2 = 0
EXAMPLE 4 ∴ x2 – x – 2 = 0 which is the required equation.
Solution :
Find the roots of x2 – 5x = 0 by using factorization.
EXAMPLE 5
Solution : Here,

x2 – 5x = 0

or, x (x – 5)= 0

Either, x = 0 or, x–5 = 0
x = 5
x = 0

∴ x = 0, 5

Solve by using factorization method x2 – 2x = 8

x2 – 2x = 8
or, x2 – 2x – 8 = 0
or, x2 – x (4 – 2) – 8 = 0

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

or, (x – 4) (x + 2) = 0

Either, x – 4 = 0 or, x + 2 = 0

i.e. x = 4 x = –2

∴ x = 4, – 2

Solve by using factorisation method.

x–2 + x+2 = 2(x + 3)
x+2 x–2 x–3

x–2 + x+2 = 2(x + 3)
x+2 x–2 x–3

or, (x – 2)2 + (x + 2)2 = 2(x + 3)
(x + 2) (x – 2) x–3

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

or, 2x2 + 8 = 2(x + 3)
x2 – 4 x–3

or, 2(x2 + 4) = 2(x + 3)
x2 – 4 x–3

or, (x2 + 4) (x – 3) = (x + 3) (x2 – 4)

or, x3 – 3x2 + 4x – 12 = x3 – 4x + 3x2 – 12

GREEN Mathematics Book-9 155

or, x3 – 3x2 + 4x – 12 – x3 + 4x – 3x2 + 12 = 0
or, –6x2 + 8x = 0
or, –2x (3x – 4) = 0

Either, – 2x = 0 or, 3x – 4 = 0

i.e. x = 0 i.e. 3x = 4
–2
∴ x = 0 ∴ x = 4
3
∴ The values of x is either 0 or 4 .
3

EXAMPLE 6 Solve : 3x – 8 = 5x – 2
Solution : x–2 x+5

3x – 8 = 5x – 2
x–2 x+5
or, (3x – 8) (x + 5) = (5x – 2) (x – 2)
or, 3x2 + 15x – 8x – 40 = 5x2 – 10x – 2x + 4
or, 3x2 + 7x – 40 – 5x2 + 12x – 4 = 0
or, –2x2 + 19x – 44 = 0
or, –(2x2 – 19x + 44) = 0
or, 2x2 – 11x – 8x + 44 = 0

–1
or, x(2x – 11) –4 (2x – 11) = 0

or, (2x – 11) (x – 4) = 0

Either, 2x – 11 = 0 or, x – 4 = 0

i.e. 2x = 11 i.e. x = 4

i.e. x = 11
2
∴ The values of x are 11 and 4.

2

EXERCISE: 10.1

A. Very Short Questions
1. Solve the following quadratic equations.

a. (3x – 1) (x + 4) = 0 b. (x – 2) (x + 3) = 0 c. (7x – 8) (2 – 7x) = 0

d. x2 = 5x e. x2 – 3x = 0 f. 9x2 – 2x = 0
g. 4x = a

x

156 GREEN Mathematics Book-9

2. Solve the following quadratic equations by factorization method.

a. x² + 6x + 8 = 0 b. 3x² – 5x – 2 = 0 c. – 8x² + 6x + 9 = 0

3. a. Find the roots of quadratic equation ax2 – bx = 0.

b. Solve 6x = 15 x2 – 1
2x c. Solve: 8 = 3

B. Short Questions
4. Solve the following quadratic equations by factorization method.

a. x² + 6x + 9 = 0 b. a² – 6a + 9 = 0 c. p² – 2p – 15 = 0

d. x² + 2x – 15 = 0 e. 6m² – m – 1 = 0

5. Solve the following equation by using factorization method.

a. 4x² = 9 b. x(x + 1) = 4x c. (x – 3) (x + 5) = 48
f. x = 4
d. (2x – 3) (x – 5) = (x – 3)2 e. x + 6 = 5
x 9x

C. Long Questions

6. Solve the following equations.

a. (x + 2)2 – 3(x + 2) + 2 = 0 b. x2 – 3ax + 2bx – 6ab = 0
c. (1 – x) (2x – 3) + 10 = (x + 1) (x – 5) x – 5 4 – 10

d. 2 = x
x–2 x–2

e. x + 3 + x – 3 = 2x – 3 f. x + 1 + x = 2 1
x+2 x–2 x–1 x x+1 6

g. 3x – 7 = x + 1 h. 4 – 5 = 3
2x – 5 x – 1 x–1 x+2 x

i. 1 + 2 = 3 j. x + 1 + x + 2 = 2x + 13
x–1 x 2–x x–1 x–2 x+1

GREEN Mathematics Book-9 157

10.3 Completing square method

In this method, we should make an expression perfect square by adding or subtracting
suitable term in both sides of expression.
Look at the following worked out examples to be more clear.

Worked Out EXAMPLES

EXAMPLE 1 Solve : x2 + 4x + 4 = 9

Solution : x2 + 4x + 4 = 9
or, x² + 4x + 4 = 9
or, (x + 2)2 = 9
or, x + 2 = 9
∴ x + 2 = ± 3
Taking square root on both sides
We get,
(x + 2) = ± 3
or, x = –2 ± 3
Taking +ve sign Taking –ve sign
x = – 2 + 3 x = – 2 – 3

= 1 = –5

∴ The values of x are 1 and –5.

EXAMPLE 2 Solve the quadratic equation, x² + 7x + 10 = 0 by completing square
method.

Solution : We have,

x² + 7x + 10 = 0

or, x² + 2. x . 7 + o 7 2 – o 7 2 + 10 = 0

22 p 2 p

or, ox + 7 2 – 49 + 10 = 0

p
24

or, ox + 7 2 – 49 – 40 = 0

p
24

or, ox + 7 2 – 9 = 0

p
24

or, ox + 7 2 = 9
4
p
2

158 GREEN Mathematics Book-9

or, ox + 7 2 = o± 32

p p
22

or, ox + 7 = ± 32
o2p
p
2

or, ox + 7 = ± 3

p
22

Taking +ve sign Taking –ve sign

x + 7 = 3 x + 7 = – 3
22 22

or, x = 3 – 7 or, x = – 3 – 7
2 2 or, x 22

or, x = 3 – 7 = –3–7
2 2

or, x = – 4 or, x = – 10
2 2

∴ x = – 2 ∴ x = – 5

∴ x = – 2 or – 5

EXAMPLE 3 Solve standard form of quadratic equation ax² – bx – c = 0, by using
completing square method.

Solution : Here,

ax² – bx – c = 0

or, ax² – bx = c

or, ax² – bx = c [Dividing both sides by 'a']
aa a
bx
or, x² – a = c
a

or, x² – 2 . x b + o b 2 o b 2 c

p– p =
2a 2a 2a a

or, ox – b 2 o b 2 + c

p = p
2a 2a a

or, ox – b 2 b² + c

p =
2a 4a² a

or, ox – b 2 b² + 4ac

p =
2a 4a²

or, x – b = ± b² + 4ac
2a 4a2

GREEN Mathematics Book-9 159

or, x = b ± b² + 4ac
2a 2a
or,
Taking +ve sign x = b ± b² + 4ac
2a
Taking – ve sign,
x = b + b² + 4ac
2a

x = b – b² + 4ac
2a

∴ Roots of quadratic equation ax² + bx + c = 0 are

b + b² + 4ac and + b – b² + 4ac
2a 2a

EXAMPLE 4 Solve : 2x – 9 – 2x – 7 = 7
2x – 7 2x – 9 12

Solution : 2x – 9 – 2x – 7 = 7
2x – 7 2x – 9 12

or, (2x – 9)2 – (2x – 7)2 = 7
(2x – 7) (2x – 9) 12

or, (4x2 – 36x + 81) – (4x2 – 28x + 49) = 7
4x2 – 18x – 14x + 63 12

or, 4x2 – 36x + 81 – 4x2 + 28x – 49 = 7
4x2 – 32x + 63 12

or, – 8x + 32 = 7
4x2 – 32x + 63 12

or, 28x2 – 224x + 441 = – 96x + 384

or, 28x2 – 224x + 441 + 96x – 384 = 0

or, 28x2 – 128x + 57 = 0

or, 28x2 – 128x + 57 = 0
28 28 28
57 = 0
or, x2 – 32x + 28
7

160 GREEN Mathematics Book-9

or, x2 – 2 . x 16 + 16 2 – 16 2 + 57 = 0
7
op op
7 7 28

or, ox – 16 2 = 256 – 57
49 28
p
7

or, ox – 16 2 = 1024 – 399
196
p
7

or, ox – 16 2 = 625
196
p
7

or, ox – 16 2 = o± 25 2

p p
7 14

or, x – 16 = ± 25
7 14

or, x = 16 ± 25
7 14

Taking +ve sign + 25 Taking –ve sign – 25
Either x = 16 14 or, x = 16 14

7 7

32 + 25 32 – 25
= 14 = 14

= 1574 = 174 = 1
∴ The value of x are 57 and 1 . 2

14 2

EXERCISE: 10.2

A. Very Short Questions

1. Solve the following equations.

a. x2 – 2 . x . 3 + o 3 2 o 3 2 = 0 b. x2 + 2 . x . 1 + o 1 2 = o 1 2 + 6

22 p – 2 p 22 p 2 p

c. x2 – 2 . x . 5 + 52 = 52 – 9 d. x2 + 2 . x . 4 + o 4 2 = o 4 2 + 9

55 p p
5 25

B. Short Questions

2. Solve the following quadratic equations by completing method.

a. x² + 4x – 5 = 0 b. x² – 5x + 4 = 0 c. x(x + 6) + 8 = 0

d. x2 + 3x – 18 = 0

GREEN Mathematics Book-9 161

C. Long Questions

3. Solve the following equation by completing square method.

a. x² + 5x – 6 = 0 b. 4x² – 5x + 1 = 0 c. x2 – 15x + 36 = 0

d. 2p2 + 3p + 1 = 0 e. 4x2 + 5x – 6 = 0 f. x2 + 7x + 12 = 0

4. Solve the following equation by square completing method.

a. x+3 = 3x – 7 b. 75xx ++ 1 = 3x + 1 c. x + 2 – 1 = 1
x+2 2x – 3 5 7x + 1 6 x+2 6

d. 3 + 2 = 8
5–x 4–x x+2

10.4 Solving quadratic equation by using formula

The standard form of quadratic equation is ax2 + bx + c = 0. These expressions don't have
simple factors. In such condition we use the quadratic formula.

To find the formula from ax2 + bx + c = 0

ax2 + bx + c = 0

or, ax2 + bx + c = 0
a aa

or, x2 + 2 . x . b + o b 2 – o b 2 + c = 0

p p
2a 2a 2a a

or, ox + b 2 – b2 + c = 0

p
2a 4a2 a

or, ox + b 2 = b2 – 4ac

p
2a 4a2

or, ox + b 2 = o± b2 – 4ac 2

p p
2a 4a2

Taking square root in both side:

ox + b = ± b2 – 4ac
4a2
p
2a

or, x = – b b2 – 4ac
2a
±
2a

∴ x = – b ± b2 – 4ac


2a

162 GREEN Mathematics Book-9

Worked Out EXAMPLES

EXAMPLE 1 Solve the equation : 2x2 – 3x – 1 = 0 by using formula.
Solution :
Here, the given equation 2x2 – 3x – 1 = 0

Comparing the given equation with ax2 + bx + c = 0

a = 2, b = – 3 and c = – 1

We have,

x = – b ± b2 – 4ac


2a

= (–3) ± (–3)2 – 4 . 2 .(–1)
2×2

= – 3 ± 9 + 8
4

= 3± 17

4

∴ The roots of equation are o 3 + 17 p and o 3 – 17 p.
44

EXAMPLE 2 Solve quadratic equation x² – 7x + 12 = 0, by using formula.

Solution : Here,

x² – 7x + 12 = 0

Comparing it with quadratic equation

ax² + bx + c = 0

We get a = 1, b = – 7, c = 12

Then, we have,

x = – b ± b² – 4ac
2a

Substituting the value of a, b and c, we have,

x = – (– 7) ± (– 7) ² – 4 × 1 × 12

2×1

= 7 ± 49 – 48
2

= 7 ± 1
2

= 7 ± 1
2

GREEN Mathematics Book-9 163

Taking +ve sign, Taking –ve sign,

x= 7+1 x= 7–1
2 2
8 6
= 2 = 2

=4 =3

∴ The value of x are 4 and 3.

EXAMPLE 3 Solve : 1 + 1 = 1
Solution : x–2 x+3 5

1 + 1 = 1
x–2 x+3 5

1(x + 3) + 1(x – 2) = 1
or, (x – 2) (x + 3) 5

x+3+x–2 = 1
or, x2 – 2x + 3x – 6 5

or, 2x + 1 = 1
x2 + x – 6 5

or, x2 + x – 6 = 10x + 5

or, x2 + x – 6 – 10x – 5 = 0

or, x2 – 9x – 11 = 0

Comparing equation with ax2 + bx + c = 0

Then, a = 1, b = – 9, c = –11

We have,

x = – b ± b2 – 4ac


2a

= 9 ± (–9)2 – 4 . 1 .(–11)
2×1

= 9 ± 81 + 44


2

= 9 ± 2 125



= 9 ± 5 5
2

∴ The roots of equation are 9 + 5 5 and 9 - 5 5 .
22

164 GREEN Mathematics Book-9

EXERCISE: 10.3

A. Very Short Questions
1. a. If a = – 1, b = 2 and c = 4, find the value of – b ± b2 – 4ac .

2a
b. If a = 3, b = 5 and c = –2, find the value of – b ± b2 – 4ac .

2a
B. Short Questions

2. Solve the following quadratic equations by using formula.

a. x² – 9x + 20 = 0 b. x² + 5x + 6 = 0 c. x² + 16x – 36 = 0

d. 2m2 + 3m + 1 = 0 e. 3x2 – 5x – 2 = 0 f. 3x2 + 5x – 12 = 0

C. Long Questions

3. Use quadratic formula to solve following quadratic equations.

a. lx2 + mx + n = 0 b. (a – b) x2 – (a + b) x + 2b = 0

c. x + 5 + x – 5 = 5 d. m – 6 – m – 12 = 5
x–5 x+5 7 m – 12 m – 6 6

e. x+1 + x = 2 1 f. 4m2 + 6m – 4 = 0
x x+1 6 h. x2 = 3 – 2x

g. 1 – 1 = –1 49 7
x + 2 x – 3 10

i. 5x + 1 = 3x + 1 j. b + a = 2
7x + 5 7x + 1 x–a x–b

k. 1 – 1 = 1
x–1 x–2 x+2

GREEN Mathematics Book-9 165

11

Triangle

Estimated Teaching Periods : 20

Pythagoras was a Greek philosopher who made important developments
in mathematics, astronomy, and the theory of music. The theorem now
known as Pythagoras's theorem was known to the Babylonians 1000 years
earlier but he may have been the first to prove it.

Contents

10.1 Triangle
10.2 Theorems
10.3 Mid-point Theorems

Objectives

At the end of this unit, students will able to :
classify the triangles according to measurement of sides and angles
know the properties of triangles
verify the theorems experimentally
prove the theorems theoretically
solve the problems related to triangle by using its properties and facts

Materials

Models of different types of triangles, chart paper, instrument box, scissor, gum, cello tape,
etc.

166 GREEN Mathematics Book-9

11.1 Triangles

Triangle is a closed plane figure formed by three line A
segments. The line segments are called sides of the
triangle, A point where two line segments meet is called Side of triangle Side of triangle
vertex.
B C
Types of triangle on the basis of side.
Side of triangle
There are three types of triangle on the basis of side. They are:
A3 cm C
i. Scalene triangle 4.5cm4cm

The triangle whose angles and sides are unequal with each B
other is a called scalene triangle. 6 cm
In the give figure, AB = 3cm, BC = 6cm and

AC = 4.5cm. AB ≠ BC ≠ AC. So, ∆ABC is a scalene triangle
where, ∠A ≠ ∠B ≠ ∠C.

ii. Isosceles triangle

If two sides and two angles of a triangle are equal then the A

triangle is called an isosceles triangle. In the given figure,

AB = 4cm, BC = 5cm and AC = 4cm. Being AB = AC, ∆ABC is an 4cm 5cm

isosceles triangle where, ∠B = ∠C = base angles. BC
5cm

iii. Equilateral triangle

The triangle with all three sides and three angles equal is called an A
equilateral triangle.

In the given figure AB = 5cm, BC = 5cm and AC = 5cm.

∴ AB = BC = AC = 5cm 5cm

So, the given triangle is an equilateral triangle. Each angle of an

equilateral triangle is 60° i.e. ∠A = ∠B = ∠C = 60°.   B 5cm C

GREEN Mathematics Book-9 167

Types of triangle on the basis of angle A
B
There are three types of triangles on the basis of angle. They are:

i. Right angled triangle

The triangle with one angle 90° is called a right angled
triangle.

In ∆ABC, ∠ABC = 90°
So, ∆ABC is a right angled triangle.

C

If two sides of right angled triangle are equal then the triangle is known as right
angled isosceles triangle.

ii. Acute angled triangle A
70°
If all the angles of a triangle are less than 90° then the
triangle is known as acute angled triangle. 50° 60°
B C
In ∆ABC, ∠ABC = 50° ∠BCA = 60° and ∠BAC =
70°. Each angle is less than 90°. So, ∆ABC is an acute

angled triangle.

iii. Obtuse angled triangle

If one angle of a triangle is greater than 90° then the P

triangle is called an obtuse angled triangle.

In ∆PQR, ∠PQR = 130° which is greater than 90°. So, 20°
∆PQR is called an obtuse angled triangle.
130° 30°
QR

168 GREEN Mathematics Book-9

11.2 Theorems

Theorem is a geometrical statement which is true and can be proved by using known facts
and axioms.
Generally, we prove geometrical theorem by two methods. One method is experimental
verification and another method is theoretical proof. We use following steps to prove a theorem
by experimental verification.
Step 1 : Draw the three appropriate figures of different measurements and label all the

figures with the same letters.
Step 2 : Explain the figures so that it is easy to understand the theorem.
Step 3 : Write down "to verify" part of the theorem.
Step 4 : Draw a table in which we write the necessary measurements.
Step 5 : Write the result of the experiment.
Step 6 : Write the conclusion from the experiment.
Let us learn some of theorems about the triangle.

Theorem - 1

Statement:

Among the straight lines drawn from an external point to a given straight line, the
perpendicular is the shortest one.

Experimental verification

X

M P Y
AP P

N

BB

NA

X YM
Y MA B N X

(i) (ii) (iii)

PA, PB, PM an PN are the straight lines drawn from a point P to the line XY and PA⊥XY.

To verify: PA is shortest (PA ⊥XY) distance.

GREEN Mathematics Book-9 169

Measurement table

Figure PA PB PM PN Remarks

i. PA < PB

ii. PA < PM
iii. PA < PN

Result : In all above figures, we found PA < PB, PA < PM and PA < PN. So, PA is the
shortest distance from XY.

Conclusion : The perpendicular drawn from a point to the straight lines is the shortest
distance from the point to the line.

Theorem - 2

Statement:
The sum of two sides of a triangle is greater than the third side.
Experimental verification

A AA

B CB C B C
(i)
(ii) (iii)

AB, BC and CA are the sides of ∆ABC.

To verify: AB + BC > AC, AB + AC > BC and BC + AC > AB

Measurement table

Figure AB BC AC AB + BC AB + AC BC + AC Remarks
i. AB + BC > AC
ii. AB + AC > BC
iii. BC + AC > AB

Result : From the above figures we found AB + BC > AC, AB + AC > BC and BC + AC
> AB.

Conclusion : The sum of two sides of a triangle is greater than the third side.

170 GREEN Mathematics Book-9

Theorem - 3

Statement:

In a triangle, the side opposite to the greater angle is longer than the side opposite to the
smaller angle.

Experimental verification

Q

R
P

Q RQ P P
R
(i) (ii)
(iii)

In ∆PQR, ∠P is greater than other angles ∠Q and ∠R.

To verify: QR > PR and QR > PQ

Measurement table

Figure QR PR PQ Remarks

i.

ii. QR > PR and QR > PQ

iii.

Result : From the above figures we found QR > PR and QR > PQ.

Conclusion : In a triangle, the side opposite to the greater angle is longer than the
side opposite to the smaller angle.

Worked Out EXAMPLES

EXAMPLE 1 Find the longest and the shortest sides of P
∆PQR in which ∠P = 85° and ∠Q = 30°. 85°

Solution : By the question, R 65° 30° Q
In DPQR

∠P = 85° and ∠Q = 30°

We can write,

∠P + ∠Q + ∠R = 180° [∵ Sum of the angle of a triangle.]

or, 85° + 30° + ∠R = 180°

or, 115° + ∠R = 180° Now,

or, ∠R = 180° – 115° \ Longest side = RQ

\ ∠R = 65° \ Shortest side = PR

GREEN Mathematics Book-9 171

EXAMPLE 2 In DLMN, LM = 6cm, MN = 10cm and LN L
= 8cm, find the greatest and the smallest 8 cm
angles. 6 cm

Solution : By the question, M 10 cm N

In DLMN

LM = 6cm, MN = 10cm and LN = 8cm

Now, MN > LN > LM

Since, the side MN is the longest, the ∠L opposite to it is the greatest
angle.

Similarly, ML is the shortest so the ∠N opposite to it is the smallest
angle.

EXERCISE 11.1

A. Very Short Questions
1. Find the longest and the shortest sides of the given triangles.

a. A b. Q c. A d. A

P 80° 40° R B 60° C 25° C
B 75°

B 60° 72° C

2. Determine the largest and the smallest angles in the following triangles

a. In ∆ABC, AB = 3cm, BC = 7 cm and AC = 5.5cm.
b. In DDEF, DE = 7.2cm. DF = 5.9 cm and EF = 8cm.
c. In DPQR, PQ = 6cm, QR = 8cm and PR = 10cm.

B. Short Questions B

3. Prove that the hypotenuse AC of a right angled triangle C
ABC is the longest side.
A

4. Find the longest side of ∆ABC in which ∠A = 60° and A
∠C = 75°.

5. In the figure along side, ABC is a triangle and AD⊥BC,
Show that AB > AD.

BC
D

172 GREEN Mathematics Book-9

6. In the figure, along side, ABC is a triangle in which AC A
> AB, KC and KB are the bisectors of ∠ACB and ∠ABC
respectively. Prove that KC > BK. K

B C
C

7. In the figure along side, ABC is a triangle in which AC > AB,
PC is bisector of angle ACB respectively. Prove that AC > AP.

AB
P

Theorem - 4

Statement:

The sum of three angles of a triangle is equal to two right angles.

Experimental verification BA

A

C

B A C (iii) B
(i) C
Remarks
(ii) Sum of ∠A, ∠B and ∠C =

ABC is the triangle and ∠A, ∠B and ∠C are its angles. 180°

To verify: ∠A + ∠B + ∠C = 180°

Measurement table

Figure ∠A ∠B ∠C ∠A + ∠B + ∠C

i.

ii.

iii.

Result : From the above figure we found the sum of ∠A +∠B + ∠C = 180°.
Conclusion : The sum of the angles of a triangle is two right angles.

GREEN Mathematics Book-9 173

Theoretical Proof A
Q
Given : ABC is a triangle where ∠BAC, ∠ACB and ∠CBA are its P
interior angles. C

To prove: ∠ABC + ∠BAC + ∠ACB = 180° B
Construction: Through A draw PQ parallel to BC.

Statements Reasons
1. PQ||BC 1. By construction
2. ∠PAB = ∠ABC 2. Pair of alternate angles formed

3. ∠QAC = ∠ACB beween parallel lines are equal.
4. ∠PAB + ∠BAC + ∠QAC = ∠PAQ 3. Same as reason no. 2.
5. ∠PAB + ∠BAC + ∠QAC = 180° 4. By whole part axiom.
5. ∠PAB = ∠ABC, ∠QAC = ∠ACB and
6. ∠ABC + ∠BAC + ∠ACB = 180°
∠PAQ is being straight angle.

6. From above statement.

∴ Therefore, the sum of the angles of a triangle is two right angles.

Theorem - 5

Statement:

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

Experimental verification S P

PR Q

P

QS R
RQ
S

(i) (ii) (iii)

PQR is a triangle. QR is produced to S. So, ∠PRS is an exterior angle of the triangle PQR.

To verify: ∠PRS = ∠RPQ +∠PQR

Measurement table

Figure ∠PRS ∠RPQ ∠PQR ∠RPQ + ∠PQR Remarks
i.
ii. ∠PRS = ∠RPQ + ∠PQR
iii.

Result : From the above figure we found that ∠PRS = ∠RPQ + ∠PQR

Conclusion : The exterior angle of a triangle is equal to the sum of the two interior opposite
non-adjacent angles.

174 GREEN Mathematics Book-9

Theoretical Proof AE

Given : In ∆ABC the side BC is produced to D. So, ∠ACD is B D
an exterior angle of DABC. C

To prove: ∠ACD = ∠CAB + ∠ABC

Construction: CE is drawn parallel to BA.

Statement Reasons

1. BA||CE 1. By construction

2. ∠CAB = ∠ACE 2. Being pair of alternate angles formed between
parallel lines BA and CE.

3. ∠ABC = ∠ECD 3. Being pair of corresponding angles formed on
parallel lines BA and CE.

4. ∠ACD = ∠ACE + ∠ECD 4. By whole part axiom.

5. ∠ACD = ∠CAB + ∠ABC 5. ∠CAB and ∠ACE, ∠ABC and ∠ACD being equal.

∴ In a triangle the exterior angle is equal to the sum of the two interior opposite non-
adjancent angles.

Theorem - 6

Statement:
Base angles of an isosceles triangle are equal. OR

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

Experimental verification P R

P

P

Q R Q R
(i) (ii)
(iii) Q
PQR is the isosceles triangle in which PQ = PR.

To verify: ∠PQR = ∠QRP

Measurement table

Figure ∠PQR ∠QRP Remarks
i. ∠PQR = ∠QRP
ii.
iii.

Result : From the above figures we found that ∠PQR = ∠QRP.

Conclusion : If two sides of a triangle are equal then the angles opposite to them are also
equal.

GREEN Mathematics Book-9 175

Theoretical Proof X

Given : ∆XYZ is an isosceles in which XY = XZ. YZ
To prove: ∠XYZ = ∠YZX A
Construction: Draw XA⊥YZ.

Statements Reasons
1. In DXYA and DXZA
1.

i. ∠XAY = ∠XAZ (R) i. Both are right angles.

ii. XY = XZ (H) ii. Given.

iii. XA = XA (S) iii. Common sides of the both triangles.
2. DXYA ≅ DXZA 2. By R.H.S. axiom.
3. ∠XYA = ∠XZA 3. Corresponding angles of the congruent triangles.
4. ∠XYZ = ∠YZX 4. From statement 3.

Proved.

Therefore the angles opposite to the equal sides of a triangle are equal.

Theorem - 7

Statement:
In a triangle, the sides opposite to the equal angles are also equal.

Experimental verification P R

P

P

Q R Q R
(i) (ii)
(iii)
Q

PQR is the triangle in which ∠Q = ∠R.
To verify: PQ = PR
Measurement table

Figure PQ PR Remarks
i. PQ = PR
ii.
iii.

Result : From the above figures we found that PQ = PR.

Conclusion : In a triangle, the sides opposite to the equal angles are equal.

176 GREEN Mathematics Book-9

Theoretical Proof X

Given : XYZ is a triangle in which ∠XYZ = ∠YZX. YZ
To prove : XY = XZ A
Construction : Draw XA⊥YZ.

Statements Reasons
1. In DXYA and DXZA
1.

i. ∠XYA = ∠AZX (A) i. Given.

ii. ∠XAY = ∠XAZ (A) ii. Both are right angles.

iii. XA = XA (S) iii. Common to the both triangles.
2. DXYA ≅ DXZA 2. By A.A.S. theorem.
3. XY = XZ 3. Corresponding sides of the congruent triangles.

Therefore, the side opposite to the equal angles of a triangle are equal.

Theorem - 8

Statement:

The bisector of the vertical angle of an isosceles triangle is perpendicular to the base and
bisects the base.

Experimental verification R

PP

PX

Q RQ R Q
XX

(i) (ii) (iii)

PQR is the triangle and PX is the bisector of ∠QPR. i.e. ∠QPX = ∠XPR.

To verify: (i) QX = XR

(ii) ∠QXP = ∠PXR = 90°

Measurement table:

Figure QX XR ∠QXP ∠PXR Remarks
i.
ii. QX = XR and
iii. ∠QXP = ∠PXR = 90°

Result : From the above figures we found that QX = XR and ∠QXP = ∠PXR = 90°.

Conclusion : The bisector of the vertical angle of an isosceles triangle is perpendicular to
the base and bisects the base.

GREEN Mathematics Book-9 177

Theorem - 9

Statement:

The line joining the mid-points of the base of an isosceles triangle to the opposite vertex is
perpendicular to the base and bisects the vertical angle.

Experimental verification R

P P

PA

Q RQ R
A A
Q
(i) (ii) (iii)

PQR is a triangle in which PQ = PR and A is the mid - point of QR.

To verify: (i) ∠QAP = ∠PAR = 90° (ii) ∠APQ = ∠RPA.
Measurement table

Figure ∠QAP ∠PAR ∠APQ ∠RPA Remarks
i.
ii. ∠APQ = ∠RPA and
iii. ∠QAP = ∠PAR = 90°

Result : From the above figures we found that ∠APQ = ∠RPA and ∠QAP = ∠PAR = 90°.

Conclusion : The line joining the mid-point of the base of a isosceles triangle to the

opposite vertex is perpendicular to the base and bisects the vertical angle.

Worked Out EXAMPLES A

EXAMPLE 1 From the given figure, find the values of x 85°
and y. Ex
C

46° M 30°

Solution : In DAED y D
∠A + ∠D + ∠E = 180° B

or, 85° + 30° + x = 180° ∵ Sum of angles of triangle.

or, 115° + x = 180°

or, x = 180° – 115°

∴ x = 65°

∠EBM + ∠EMB = ∠AED ∵ Exterior angle of a triangle is
Again, y + 46° = x equal to the sum of opposite interi-
or, y = 65° – 46° or angles.

∴ y = 19°

Therefore, in the given figure the values of x and y are 65° and 19°

respectively.

178 GREEN Mathematics Book-9

EXAMPLE 2 In DXYZ, XZ is produced to A. If Y
XY = YZ and ∠XYZ = 66°, find the measure
of ∠YZA.

XA
Z

Solution : i. ∠ZXY = ∠YZX ∵ Opposite angles to equal sides being equal.

ii. ∠XYZ + ∠YZX + ∠ZXY = 180° ∵ Sum of angles of triangle.

or, 66° + ∠YZX + ∠YZX = 180° ∴∠YXZ = ∠XZY

or, 2 ∠YZX = 180° – 66°

or, ∠YZX = 114°
2

∴ ∠YZX = 57°

iii. Again, Sum of two interior angles is equal to one
∠ YZA = ∠ZXY + ∠XYZ exterior angle.

= 57° + 66°

∴ ∠YZA = 123°.

Therefore, in the given figure measure of ∠YZA is 123°.

EXAMPLE 3 In the adjoining figure, QS = SP = PR, T
Solution : ∠SPQ = 37°, find the measure of ∠TPR. P
37°

QR

S

i. ∠SQP = ∠QPS ∵ Base angles of an isoceles triangle.

\ ∠SQP = 37°

ii. ∠PSR = ∠SPQ + ∠PQS ∵ An exterior angle of a triangle
is equal to sum of the opposite

= 37° + 37° interior angles.

∴ ∠PSR = 74°

iii. ∠SRP = ∠PSR ∵ Base angles of an isosceles

∠SRP = 74° triangle.

iv. Now,

∠TPR = ∠PQR + ∠QRP

∵ An exterior angle of a

= 37° + 74° triangle is equal to sum of the

opposite interior angles.

∴ ∠TPR = 111°

Therefore, in the given figure measure of ∠TPR is 111°.



GREEN Mathematics Book-9 179

EXAMPLE 4 In the given figure, ∠XAB = ∠YAC and AX = A
AY. Prove that ABC is an isosceles triangle.

Solution : Given : ∠XAB = ∠YAC and AX = AY

To prove : ABC is an isosceles triangle. (AB = AC) XY
BC

Proof :

S.N. Statements S.N. Reasons

1. ∠AXY = ∠AYX 1. In DAXY, the sides opposite to
ie. ∠AXB = ∠AYC them being equal.

2. In DAXB and DAYC 2.

i. ∠XAB = ∠YAC i. Given.

ii. AX = AY ii. Given.

iii. ∠AXB = ∠AYC iii. From statement 1.

3. DAXB ≅ DAYC 3. By A.S.A. axiom.

4. AB = AC 4. Corresponding sides of
congruent triangles.

5. DABC is an isosceles 5. AB and AC being equal.

Proved

EXAMPLE 5 In the adjoining figure, PQ||RS. If BM and CM are bisector of ∠PBC
Solution :
and ∠BCR respectively, prove that ∠BMC = a right angle.
A

Given : In the given figure, PQ||RS, P BQ
∠PBM = ∠MBC and ∠MCB = ∠MCR. M

RC S
D
To prove : ∠BMC a right angle.
Proof :

Statements Reasons
1. ∠PBM = ∠MBC 1. Given

∠MCB = ∠MCR

2. ∠PBC + ∠BCR = 180° 2. The sum of co-interior
angles between two
parallel lines

3. ∠PBM + ∠MBC + ∠BCM + 3. By whole part axioms
∠MCR = 180°

180 GREEN Mathematics Book-9

4. ∠MBC + ∠MBC + ∠BCM + 4. From statement 1.
∠BCM = 180°

or, 2∠MBC + 2∠BCM = 180°

i.e. ∠MBC + ∠BCM = 90°

5. ∠MBC + ∠BCM + ∠BMC = 5. From statement (1) and
180° (3).

6. 90° + ∠BMC = 180° 6. From statement 5.

7. ∠BMC = 180° – 90° = 90° 7. From statement 6.

Proved

P

EXAMPLE 6 In the adjoining figure ∠QPM = ∠MPR and
Solution : QM = MR. Prove that PQ = PR.

Q R
M
Given : ∠QPM = ∠MPR, and QM = MR. P

To prove : PQ = PR

Construction : PM is produced to A such that Q R

PM = MA and join AR. M

Proof : A

Statements Reasons
1. In DPQM and DMAR
1.

i. PM = MA i. By construction

ii. ∠PMQ = ∠AMR ii. Vertically opposite angles

iii. QM = MR iii. Given

2. \ DPQM ≅ DMAR 2. By S.A.S axiom

3. PQ = AR and ∠QPM = ∠MAR 3. Corresponding parts of
congruent triangles.

4. ∠QPM = ∠MPR 4. Given

5. ∠MAR = ∠MPR 5. From statements (3) and (4).

6. PR = AR 6. From statement 5.

7. PQ = PR 7. From statement (3) and (6).

Proved

GREEN Mathematics Book-9 181

EXAMPLE 7 In the figure given alongside ∆ABD and D A E
Solution : ∆ACE are two equilateral triangles. Prove C
that BE = CD.

Given : ∆ABD and ∆ACE are equilateral B

triangles.

To prove : CD = BE

Proof :

Statements Reasons
1. In DABD, ∠ADB = ∠BAD = 60°
1. Each angle of on equilateral
triangle is 60°.

2. In DACE, ∠EAC = ∠ACE = 2. Same as statement 1.
∠AEC = 60°

3. ∠BAD+∠BAC = ∠EAC + ∠BAC 3. Adding ∠BAC in both of equal
angles in st. (1) and (2).

4. ∠DAC = ∠BAE 4. From statement 3, by addition
axiom.

5. In DDAC and DBAC 5.

i. AD = AB (S) i. Sides of equilateral triangle.

ii. ∠DAC = ∠BAE (A) ii. From statement 4.

iii. AC = AE (S) iii. Sides of eqilateral triangle.
6. DDAC ≅ DBAE
7. CD = BE 6. From S.A.S. axiom

7. Corresponding sides of
congruent triangles.

Proved

EXERCISE 11.2

A. Very Short Questions
1. Find the unknown sizes of angles in the following figures.

a. A b. c. P

P

110°

B 60° CQ 45° R Q R

A A

d. 68° D e. y

44° C B 70° xC

B 22°

182 GREEN Mathematics Book-9

2. Find the value of x in each of the following triangles.

a. A b. c. P
Qx
P

7x

B 3x 2x C Q 2x 3x R R

A D e. A f. P
d. 68°
x C 2x
2x 40° 48° S
Bx C 80° y 57° R
BD Qx
a
B. Short Questions P S

3. Calculate the value of a in the following figures. 30° R

A

A 30°

a. 57° b. D c.

a a CQ

B 43° DB a

C

P e. A 30° B M
N 40° K
d. 80° a f.
50° L
O D
a RC

Q

g. A a B h. BA
6a
D
O 7a E
CD 5a F
A
60° x
C

4. In the given figure, ∠ABP = 110° and ∠ACQ = 125°. Find 110° 125°
C
the value of x. P B Q

P Q
80°
S
5. In the given figure, if PQ||ST, ∠PQR = 80° and 130° T

∠RST = 130°, find the value of a. a

R
GREEN Mathematics Book-9 183

AC

6. In the given figure, AB||CD. ∠APQ = 70° and RP 70° Q R
∠QDR = 30°, find the value of x. B B
x
C. Long Questions A
7. In the adjoining figure, RT and ST are angular bisectors D 30°

of ∠BRS and ∠DSR respectively. Prove that ∠RTS is a P
right angle.
R
8. In the figure alongside, if PQ = PR, AQ = BR and
∠APB = 25°, calculate ∠PAB and prove that PAB is an S T
isosceles triangle. C D

Q
P

QR
AB

A

9. In DABC, AB = AC, BD and CD are the bisectors of D C
∠ABC and ∠ACB respectively. Prove that BD = CD.
B
10. In the figure alongside, AB = AC and AD bisects ∠EAC. E
Prove that AD||BC.
AD

B C
D P
C
11. In the adjoining figure ABCD is a parallelogram and A
PA and PB are angular bisectors of ∠BAD and ∠ABC B
respectively. Prove that ∠APB = 90°.

A

12. In the given figure, BD and CE are the bisectors of ∠B ED
and ∠C of an isosceles triangle ABC with AB = AC. P
Prove that PE = PD.
B
C

184 GREEN Mathematics Book-9

P

75°

13. In the given figure, PQR is an equilateral triangle. If

∠RPT = 75° and ∠QSR = 45°. Prove that QS = PT. T Q R
45°
M S
A
N
14. In the adjoining figure, AMB and ANC are equilateral
triangles, prove that MC = BN.

B C
B

15. In the adjoining figure, AB = DC and AC = DB, prove A O B
that AO = OD.
C
D P

16. In the adjoining figure, ∠BPR = ∠APQ, ∠PAR = ∠PBQ A
and AP = BP, prove that DPQR is an isosceles triangle.

QR

A

17. In the adjoining figure RP⊥AB, RQ⊥AC, PR = RQ and

AB = AC, prove that AP = AQ. P Q

B C
R
P
A
18. In the given figure, PQ = QB ∠PQA = ∠RQB, R

∠QAB = ∠PRQ, prove that AB = PR and QA = QR. P

Q

19. Prove that the bisector of vertical angle of

triangle bisect its base, if it is an isoscles B

triange. A

20. In the adjoining figure, AB = BC, PB = BQ and B
∠ABC = ∠PBQ = 90°. Prove that CP = AQ.

C

Q

A

21. In the given figure, BX = XC, XY⊥AB and XZ⊥AC, prove that

AY = AZ. YZ

P BC
X
22. In the given figure, ∠P = ∠S and PQ = QS.

Prove that PT = RS. T

Q RS

GREEN Mathematics Book-9 185

12

Parallelogram

Estimated Teaching Periods : 14

Pythagoras was a Greek philosopher who made important developments
in mathematics, astronomy, and the theory of music. The theorem now
known as Pythagoras's theorem was known to the Babylonians 1000 years
earlier but he may have been the first to prove it.

Contents

12.1 Quadrilateral
12.2 Trapezium
12.3 Mid-point theorem
12.4 Pythagoras theorem

Objectives

At the end of this unit, students will be able to:
state different types of quadrilateral.
know the properties of parallelogram, rectangle, rhombus, square, kite and trapezium.
verify theorem experiementaly.
prove the theorem theoretically.
solve the problems related to parallelogram.
find the area of different types of quadrilateral theoretically and by project work.
mid-point theorem and its uses.

Materials

Geometry box, geo board, chart paper, etc.

186 GREEN Mathematics Book-9

12.1 Quadrilateral

Quadrilateral

Quadrilateral is a plane figure bounded by four line segments. A D
The given figure ABCD is a quadrilateral which is written as B C
quad. ABCD or ABCD.
In the quadrilateral ABCD, its different parts are as follows:
i. Vertices : the points A, B, C and D .
ii. Sides : AB, BC, CD and DA .
iii. Diagonals : AC and BD (line segment joining the opposite vertices)

Area of quadrilateral

In the quadrilateral ABCD, AC is a diagonal, BM and DN are
drawn perpendiculars to AC . Let AC = d, BM = P1 and DN = P2.

Area of quad ABCD = Area of D ABC + Area of DADC A
M P2 D
= 21. AC. BM + 12AC. DN
= 21.d . P1 + 12. d. P2 P1 N

= 1 . d (P1 + P2) B
2 C
∴ Area of quadrilateral = 12. diagonal × (Sum of
perpendiculars drawn to the diagonal from the opposite vertices)

Types of quadrilateral

1. Parallelogram

Parallelogram is a Properties of parallelogram
1q. uaPdarrialallteelroaglrwamhose Opposite sides of a parallelogram are equal.
opposite sides are parallel
and equal. (i.e. AB = CD, AD = BC)
Opposite angles of a parallelogram are equal.
A D
B O (i.e. ∠A = ∠C and ∠B = ∠D)
Diagonals of a parallelogram bisect each other.
C
(i.e. AO = OC, BO = OD)
Opposite sides of a prallelogram are parallel
(i.e. AD||BC and AB||CD.

GREEN Mathematics Book-9 187

2. Rectangle Properties of rectangle

Rectangle is a Oppo. asintgelaensgalneds asniddesaidreeeaqrueaelq. ual.
parallelogram whose
each of the angle is Diagonals bisect each other.
90°.
Each angle is 90° (i.e. ∠A = ∠B = ∠C = ∠D = 90°.)
ITnhtehe fgigivuerne ABfCigDuries
Aa BreCcDtanisglae rwechtiacnhglies Diagonals are equal. [i.e. AC = BD]
AwCrittenanads rBecDtangalree.
dAiBaCgoDn.alAs.C and BD A AD D
are diagonals.
O
B BC C

3. Square PPrrooppeerrttiieess ooff rseqcutaarnegle
EEaacchh aannggllee iiss 9900°° ((ii..ee.. ∠∠PP == ∠∠QQ == ∠∠RR == ∠∠SS == 9900°°))
A rectangle having aanndd aallll tthhee ssiiddeess aarree eeqquuaall..
adjacent sides equal DDiiaaggoonnaallss aarree eeqquuaall.. ((ii..ee.. PPRR == QQSS))
is called a square. DDiiaaggoonnaallss bbiisseecctt eeaacchh ootthheerr aatt 9900°°

PA SD ((ii..ee.. PPOO==OORR, ,QQOO, Q=OO, OS Sanandd∠∠PPOOSS==∠∠QQOORR ==
∠∠PPQORQ == ∠∠SSOORR== 9900°° ..
O RC
QB AA ddiiaaggoonnaall bthiseecvtserttihcael vaenrgtilcea. l[i.aen. g∠lePsQ. O[i.e=.
∠∠PAQQOOS= =∠4O5Q° Ran=d4s5o° oann]d so on]

43. RSqhuoamrebus

iAasdAarchjlalaropcselmlaeicedrntdbaaetulnsalsesges.illq doqe ugeuasrahalramieesvq. ciuhnaaalgllv e idnga Properties of rhecotmanbgules

AA D D EAallchsidanesglaereise9q0u°a(li..e(i..e∠.PA=B∠= QBC= =∠CRD= =∠DS A= )9 0 °)
aDniadgaolnl athlsebsiisdeecst aeraecheqoutahle.r at right angles. (i.e.
O DAiOag=oOnaCls, BaOre =eqOuDala. n(id.e.(APRC⊥=BQDS))
BB CC
DAidagiaognoanlsalbbisiseeccttesatchheoatnhgelrea. t 90°

([i∠.eB. APOC = O∠RD,AQCO, ,∠QAOC,BO=S∠anAdC∠DPaOnSd=so∠oQnO] R =
∠EvPeQryRs=q∠uaSrOeRis=a9r0h°o.mbus.
A diagonal the vertical angle. [i.e. ∠PQO =
∠AQOS = 45° and so on]

188 GREEN Mathematics Book-9

5. Kite

A quadrilateral whose two Properties of kite
pairs of adjacent sides are
equal is known as a kite Longer diagonal bisects the shorter diagonal at
90° . [i.e. OB = OD and AC⊥ BD]
D
Longer diagonal bisects the kite. [i.e. ΔABC =
AO C ΔADC].

B Shorter diagonal divides the kite into two
isosceles Δs. i.e. ΔABD and ΔBCD

12.2 Trapezium

Trapezium is a quadrilateral whose any two opposite sides are A D
parallel. In the figure two parallel sides BC and AD are bases or N
C
legs. When non-parallel sides are equal, the trapezium is called M

isosceles trapezium, The line segment joining the mid-point
of non-parallel sides is called median. In the figure MN is the B

median.

Theorem - 10 Q
S
The straight line segments that join the ends of two equal and P
parallel line segments towards the same sides are also equal and
parallel.
Given : PQ = RS and PQ||RS. The ends of same sides are P, R and R
Q, S are joined.

To prove: PR = QS and PR||QS .

Construction: P and S are joined.

Proof:

S.N. Statements S.N. Reasons

1. In D PRS and DPQS, 1.

i. RS = PQ (S) i. Given

ii. ∠PSR = ∠SPQ (A) ii. Being alternate angles equal

iii. PS = PS (S) iii. Common side of both Ds

2. ΔPRS ≅ ΔPQS 2. By S.A.S. axiom

3. PR = QS 3. Corresponding sides of congruent triangles

4. ∠SPR = ∠PSQ 4. Corresponding angles of congruent triangles

5. PR||QS 5. From statement (4); alternate angles being equal

Proved

GREEN Mathematics Book-9 189

Theorem - 11

The straight line segments joining the ends of two equal and parallel A O B
line segments towards the opposite ends bisect each other. D

Given : AB = CD and AB||CD. AD and BC are joined which intersect

each other at O. C

To prove : AO = OD and BO = OC

Proof:

S.N. Statements S.N. Reasons
1. In D AOB and DCOD, 1.
i. ∠BOA = ∠COD (A)
ii. ∠ABO = ∠DCO (A) i. Being the vertically pposite angles
iii. AB = CD (S)
2. ΔAOB ≅ ΔDOC ii. Being Alternate angles equal
3. AO = OD, and BO = OC
iii. Given
Theorem - 12 2. By A.A.S. theorem
3. Corresponding sides of congruent triangles

Proved.

Opposite sides and angles of a parallelogram are equal. A D
Given : ABCD is a parallelogram. B C
To prove : (i) AB = DC, BC = AD
(ii) ∠A = ∠C, ∠B = ∠D
Construction : Joining AC.
Proof:

S.N. Statements S.N. Reasons
1. In DABC and DADC, 1.

i. ∠BAC = ∠DCA (A) i. Alternate angles are equal
ii. AC = AC (S) ii. Common side
iii. ∠BCA = ∠CAD (A) iii. Being alternate angles equal
2. ΔABC ≅ ΔADC 2. By A.S.A. axiom
3. AB = DC and BC = AD 3. Corresponding sides of congruent triangles
4. ∠ABC = ∠ADC 4. Corresponding angles of congruent triangles
5. ∠BAC + ∠CAD 5. Adding statement 1(i) and 1 (iii)
= ∠ACB + ∠DCA
6. ∠BAD = ∠BCD 6. Whole- parts axiom.

Proved.

190 GREEN Mathematics Book-9

Converse I of theorem 12

If a quadrilateral has its opposite sides equal, prove that it is a parallelogram.

A D
C
Given : In the given quadrilateral ABCD, AB = DC and BC = AD.

To prove : ABCD is a parallelogram B
Construction: A and C are joined.

Proof:

S.N. Statement S.N. Reasons
1. In DABC and DADC, 1.

i. AB = DC (S) i. Given

ii. BC = AD (S) ii. Given

iii. AC = AC (S) iii. Common side of both ∆.

2. ΔABC ≅ ΔADC 2. By S.S.S. theorem

3. ∠BAC = ∠DCA, ∠ACB = 3. Corresponding angles of congruent triangles
∠DAC

4. AB||DC, BC ||AD 4. From statement 3, alternate angles are equal.

5. ABCD is a parallelogram 5. From statement 4.

Proved

Converse II of theorem 12

If a quadrilateral has opposite angles equal, prove that it is a parallelogram.

Given : In the quadrilateral ABCD, ∠A = ∠C and ∠B = ∠D. A D
C
To prove : ABCD is parallelogram

Proof: B

S.N. Statement S.N. Reasons
1. ∠A + ∠B + ∠C + ∠D = 360° 1. Sum of angles of a quadrilateral.
2. ∠A + ∠B + ∠A + ∠B = 360° 2. Being ∠A = ∠C and ∠B = ∠D
3. 2(∠A + ∠B) = 360° 3. From statement 2.
i.e. ∠A + ∠B = 180°
4. BC||AD 4. From statement 3.
5 ∠B + ∠C = 180° 5. Same as above.
6. AB||DC 6. From statement 5
7. ABCD is a parallelogram 7. Form statements 4 and 6

Proved.

GREEN Mathematics Book-9 191

Theorem - 13 AD

The diagonals of a parallelogram bisect each other. O
C
B

Given : ABCD is a parallelogram in which diagonals AC and
BD intersect each other at O.

To prove : AO = OC, BO = OD

Proof

S.N. Statements S.N. Reasons
1. In DAOD and DBOC, 1.
i. Alternate angles are equal
i. ∠DAO = ∠BCO (A) ii. Opposite sides of a parallelogram
ii. AD = BC (S) iii. Alternate angles are equal
2. By A.S.A. axiom
iii. ∠ADO = ∠CBO (A) 3. Corresponding sides of congruent triangles
2. ΔAOD ≅ ΔBOC
3. AO = OC, BO = OD Proved.

Converse of theorem 13

If the diagonals of a quadrilateral bisect each other, the A O D
quadrilateral is a parallelogram. C

Given : ABCD is a quadrilateral in which diagonals AC and BD B
bisect each other at O, ie. AO = OC, BO = OD

To prove : ABCD is a parallelogram

Proof

S.N. Statement S.N. Reasons
1 In DAOD and DBOC, 1

i. AO = CO (S) i. Given
ii. ∠AOD = ∠BOC (A)
ii. Vertically opposite angles
iii. OD = OB (S)
2. ΔAOD ≅ ΔBOC iii. Given
3. AD = BC, ∠DAO = ∠BCO 2. By S.A.S axiom
4. AD = BC, AD||BC 3. Corresponding parts of congruent triangles
5. AB = DC, AB||DC 4. From statement 3
6. ABCD is a parallelogram 5. From statement 4
6. From statements 4 and 5

Proved.

192 GREEN Mathematics Book-9

Worked Out EXAMPLES A B
C
EXAMPLE 1 In the adjoining figure ABCD is a square BD F
and AE are intersecting at F. If ∠AED = 55°, find D 55°
the value of ∠BFE.
E
Solution : Here,

∠BDC = 90° = 45°
2

and ∠AED = 55° In DEF, an exterior angle of
Now, ∠BFE = ∠FDC + ∠DEF a triangle is equal to its two
= 45° + 55° non-adjacent interior angles.
= 100°

EXAMPLE 2 In the figure, PQRS is a parallelogram. The diag- X P S
onal PR is produced to the points X and Y such

that PX = RY. Prove that QX||YS.

O

Solution : Given : PQRS is a parallelogram and diagonal PR Q R

is produced to the point X and Y such that PX = RY. Y

To prove QX||YS :

Statements Reasons
1. OP = OR 1. Diagonals of a parallelogram

2. PX = RY bisect each other.
3. OP + PX = OR + RY 2. Given
4. OX = OY 3. Adding statements 1 and 2
5. OQ = OS 4. From statement 3.
4. Diagonals of a parallelogram
6. XQYS is a parallelogram
7. QX||YS bisect each other
6. From statement 4 and 5
7. Opposite sides of a parallelogram.

Proved.

EXAMPLE 3 In the adjoining figure, AB = AC, PB = CM A C
and PQ||AM. Prove that PM and QC bisect M
each other. P
BQ
Solution : Given : AB = AC, PB = CM, PQ||AM
To prove : PM and QC bisect each other

GREEN Mathematics Book-9 193

Proof :

Statements Reasons

1. ∠ABC = ∠ACB 1. AB = AC, base angles of an
isosceles triangle ABC.

2. ∠PQB = ∠ACB 2. Corresponding angles; PQ||AM

3. ∠ABC = ∠PQB 3. From statements 3

i.e. ∠PBQ = ∠PQB 3. From statements 1 and 2

4. PB = PQ 4. Opposite sides of equal angles of
∆PBQ.

5. PB = CM 5. Given

6. PQ = CM, PQ||CM 6. From statements 4 and 5 and given

7. PC = QM, PC||QM 7. PQ = CM, PQ||CM

8. PQMC is a parallelogram 8. From statements 6 and 7

9. PM and QC bisect each 9. Diagonals of a parallelogram

other. bisect each other.

Proved.

EXAMPLE 4 In the figure P, Q, R and S are the mid- A PB
points of AB, BC, CD and AD respectively. QS

Prove that PQRS is a parallelogram.

Solution : Given : P, Q, R and S are the mid-points of C D
AB, BC, CD and AD respectively. R

To prove : PQRS is a parallelogram.

Construction : Join AC.

To proof:

Statements Reasons

1. In ΔABC, PQ = 1 of AC, 1. Straight lines joining the mid-points
2 of two sides of a triangle is half of

PQ||AC the third side and is parallel to it.

2. In ΔACD, RS = 1 of AC, 2. Same as reason 1.
2

RS||AC

3. PQ = RS, PQ||RS 3. From statements 1 and 2

4. QR = PS, QR||PS 4. From statements 3

5. PQRS is parallelogram 5. From statements 3 and 4

Proved.

194 GREEN Mathematics Book-9