Vedanta Excel in Mathematics Book 7 Final (2078)

Laws of Indices

5. Let's find the products.

a) x (2x + 4) b) 2x (3x – 5) c) 3p (2p2 – 1)

d) – 3p (p2 – 2p + 3) e) – 3t2 (2t2 + t – 2) f) xyz (x2 – y2 – z2)

6. Let's simplify.

a) x (x + 2) + 3 (x + 2) b) 2x (x – 3) + 5 (x – 3)
c) 3x (x + y) + y (x + y) d) 2a (2a + b) – b (2a + b)
e) x (x2 – xy + y2) + y (x2 – xy + y2) f) a (a2 + ab + b2) – b (a2 + ab + b2)

7. Let's multiply.

a) (a + b) (a + ) b) (a – b) (a – ) c) (a + b) (a – )
d) (x + 2) (x +3) e) (y + 4) (y –5) f) (2a 5) (3a 4)
g) (4p – 3) (p 2) h) (3c 2d) (2c – 3d ) i) (x + y) (x – y)
j) (2x + 3y) (2x 3y) k) (7m 3n) (7m 3n) l) ( x + 2) (x2 2x + 3)
m) (x 3) (2x2 – 4x + 5) n) (a b) (a2 ab + b2) o) (x –y) (x2 xy + y2)

Creative Section - B c) x
y
8. Let's find the areas of the following rectangles. p
a) b)
nx

m ab

d) e) a f) 3x
2a b 8x y
2y
3a b 2a b

9. a) The length of a rectangular garden is (2x + 3y)m and its breadth is
(4x – y)m . Find the area of the garden.

b) A rectangular piece of carpet has length (3x + y)m and breadth (2x – y)m.
Find its area.

c) The floor of a bedroom is (5a –3)m long and (3b +2)m wide.

(i) Find the area of the floor.

(ii) Find its actual area if a = 2 and b = 1

d) The length of a play ground is (7p + 2q)m and width is (5p – 3q)m.
(i) Find the area of playground

(ii) If p = 5 and q = 2, find its actual area.

10. a) If x = (a + 7) and y = (a – 7), show that a2 = xy + 49

b) If p = (3x + 1) and q = (3x – 1), show that: pq 1 = x2
c) If a = (2x + 1) and b = (4x2 – 2x + 1) show 9
: ab – 1= x3
that 8

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It's your time - Project work!

11. a) Let's write any three pairs of monomial expressions and find the product
of each pair.

b) Let's write any three pairs of binomial expressions and find the product of
each pair.

c) Let's write any three pairs of monomial expressions such that the product
of each pair is 12x4y4.

10.3 Some special products and formulae

Da bC
ab a + b
(i) The product of (a + b) and (a + b) a a2
(square of binomials)

Let’s multiply (a + b) by (a + b) b ab b2 b

(a + b) u (a + b) = a (a + b) + b (a + b) A a+b b B
(a + b)2 = a2 + ab + ab + b2
= a2 + 2ab + b2 In the figure, a be the length of each side

Thus, (a + b)2 = a2 + 2ab + b2 of a smaller square. When the length is

Here, if (a + b)2 = a2 + 2ab + b2, then, increased by b, each side of the bigger
a2 + b2 = (a + b)2 – 2ab
square ABCD is (a + b).

Now, area of ABCD = a2 + ab + ab + b2

(a + b)u(a + b)=(a + b)2 = a2 + 2ab + b2

(ii) The product of (a – b) and (a – b) D a–b b C

Let’s multiply (a – b) by (a – b) (a – b)2 b(a – b) a–b
a
(a – b) u (a – b) = a (a – b) – b (a – b)
(a – b)2 = a2 – ab – ab + b2 b(a – b) b2 b
= a2 – 2ab + b2
A a–b b B
Thus, (a – b)2 = a2 – 2ab + b2
In the figure, the length and breadth of
Here, if (a – b)2 = a2 – 2ab + b2, then, the square ABCD are decreased by b.
a2 + b2 = (a – b)2 + 2ab Area of square

= (a – b)2 + b (a – b) + b (a – b) + b2
a2 = (a – b)2 + ab – b2 + ab – b2 + b2
a2 = (a – b)2 + 2ab – b2
? (a – b)2 = a2 – 2ab + b2

Now, let's recall the following important formulas,

1. (a + b)2 = a2 + 2ab + b2 = a2 – 2ab + b2 + 4ab = (a – b)2 + 4ab

2. (a – b)2 = a2 – 2ab + b2 = a2 + 2ab + b2 – 4ab = (a + b)2 – 4ab

Worked-out examples

Example 1: Find the squares of (x + 2) a) by geometrically

b) without using formula c) using formula

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Solution: Dx 1 1C
a) By geometrical process: x x2
1.x
The square of (x + 2) 1.x
(x + 2)2 = x2 + x + x + x + x + 1 + 1 + 1 + 1 x+2

= x2 + 4x + 4 1 1.x 1 1
b) Without using formula: 1 1.x 1 1
A x+2 B
The square of (x 2) = (x 2)2
= (x 2) (x 2)
= x (x 2) 2 (x 2)

= x2 + 2x + 2x + 4 = x2 + 4x + 4

c) By using formula: Consider, x = a and 2 = b
The square of (x 2) = (x 2)2 Now, (a + b)2 = a2 + 2ab + b2
? (x + 2)2 = (x)2 + 2.x.2 + (2)2
= (x)2 + 2. x . 2 + (2)2

= x2 + 4x + 4

Example 2: Expand: a) (2p2 3q2)2 b) 3x 12
Solution: 3x
Here,
a) (2p2 + 3q2)2 Let, 2p2 = a and 3q2 = b, then,
(a + b)2 = a2 + 2ab + b2
= (2p2)2 + 2. 2p2 . 3q2 + (3q2)2 ? (2p2 + 3q2)2 = (2p2)2 + 2. 2p2 . 3q2 + (3q2)2

= 4p4 + 12p2q2 + 9q4

b) 3x 12 = (3x)2 – 2 . 3x . 1 12 It is in the form of ( )2. So, using
3x 3x 3x the formula, (a – b)2 = a2 – 2ab + b2,

= 9x2 – 2 + 1 3x 1 2 = (3x)2 – 2 . 3x. 1 + 12
9x2 3x 3x 3x

Example 3: Simplify (2x 3y)2 – (2x 3y)2.

Solution:

(2x + 3y)2 – (2x – 3y)2 = (2x)2 2. 2x.3y + (3y)2 – [(2x)2 2. 2x.3y + (3y)2]
= 4x2 12xy + 9y2 – (4x2 12xy + 9y2)

= 4x2 12xy + 9y2 – 4x2 12xy – 9y2 = 24xy

Example 4: Express 9p2 + 24pq + 16q2 as a perfect square.

Solution: 9p2 24pq + 16q2
9p2 24pq + 16q2 (3p)2 ………. + (4q)2
= (3p)2 2. 3p. 4q + (4q)2 (3p)2 2. 3p. 4q. + (4q)2
= (3p + 4q)2 It is in the form a2 2ab + b2 which is equal to (a b)2.

Example 5: Find the squares of a) 99 b) 101
Solution:
a) Square of 99 = 992 = (100 – 1)2 = (100)2 – 2.100.1 + 12 = 10000 – 200+1 = 9801
b) Square of 101 = (101)2 = (100 + 1)2 = (100)2 + 2. 100.1 + 12

=10000 + 200 + 1 = 10201

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Example 6: If x + y = 5 and xy = 6, find the value of x2 + y2.

Solution:
Here, x + y = 5
On squaring both sides, we get

(x + y)2 = (5)2
or, x2 + 2xy + y2 = 25 Using (a + b)2 = a2 + 2ab + b2

or, x2 + 2 × 6 + y2 = 25 Putting xy = 6 [ xy = 6]

or, x2 + 12 + y2 = 25

or, x2 + y2 = 25 – 12

? x2 + y2 = 13 1
p2
Example 7: If p – 1 = 3, find the value of p2 – .
Solution: p

Here, p – 1 = 3
p

On squaring both sides, we get,

p 12 = (3)2
p

or, p2 – 2 × p × 1 + 1 =9 Using (a – b)2 = a2 – 2ab + b2
p p2

or, p2 + 1 =9+2
p2

? p2 +p12 = 11

Example 9: If m – 1 = 7, prove that : a) m2 1 = 51 b) m 1 2 = 53
Solution: m m2 m
Here, a)
m – 1 = 7
m

or, m 1 2 = 72 Squaring on the sides
m

or, m2 – 2 × m × 1 + 1 = 49
m m2

? m2 + 1 = 49 + 2 = 51 Proved.
m2

b) m 1 2 = m 1 2 + 4m × 1 (a + b)2 = (a – b)2 + 4ab
m m m

= 72 + 4
= 49 + 4

? m 1 2 = 53 Proved
m

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EXERCISE 10.3
General Section – Classwork
1. Let's tell and write the answers in the expanded forms.

a) (m + n)2 = .................................... b) (m – n)2 = ....................................
c) (p + q)2 = .................................... d) (p – q)2 = ....................................
e) (a + 1)2 = .................................... f) (a – 1)2 = ....................................
g) (x + 2)2 = .................................... h) (x – 2)2 = ....................................

2. Let's tell and write these expressions in their square forms.

a) x2 + 2xy + y2 = ......................... b) t2 – 2tr + r2 = .........................

c) x2 + 2.x.3 + 32 = ......................... d) y2 – 2.x.3 + 32 = .........................

e) a2 + 2.a.5 + 52 = ......................... f) b2 – 2.b.7 + 72 = .........................

g) p2 – 4p + 4 = ......................... h) r2 + 4r + 4 = .........................

Creative Section - A

3. Let’s study these diagrams and write the areas in algebraic expression forms
as shown in the example.

Dx 1C

x x2 1.x x+1 Area of ABCD = x2 + 1.x + 1.x + 12
(x + 1)2 = x2 + 2x + 1
1 1.x 12
A x+1 B

D x 2C S x 3R M x yE
x2 2.x x2 x.y
a) x+2 b) c) x+y

x x x

2 2.x y x.y
A x+2 B
PQ P x+y R
Area of ABCD
Area of PQRS Area of PREM

4. Let's find the squares of the following expressions (i) by geometrical process
(ii) without using formula (iii) using formula.

a) (x + 1) b) (x + 2) c) (x + 3)

5. Let's find the squares of the following expressions.

a) (x + 3) b) (x – 4) c) (2y + 3) d) (x – 5y) e) (6m – 5n)
f) (4x + 3y ) g) (a2 + x2) 1 1
h) (x + 1 ) i) y y j) a 2a
x

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6. Let's expand.

a) (a + 3) 2 b) (2a – 5)2 c) (2x + 3y)2

d) 2t + 1 2 e) 3x + 1 2 f) c2 1 2
2t 3x c

7. Let's simplify.

a) (a + b)2 – 2ab b) (p – q)2 + 2pq c) (x + 3)2 – 3 (2x + 3)

d) (3m – 4)2 + 8(3m – 2) e) (a – t)2 + (a + t)2 f) (x + y)2 – (x – y)2

g) (2x – 3y)2 – (2x + 3y)2 h) (4b + 5c)2 – (4b – 5c)2 i) (3a – 5b)2 – (5a + 3b)2

8. Let's express the following expressions in the square forms.

a) x2 + 6x + 9 b) x2 – 8x + 16 c) 4a2 + 4ab + b2

d) p2 – 6pq + 9q2 e) 4x2 + 12xy + 9y2 f) 25x2 – 40xy + 16y2

g) 49a2 – 42ab + 9b2 h) x2 + 2 + 1 i) 4p2 – 2 + 1
x2 4p2

9. Find the squares of the following numbers by using the formula of (a + b)2 or
(a–b)2.

a) 49 b) 51 c) 99 d) 101 e) 999 f) 1001

Creative Section - B

10. a) If a + b = 3 and ab = 2, find the value of a2 + b2.

b) If p + q = 7 and pq = 12, find the value of p2 + q2.

c) If x – y = 1 and xy = 6, find the value of x2 + y2.

d) If m – n = 5 and mn = 14, find the value of m2 – n2.
11. a)
If p + 1 = 4, find the value of p2 + 1 and p– 1 2.
b) p p2 p

c) If m + 1 = 5, find the value of m2 + 1 and m– 1 2.
m m2 m

If x 1 = 3, find the value of x2 + 1 and x 1 2.
x x2 x

d) If y 1 = 6, find the value of y2 + 1 and y 1 2.
y y2 y

It's your time – Project Work!

12. a) Let's write the formulae you learnt in this lesson in a chart paper and
b) compare it to your friends' work.
c)
Let's write any three binomial expressions in the form of (a + b) and find
the squares of your expressions.

Let's write any three binomial expressions in the form of (a – b) and find
the square of your expressions.

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(iii) The product of (a + b) and (a – b)
Lets multiply (a – b) by (a + b).
(a + b) u (a – b) = a (a – b) + b (a – b) = a2 – ab + ab – b2 = a2 – b2

Thus, (a + b) (a – b) = a2 – b2

Da CD C Da bC

a–b (a + b) (a – b)
a–b
ab a2 – b2
b A a+b B
EF
AB Area of ABCD
Area of ABCD AB a2 – b2 = (a + b) (a – b)
= a2 Area of ABCDEF
= a2 – b2

(iv) The product of (a + b), (a + b) and (a + b) (cube of binomials)
Let’s find the product of (a + b)3.
(a + b)3 = (a + b) (a + b) (a + b)
= (a + b) (a + b)2
= (a + b) (a2 + 2ab + b2)
= a (a2 + 2ab + b2) + b (a2 + 2ab + b2)
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
= a3 + 3a2b + 3ab2 + b3
? (a + b)3 = a3 + 3a2b + 3ab2 + b3

Geometrical interpretation:
Let's take a cube of each side (a b). Then volume of cube = (a b)3. Which is
shown in the diagram.

a+b a a b
a bb
a+b a+b a a a
(a + b)3 = b bb

a3 + 3a2b + 3ab2 + b3

(v) The product of (a – b), (a – b) and (a – b)

Let’s find the product of (a – b)3.

(a – b)3 = (a – b) (a – b) (a – b)]
= (a – b) (a – b)2
= (a – b) (a2 – 2ab + b2)
= a (a2 – 2ab + b2) – b (a2 – 2ab + b2)
= a3 – 3a2b + ab2 – a2b + 2ab2 – b3
= a3 – 3a2b + 3ab2 – b3

? (a – 3)3 = a3 – 3a2b + 3ab2 – b3

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Worked-out examples

Example 1: Find the products of the following expressions.

a) (4x + 5) (4x – 5) b) (a2 + b2) (a2 – b2)

Solution:

a) (4x + 5) (4x – 5) Consider, 4x = a and 5 = b
= (4x)2 – 52 Now, (a + b) (a – b) = a2 – b2
= 16x2 – 25 ?(4x + 5) (4x – 5) = (4x)2 – 52.

b) (a2 + b2) (a2 – b2) Consider, a2 = a and b2 = b
= (a2)2 – (b2)2 Now, (a + b) (a – b) = a2 – b2
= a4 – b4
?(a2 + b2) (a2 – b2) = (a2)2 – (b2)2

Example 2: Find the product of (2x + 3y) (2x – 3y) (4x2 + 9y2).
Solution:

(2x + 3y) (2x – 3y) (4x2 + 9y2) = [(2x)2 – (3y2)] (4x2 + 9y2)
= (4x2 – 9y2) (4x2 + 9y2)
= (4x2)2 – (9y2)2 = 16x4 – 81y4

Example 3: Find the product of

a) 51 u 49 b) 102 u 98 by using the formula (a + b) (a – b) = a2 – b2.
Solution:

a) 51 u 49 = (50 + 1) (50 – 1) b) 102 u 98 = (100 + 2) (100 – 2)

= (50)2 – (1)2 = (100)2 – (2)2

= 2500 – 1 = 10000 – 4

= 2499 = 9996

Example 4: Simplify 2.6 × 2.6 1.4 × 1.4
Solution: 2.6 1.4

2.6 × 2.6 – 1.4 × 1.4 = (2.6)2 – (1.4)2 = (2.6 + 1.4) (2.6 – 1.4) = 2.6 + 1.4 = 4
2.6 1.4 2.6 1.4 2.6 – 1.4

Example 5: Find the cubes of a) (x + 2) b) (2m + n) c) 3p – 1
3p
Solution:
a) Cube of (x + 2)

= (x + 2)3

= x3 + 3 . x2 . 2 + 3 . x . 22 + 23 Using (a + b)3 = a3 + 3a2b + 3ab2 + b3

= x3 + 6x2 + 12x + 8

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b) Cube of (2m + n) Consider, 2m = a and n = b
= (2m + n)3 Now, (a + b)3 = a3 + 3a2b + 3ab2 + b3
? (2m + n)3 = (2m)3 + 3.(2m)2.n + 3.2m.n2 + n3

= (2m)3 + 3. (2m)2. n+ 3. 2m. n2 + n3

= 8m3 + 12m2n + 6mn2 + n3

c) Cube of 3p – 1 1
3p 3p
Consider, 3p = a and =b
= 3p – 13
3p Now, (a – b)3 = a3 – 3a2b + 3ab2 – b3

= (3p)3 – 3. (3p)2. 31p+ 3 . 3p 1 2 – 13 1 3 1
3p 3p 3p 3p
? 3p + = (3p)3 – 3. (3p)2.

= 27p3 – 3. 3p + 3. 3p. 1 – 1 + 3. 3p 1 2– 13
9p2 27p3 3p 3p
1 1
= 27p3 – 9p + p – 27p3

Example 6: Express x3 + 12x2 + 48x + 64 as a cube of a binomial.

Solution: x3 + 12x2 + 48x + 64
x3 + 12x2 + 48x + 64
= x3 + … + … + 43
= x3 + 3. x2. 4 + 3. x. 42 + 43
= (x + 4)3 = x3 + 3. x2. 4 + 3. x. 42 + 43.

It is in the form a3 + 3a2b + 3ab2 + b3 which is
equal to (a + b)3.

Example 7: If (a + b) = 5, find the value of a3 + b3 + 15ab.

Solution:

Here, (a + b) = 5

On cubing both sides, we get

? (a + b)3 = 53

or, a3 + 3a2b + 3ab2 + b3 = 125

or, a3 + b3 + 3ab (a + b) = 125

or, a3 + b3 + 3ab u 5 = 125

or, a3 + b3 + 15ab = 125

So, the required value of a3 + b3 + 15ab is 125.

Example 8: a) If x + y = 7 and xy = 6, find the value of x3 + y3.
Solution:
b) If a – 1 = 3, find the value of a3 – 1 .
a a3

Here, b) a– 1 =3
a
a) x + y = 7 and xy = 6
Now, x + y = 7 on cubing both side, we get,
on cubing both side, we get

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(x + y)3 = 73 a– 1 3 = 33
or, x3 + 3x2y 3xy2 y3 = 343 a
1 1 1a3
or, x3 3xy (x y) + y3 = 343 or, a3 3a2 u a 3a u a2 = 27

or, x3 3 u 6 u 7 y3 = 343 or, a3 3 a – 1 1 = 27
or, x3 126 y3 = 343 a a3
? x3 y3 = 343 126 = 217
or, a3 3 u 3 1 = 27
a3
1
? a3 a3 = 27 9 = 36

EXERCISE 10.4

General Section - Classwork

1. Let's express as the product of two binomials.

a) x2 – y2 = ....................... b) p2 – q2 = .......................
= .......................
c) a2 – 4 = ....................... d) 9 – b2

2. Let's tell and write the products as quickly as possible.

a) (a + x) (a x) = ....................... b) (x + 1) (x – 1) = .......................
c) (a – 3) (a + 3) = ....................... d) (2 + p) (2 p) = .......................

3. Let's tell and write the expanded forms of the following cubes.

a) (a + x)3 = ........................................................................................

b) (a – x)3 = ........................................................................................

c) (m + n)3 = ........................................................................................

d) (m n) 3 = ........................................................................................

e) (x 1) 3 = ........................................................................................

f) (x 1) 3 = ........................................................................................

Creative Section - A

4. Let's find the area of the following rectangles.

a) b) c)

x–2 x+3 a–5

x+2 a+5

x–3
5. Let's find the products of the following expression by using formula.

a) (x + 2) (x –2) b) (a +3) (a –3)

c) (2p + 3) (2p – 3) d) (2a + 5b) (2a – 5b)

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e) x 1 x– 1 f) y 2 y– 2
2 2 3 3

g) (x2 + y2)(x2 –y2) h) (x2 + 9)(x2 – 9)

6. Let's simplify.

a) (a + 1) (a – 1) (a2 + 1) b) (x + 2) (x – 2) (x2 + 4)

c) (p + 3) (p –3) (p2 + 9) d) (x +y) (x –y) (x2 + y2)

e) (a + 2y) (a – 2y) (a2 + 4y2) f) (2x + 3y) (2x – 3y) (4x2 + 9y2)

7. Let's find the products by using the formula (a + b) (a – b) = a2 – b2

a) 11 × 9 b) 41 × 39 c) 52 × 48 d) 102 × 98
8. Let's simplify.

a) 1.8 × 1.8 – 1.2 × 1.2 b) 1.6 × 1.6 – 0.9 × 0.9
1.8 + 1.2 1.6 + 0.9

c) 2.4 × 2.4 – 1.6 × 1.6 d) 4.6 × 4.6 – 2.5 × 2.5
2.4 – 1.6 4.6 – 2.5

e) 0.7 × 0.7 – 0.3 × 0.3 f) 0.009 × 0.009 – 0.006 × 0.006
0.7 + 0.3 0.009 – 0.006

9. Let's find the cubes of the following expressions.
a) ( x +3) b) (2a –1) c) (2x + 3) d) (p – 4q)
e) (5 + 2b) f) (2 – 3y) 1 1
g) (a + a ) h) ( 2p – 2p )

10. Let's expand the following cubes. c) (2a + 3b)3
a) (2x + 1)3 b) (x 3y)3 f) 3x –31x 3
d) (4 m)3 e) (1 2z)3

11. Let's simplify.

a) (a + 2)3 – 6a (a + 2) b) (a – 2)3 + 6a (a – 2) c) (m + 3)3 – 9m (m 3)

d) (m 3)3 9m (m 3) e) (x y)3 x3 – y3 f) (x + y)3 (x y)3

12. Let's express the following expressions as cubes of binomials.

a) a3 + 3a2 + 3a + 1 b) x3 + 6x2 + 12x + 8

c) p3 – 9p2 + 27p – 27 d) m3 – 12m2 + 48m – 64

e) 8a3 – 12a2b + 6ab2 – b3 f) 27x3+54x2y + 36xy2 + 8y3

Creative Section - B

13. a) If x + y = 3 and xy = 2, find the value of x3 + y3.

b) If p + q = 4 and pq = 3, find the value of p3 + q3.

c) If a – b = 2 and ab = 3, find the value of a3 – b3.

d) If m – n = 3 and mn = 4, find the value of m3 – n3.

14. a) If a 1 = 2, find the value of a3 + 1 .
a a3

b) If p 1 = 5, find the value of p3 + 1 .
p p3

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c) If m – 1 = 3, find the value of m3 – 1 .
m m3

d) If x– 1 = 6, find the value of x3 – 1 .
x x3

It's your time - Project work!

15. a) Let’s take a few rectangular sheet of papers (photocopy paper). Then fold
them to get square sheet of papers as shown in the diagrams.

b) Again, let’s fold each square sheet of paper as shown in the diagram and complete
the sums.

(i) x x x x1

x x2 x2 x x2–12 x–1 x2–12 = ....x.2..–...1..2.... = ..........................
x

(ii) x 12 1 x x+1
1 xy
x

x x2 x2 x x2–y2 x–1 x2–y2 = ................ = ..........................
x

y2 y x+y

y

16. a) Let's find the products of any three pairs of binomials of the form (a + b)

(a – b). Then show that (a + b) (a – b) is always a2 – b2.

b) Let's find the product of any two sets of three binomials of the form (a + b)
(a + b) (a + b).

c) Let's find the products of any two sets of three binomials of the form (a – b)
(a – b) (a – b).

10.4 Division of algebraic expressions
(i) Division of monomials by monomials

While dividing a monomial by another monomial, we divide the coefficient of
dividend by the coefficient of divisor. Then, subtract the power of the base of
divisor from the power of the same base of dividend. For example:

Example 1: Divide a) 28x4y3 by 7x2y2 b) 42a5b4 by 6a4b4

Solution: =14278xx24yy23 = 4x4 – 2 y3 – 2 = 4x2y
a) 28x4y3 ÷ 7x2y2

7

b) 42a5b4 ÷ 6a4b4 = 42a5b4 = 7a5 – 4 b4 – 4 = 7abq = 7a
16a4b4

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(ii) Division of polynomials by monomials
While dividing a polynomial by a monomial, we divide each term of the
polynomial by the monomial separately. For example,

Example 2: Divide a) 8x4 – 6x3 by 2x2 b) 35a3b2 + 15a2b3 by 5ab
Solution:

a) (8x4 – 6x3) ÷ 2x2 Alternative process Checking:
2x2 ) 8x4 – 6x3 ( 4x2 – 3x 2x2 (4x2 3x)
= 8x4 – 6x3 = 2x2 u 4x2 2x2 u 3x
2x2 2x2 + 8x4 = 8x4 6x3
6x3
= 4x4 – 2 – 3x3 – 2 6x3
0
= 4x2 – 3x

? Quotient = (4x2 – 3x)

b) (35a3b2 + 15a2b3) ÷ 5ab 5ab ) 35a3b2 + 15a2b3 (7a2b + 3ab2
35a3b2 15a2b3
= 5ab + 5ab ±35a3b2 Checking:
5ab (7a2b 3ab2)
= 7a3 – 1 b2 – 1 + 3a2 – 1 b3 – 1 15a2b3 = 5ab u 7a2b 5ab u 3ab2
± 15a2b3 = 35a3b2 15a2b3
= 7a2b + 3ab2
0

? Quotient 7a2b + 3ab2

(iii) Division of polynomials by a binomial

While dividing a polynomial by a binomial at first we should arrange the
terms of divisor and dividend in descending (or ascending) order of power
of common bases. Then we should start the process of division dividing the
term of dividend with the highest power. For example,

Example 3: Divide (x2 + 2x – 15) by (x – 3).

Solution:

x – 3 ) x2 + 2x – 15 ( x + 5± Divide x2 by x o x2 ÷ x = x (quotient)
+x2 – 3x Multiply the divisor (x – 3) by the quotient x.
–+ x (x – 3) = x2 – 3x
Subtract the product from the dividend.
5x – 15
±5x 15 x2 + 2x – 15

0 ±x2 3x
5x – 15

Again, the remainder is the new dividend and repeat
the process till the remainder is not divisible by divisor.

Example 4: The length of a rectangular field is (x + 10) m and its area is
(2x2 + 17x – 30) sq.m.

(i) Find the breadth of the field.

(ii) If x = 10 m, find the actual length, breadth and area of the field.

Vedanta Excel in Mathematics - Book 7 168

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Solution:

Here,
Length of a rectangular field (l) = (x + 10) m
Area of the field (A) = (2x2 + 17x – 30) sq.m
Breadth (b) = ?
Now, breadth = Area ÷ length

So, x + 10) 2x2 + 17x – 30 (2x – 3 (i) 2x2 ÷ x = 2x
– 2x2 +– 20x (ii) 2x(x + 10) = 2x2 + 20x
– 3x – 30 (iii) 2x2 + 17x – 2x2 – 20x = – 3x
+– 3x +– 30 (iv) – 3x ÷ x = – 3
(v) – 3(x + 10) = – 3x – 30
0 (vi) – 3x – 30 + 3x + 30 = 0

? Breadth (b) = (2x – 3)m
Again,
when x = 10 m, then actual length (l) = (x + 10)m = (10 + 10)m = 20m

breadth (b) = (2x – 3)m = (2 × 10 – 3) m = 17m
and area (A) = (2x2 + 17x – 30)m2

= (2 × 102 + 17 × 10 – 30) m2
= (200 + 170 – 30) m2 = 340m2 l × b = 20 m × 17 m = 340 m2

Example 5: Divide (x4 – y4) by (x y) Then, x4 ÷ x = x3
And, x3(x − y) = x4 − x3y
Solution: Again, x4 – y4 – (x4 – x3y) = +x3y – y4
Then,
x y ) x4 – y4 ( x3 x2y + xy2 y3 And, x3y ÷ x = x2y,
± x4 x3y Again, x2y(x – y) = x3y – x2y2
+ x3y – y4 Then, x3y – y4 – (x3y – x2y2) = x2y2 – y4
± x3y x2y2 And,
Again, x2y2 ÷ x = xy2
+ x2y2 – y4 Then, xy2 (x – y) = x2y2 – xy3
± x2y2 xy3 And, x2y2 – y4 – (x2y2 – xy3) = xy3 – y4

+ xy3 – y4 xy3 ÷ x = y3
± xy3 y4 y3(x – y) = xy3 – y4
xy3 – y4 – (xy3 – y4) = 0
0

EXERCISE 10.5

General Section A – Classwork

Let's tell and write the quotients as quickly as possible.

1. a) x2 ÷ x = ........ b) y2 ÷ y2 = ........ c) a4 ÷ a = ........
= ........
d) p4 ÷ p2 = ........ e) b3 ÷ b2 = ........ f) m4 ÷ m3

169 Vedanta Excel in Mathematics - Book 7

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2. a) x2y2 ÷ xy = ........ b) a3b3 ÷ ab = ........ c) p3q3 ÷ p2q2 = ........
d) p2q3 ÷ pq2 = ........ e) m4n4 ÷ mn2 = ........ f) x4y3 ÷ x2y3 = ........

3. a) 4x2 ÷ 2x = ........ b) 10y3 ÷ 5y = ........ c) 25a5 ÷ 7a3 = ........
d) 40a3b2 ÷ 4ab = ........ e) 20c2d3 ÷ 5c2d3= ........ f) 55x4y3 ÷ 11x2y2 = ........

Creative Section - A

4. Let's find the quotients.

a) 10x3 ÷ 5x b) 24x2 ÷ 6x2 c)18a2 ÷ 9a2
d) 32a3b2 ÷ 8ab e) 66x4y5 ÷ 11x3y2 f) – 20p4q5 ÷ 5p3q4

g) 45m7n6 ÷ ( – 9m2n4) h) –28a3t8 ÷ (– 7at5) i) 56a2b3c ÷ ( 8ab3c)

5. Let's divide and find the quotients.

a) (6x2 + 9x ) ÷ 3x b) (8a3 – 12a2) ÷ 4a

c) (25y6 – 30y5) ÷ 5y4 d) (2p + 4p2 – 6p3) ÷ 2p

e) (12x3 – 9x2 + 15x) ÷ 3x f) (14m4 – 21m3 – 28m2) ÷ 7m2

g) (10a4b3 – 15a3b4 + 20a2b2) ÷ 5ab

h) (36x5y4 + 45x4y5 – 63x3y3) ÷ (–9x3y3)

i) (44p3q4 – 55p4q3 + 66p5q2) ÷ ( – 11p3q2)

j) (6xyz – 9x2y2z2 – 12x3y3z3) ÷ (– 3xyz)

6. Let's find the quotients.

a) (a2 – 4) ÷ (a + 2) b) (a2 – 4) ÷ (a – 2)

c) (9x2 – 1) ÷ (3x + 1) d) (9x2 – 1) ÷ (3x – 1)

e) (x2 + 5x + 6) ÷ ( x + 2) f) (x2 + 6x + 8) ÷ (x + 4)

g) (y2 + y – 12) ÷ (y – 3) h) (y2 – 11y + 28) ÷ (y – 7)

i) (p2 – 3p – 10) ÷ (p + 2) j) (p2 – 5p – 24) ÷ (p + 3)

k) (4m2 + 5m – 6) ÷ (m + 2) l) (5m2 + 13m – 6) ÷ (m + 3)

m) (6x2 + 5x – 6) ÷ (3x – 2) n) (6x2 + x – 12) ÷ (2x + 3)

o) (4p2 + 12pq + 9q2 ) ÷ ( 2p + 3q) p) (16p2 – 40pq + 25q2 ) ÷ (4p – 5q)

7. a) The product of two algebraic expressions is 4x2 – 9. If one of the expressions is
(2x + 3), find the other expression.

b) The product of two algebraic expressions is (x2 – 4x – 32). If one of the
expressions is (x + 4), find the other expression.

c) The area of a rectangle is (x2 – 16) sq. cm. If its breadth is (x – 4) cm, find its
length.

d) The area of a rectangular field is (a2 – 25) sq. m. If its length is (a + 5) m, find
its breadth.
8. a) If a = 4x3y2, b = 3x2y3 and c = 6xy, find the value of ab .
c
21a4a4b64b,6s, hshowowththatatzxp+rq
b) If p = 7a3b3, q = 6a4b4 and r = = 3ab.
c) If x = 8a4b5, y = 4a5b4 and z =
y = 4b + 2a.
z

Vedanta Excel in Mathematics - Book 7 170

Laws of Indices

Creative Section - B

9. Let's divide and find the quotients.

a) (2x3 – 5x2 – 24x – 18) ÷ (2x + 3) b) (3x2 – 10x2 + 7x + 10) ÷ (3x + 2)

c) (a4 – b4) ÷ (a + b) d) (a4 – b4) ÷ (a – b)

e) (a5 + b5) ÷ (a + b) f) (a5 – b5) ÷ (a – b)

g) (x6 – y6) ÷ (x2 – y2) h) (x6 – y6) ÷ (x3 – y3)

10.a) The length of a rectangular carpet is (x – 5)m and its area is (x2 – 12x + 35) sq. m.
(i) Find its breadth
(ii) If x 10m, find its actual length, breadth and area of the carpet.

b) The area of a rectangular garden is (x2 – 4x – 32) sq. m and its breadth is (x – 8)m.
(i) Find its length.
(ii) If x = 15m, find the actual length, breadth, and area of the garden.

11. a) Sunayana distributed (3x2 + 20x + 25) sweets equally among (3x + 5) friends
on her birthday.
(i) How many sweets did each of the friends receive?
(ii) If x = 5, find the total number of sweets distributed among friends, actual
number of friends and share of sweets.

b) On Children's Day Hari Narayan distributed (2x2 + 13x – 24) copies equally
among (2x – 3) number of children of a child care centre.

(i) Find the number of copies received by each child.

(ii) If x = 20, find the actual number of copies, students, and share of each
child.

It's your time - Project work!

12. a) Let's write any three pairs of monomial expressions with the same bases with
higher power of dividend. Divide the dividend by the divisor in each pair
and show that: Dividend = Quotient × Divisor + Remainder

b) Let's write any three pairs of binomial dividend and divisor in the forms of
(a2 – b2) and (a + b) or (a2 – b2) and (a – b).

Then, divide the dividend by the divisor and find quotient in each pair.

c) Let's find the products of any three pairs of binomial expressions. Then,
divide each product by one of the expressions in each pair and get another
expression.

E.g.: find o (x + 2) (x + 3) = x2 + 5x + 6, then find

(x2 + 5x + 6) ÷ (x + 2) = x + 3

10.5 Factors and factorisation – Introduction

We know that, in 2 × 5 = 10 o 10 is the product, 2, and 5 are the factors.

in 3 × 4 = 12 o 12 is the product, 3, and 4 are the factors.

Similarly, in x × x = x2 o x2 is the product, x, and x are the factors.

in 2a(a + b) = 2a2 + 2ab o 2a2 + 2ab is the product, 2a and (a + b)
are factors.

171 Vedanta Excel in Mathematics - Book 7

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Thus, when two or more algebraic expressions are multiplied, the result is called
the product and each expression is called the factor of the product. The process of
finding out factors of an algebraic expression is known as factorisation. It is also
known as resolution of the expression into its factors.

Of course, the process of factorisation is the reverse process of multiplication. For
example,

Multiplication of 2x(x – 3) = 2x2 – 6x and factorisation of 2x2 – 6x = 2x (x – 3)

Multiplication of (a + b) (a – b) = a2 – b2 and factorisation a2 – b2 = (a + b) (a – b)

Now, let’s learn the process of factorisation of different types of expressions.

(i) Factorisation of expressions which have a common factor in each of its
term
In this type of expression, the factor which is present in each term is taken as
common and each term of the expression is divided by the common factor. The
product form of the common factor and the quotient represents the factorisation
of the expression.

Worked-out examples

Example 1: Factorise a) ax + bx b) 2px2 – 6p2x c) 4ax2 + 6a2x – 8ax

Solution:

a) ax + bx x is present in each term. So, it is the common factor.

= x (a + b) x is taken as common, then ax ÷ x = a and bx ÷ x = b

b) 2px2 – 6p2x 2px2 – 6p2x Direct process
= 2px u x – 2 u 3p u px = 2px (x – 3p) 2px2 ÷ 2px = x and 6p2x ÷ 2px = 3p

= 2px (x – 3p)

c) 4ax2 + 6a2x – 8ax Remember!
= 2ax (2x + 3a – 4) Among the factors which are present in each term, the
factor with the least power is taken as a common factor.

Example 2: Resolve into factors 3a (x + y) – 4b (x + y)

Solution: (x + y) is the common factor.
3a (x + y) – 4b (x + y)

= (x + y) (3a – 4b) 3a(x + y) ÷ (x + y) = 3a and 4b(x + y) ÷ (x + y) = 4b

(ii) Factorisation of expressions having a common factor in the groups of
terms

In this case, terms of the given expression are arranged in groups in such a way
that each group has a common factor. For example:

Vedanta Excel in Mathematics - Book 7 172

Laws of Indices

Example 3: Factorise a) a2 + ab + ca + bc b) x2 – 3a + 3x – ax

Solution: Terms are arranged in groups.
a) a2 + ab + ca + bc

= a (a + b) + c (a + b) a is common in a2 + ab and c is common in ca + bc

= (a + b) (a + c) (a + b) is common.

b) x2 – 3a + 3x – ax Arranging the terms in suitable groups
= x2 + 3x – ax – 3a

= x (x + 3) – a (x + 3) x is common in x2 + 3x and a is common in ax – 3a

= (x + 3) (x – a)

(iii) Factorisation of expressions having the difference of two squared terms

The algebraic expression a2 – b2 is the difference of two squared terms. Here,
a2 and b2 are the squared terms and a and b are their square roots respectively.

We have learnt that a2 – b2 is the product of (a + b) and (a – b)

? a2 – b2 = (a + b) (a – b) Square root of a2
a2 – b2 = (a + b) (a – b)
Thus, (a + b) and (a – b) are the factors of a2 – b2.
Square root of b2
To factorise such expressions, we should re-write
the given terms in the form of a2 – b2.

Then, a2 – b2 = (a + b) (a – b) represents the factorisation of the expression.

Example 4: Factorise a) 4x2 – 9y2 b) 81x4 – 16y4.
Solution: Square root of 4x2 = 2x and square root of 9y2 = 3y
a) 4x2 – 9y2

= (2x)2 – (3y)2

= (2x + 3y) (2x – 3y) Using a2 – b2 = (a + b) (a – b)

b) 81x4 – 16y4 Square root of 81x4 and 16y4 are 9x2 and 4y2
= (9x2)2 – (4y2)2

= (9x2 + 4y2) (9x2 – 4y2) Using a2 – b2 = (a + b) (a – b)

= (9x2 + 4y2) [(3x)2 – (2y)2] 9x2 – 4y2 is still in the form a2 – b2.
So, it is further factorised.

= (9x2 + 4y2) (3x + 2y) (3x – 2y)

173 Vedanta Excel in Mathematics - Book 7

Laws of Indices

Example 5: Simplify by factorisation process.

a) 24 u 40 – 20 u 24 b) 652 – 552 c) 49 u 51

Solution:

a) 24 u 40 – 20 u 24 b) 652 – 552 c) 49 u 51

= 24 (40 – 20) = (65 + 55) (65 – 55) = (50 – 1) (50 + 1)

= 24 u 20 = 120 u 10 = 502 – 12
= 480 = 1200 = 2500 – 1 = 2499

(iv) Factorisation of expressions of the form x2 + px + q

While factorising a trinomial expression of the form x2 + px + q, we should
search any two numbers a and b such that a + b = p and ab = q. Clearly, a
and b must be the factors of q. Then px is expanded in the form ax + bx and
factorisation is performed by grouping.

Example 6: Factorise a) x2 + 5x + 6 b) a2 – 3a – 28

Solution: b) a2 – 3a – 28 = a2 – (7 – 4)a – 28
a) x2 + 5x + 6 = x2 + (2 + 3)x + 6 = a2 – 7a + 4a – 28
= a(a – 7) + 4(a – 7)
= x2 + 2x + 3x + 6 = (a – 7) (a + 4)
= x(x + 2) + 3(x + 2)
= (x + 2) (x + 3)

EXERCISE 10.6
General Section - Classwork
1. Let's tell and write the factors and products of these expressions as quickly as

possible.

a) a × a2 factors are .............................................. product is .........................

b) 2x2 × x factors are ............................................ product is .........................

c) 3p (p + 1) factors are ........................................... product is .........................

d) y2 (y – 2 ) factors are ............................................ product is .........................

2. Let's factorise, tell and write the factors as quickly as possible.

a) ax + ay = ............................. b) ax – ay = .............................

(c) ak + bk = ............................. (d) ak – bk = .............................

(e) a2 + a = ............................. (f) x2 – x = .............................

(g) x2 + xy = ............................. (h) p3 – p2q = .............................

Vedanta Excel in Mathematics - Book 7 174

Laws of Indices

3. a) a2 – b2 = ............................. b) b2 – c2 = .............................

c) x2 – 22 = ............................. d) y2 – 32 = .............................
e) p2 – 16 = ............................. f) r2 – 25 = .............................

4. Let's study the following tricky ways of factorisation carefully.

3+6 3×6 –3–6 –3×(–6)

x2 + 9x + 18 = (x + 3) (x + 6) x2 – 9x + 18 = (x – 3) (x – 6)

6–3 6×(-3) 3–6 3×(–6)

x2 + 3x – 18 = (x + 6) (x – 3) x2 – 3x – 18 = (x + 3 ) (x – 6)

Could you investigate the idea of tricky ways of factorisation. Apply your
investigation, tell and write the factors as quickly as possible.

a) x2 + 5x + 6 = ........................ b) x2 – 5x + 6 = ........................

c) x2 + x – 12 = ........................ d) x2 – x – 6 = ........................

Creative Section
5. Let's factorise.

a) 2ax + 2bx b) 2x2y – 2xy2 c) 2abx – 4aby

d) 3mx2 – 6mx e) ax2 + a2x – ax f) 4x5y4 – 2x4y5 – 8x3y3

g) a(x – y) + b(x – y) h) x (a + b) – y(a + b) i) 2x (x – y) + y(x – y)

6. Let's resolve into the factors.

a) ax + bx + ay + by b) ma + na + mb + nb c) xy + xz + y2 + yz
f) x2 – zx + xy – yz
d) 3ax – bx – 3ay + by e) 4ax – 3bx – 8ay + 6by i) x2 + 4a + 4x + ax

g) a2b + ca – ab2c – bc2 h) a2bc + c2a – ab2 – bc

7. Let's factorise. b) x2 – 9 c) m2 – 16 d) p2 – 25
a) a2 – 4 f) 25x2 – 36y2 g) 49p2 – 81q2 h) 64b2 – 9c2
e) 4a2 – 9b2 j) 8x2y2 – 18 k) 75a2b2 – 27x2y2 l) 80a3 – 5ab2
i) 100m2 – 49n2

8. Let's resolve into the factors.

a) a4 – 16 b) x4 – 81 c) y4 – 625 d) 16p4 – q4
g) a4 – b4 h) 16p4 – 81q4
e) 81m4 – n4 f) x4 – 16y4

9. Let's factorise. b) a2 + 7a + 12 c) p2 + 6p + 8 d) x2 + 7x + 10
a) x2 + 3x + 2 f) x2 – 9x + 20 g) m2 – 9m + 14 h) b2 – 11b + 30
e) y2 – 8y + 15 j) x2 – 3x - 28 k) a2 + 4a – 12 l) x2 + 5x – 24
i) x2 – 2x - 15

175 Vedanta Excel in Mathematics - Book 7

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10. Let's simplify by factorisation process.

a)13 × 50 – 40 × 13 b) 28 × 60 – 30 × 28 c)36 × 24 + 26 × 36
f)752 – 552
d)152 – 52 e)272 – 172 i)101 × 99

g)21 × 19 h)79 × 81

It's your time - Project work!

11. a) Let's write any five pairs of difference of square numbers in the form
a2 – b2, where a2 > b2. Then, factorise and simplify the each pair.
[E.g. 52 – 32 = (5 + 3) (5 – 3) = 8 × 2 = 16]

b) Let's write any five binomial expressions of the form x2 – a2, where a2 is a
square number. Then factorise your expressions.

12. a) Let's find the products of any four pairs of binomials of the forms
(i) (x + a) (x + b) (ii) (x + a) (x – b) (iii) (x – a) (x + b) (iv) (x – a) (x – b),
where a and b are any natural numbers and a > b.

b) Again, factorise each product to get the factors.

[E.g. (x + 1) (x + 2) = x2 + 3x + 2 and

x2 + 3x + 2 = x2 + (1 + 2)x + 2 = x2 + x + 2x + 2 = x(x + 1) + 2(x + 1)
= (x + 1) (x + 2)

10.6 Simplification of rational expressions

We know that 1, 2, 1 ,– 3 , etc. are the rational numbers. Similarly,
2 5
x x2 x + 2
3 , y2 , x – 7 , etc. are called rational expressions. Here, we shall discuss about

addition, subtraction, multiplication, and division of rational expressions.

(i) Multiplication and division of rational expressions

In multiplication, we simplify the numerical coefficients as in case of
multiplication of fraction. In the case of variables we apply the product and
quotient rules of indices. In division, we should multiply the dividend by the
reciprocal of divisor.

Worked-out examples

Example 1: Multiply 4a3b2 × 10x4y3 .
5x3y2 12a4b3

Solution:

4a3b2 10x4y3 = 14a3b2 × 120 x4y3 = 2x4 – 3 × y3 – 2 = 2xy
5x3y2 u 12a4b3 5x3y2 × 12a4b3 3a4 – 3 b3 – 2 3ab
13

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Laws of Indices

Example 2: Divide: 5x2y ÷ 15a3y3
Solution: 8ax3 16a2x2

5x2y ÷ 15a3y3 = 15x2y × 126a2x2 = 2a2x4y = 2x4 – 3 = 2x
8ax3 16a2x2 81ax3 135a3y3 3a4x3y3 3a4 – 2 y3 – 1 3a2y2

Example 3: Simplify: 6y2 × 2x3 ÷ 5xy
Solution: 4x2 3y3 6ab

6y2 × 2x3 ÷ 5xy = 261y2 × 12x3 × 6ab = 6abx3y2 = 6ab = 6ab
4x2 3y3 6ab 4x2 3y3 5xy 5x3y4 5y4 – 2 5y2
21 1

EXERCISE 10.7

General Section - Classwork

1. Let's tell and write the answers as quickly as possible.

a) x2 × a = ................................... b) x3 × a = ...................................
a2 x a3 x

c) x2 × y = ................................... d) x3 × xy22= ...................................
y2 x y3

2. a) a2 ÷ a = ................................... b) a3 ÷ a2 = ...................................
x2 x x3 x2

c) x2 ÷ x = ................................... d) x4 ÷ y2 = ...................................
y3 y y4 x2

Creative Section

3. Simplify.

a) 2x3 × 9a2 b) 4y4 × 10z4 c) x4y3 × 15ab2
3a3 8x2 5z3 12y3 5a3b3 9x2y

d) 4a4 ÷ 8a3 e) 6y5 ÷ 18y3 f) x3y2 ÷ x2y
3b3 9b2 5x4 10x6 a3b2 a2b

4. Simplify.

a) 2x2 × 10y4 × 9x4 b) 4a3 × 9b6 × 6a c) x2y3 × a2b3 ÷ ab
3y3 6x3 4y4 3b2 8a5 15b3 a3b2 x3y2 xy

d) 3a4b3 ÷ 6a2b2 × 8x2y e) x2y2 × abxy ÷ xyz f) a3 × c3x ÷ a2c2
5x3y4 15x2 9ab2 a3b3 x3y3 a2b2 x3y2 ab2 b2xyz

ii) Addition and subtraction of rational expressions with the same
denominators.

In this case, we should add or subtract the numerators and the common
denominator is written as the denominator of the new rational expression.
For example:

177 Vedanta Excel in Mathematics - Book 7

Laws of Indices

Example 1: Simplify a) 3x + y + 3x – y b) a2 – b2
Solution: 4a 4a a+b a+b

a) 3x + y 3x – y = 3x + y + 3x – y =2346ax 3x It’s easier, it is same as the
4a + 4a 4a = 2a addition of like fractions.

such as 2 + 1 = 2+1 !
5 5 5

b) a2 – b2 = a2 – b2 = (a + b) (a – b) = (a – b)
a+b a+b a+b (a + b)

EXERCISE 10.8
General Section - Classwork

Tell and write the answers as quickly as possible.

1. a) 1 + 1 = ........................ b) 2 + 3 = ........................
x x y y

c) 3a + 4a = ........................ d) a + b = ........................
b b x x

e) 3 + 5 = ........................ f) x + y = ........................
a–b a–b a+ b a+b

2. a) 2 – 1 = ........................ b) 5 – 3 = ........................
x x = ........................ a a
= ........................
c) 5x – 2x d) a – b = ........................
a a 2y 2y

e) x – y f) 2y + z = ........................
a+ b a+b p–q p–q

Creative Section

3. Let's simplify.

(a) 5x + 3x (b) 2a + 3a + 5a (c) 3y + 5y – 9y
2a 4a 3x 2x 6x 4b 2b 8b

(d) x – 2x + 5x (e) ax + 3ax – 7ax (f) 4by + 3by – 5by
6y 3y 9y 2b 5b 10b 7ax 14ax 28ax

4. Let's simplify.

(a) 2x + y + x + 2y (b) 4a + 3b + 2a + b (c) 2x + y + 2x – y
3 3 2 2 4a 4a

(d) x+y + x–y (e) x+y – x–y (f) x2 + y2
2xy 2xy 2xy 2xy x+y x+y

(g) a2 + 2a – b2 + 2a (h) x + y (i) x – y
a2 + b2 a2 + b2 x2 – y2 x2 – y2 x2 – y2 x2 – y2

Vedanta Excel in Mathematics - Book 7 178

Unit Geometry: Angles

13

13.1 Angels – Looking back

Classroom - Exercise

1. Let's tell and write the names, vertices and arms of these angles.

a) b)
Q Name ............................ A Name ............................
B

Vertex ........................... Vertex ...........................

O P Arms ............................. C Arms .............................

2. Let's tell and write the names and types of these angles as acute, right, obtuse,
straight, or reflex angles.
a) B b) c) Z

120° 180° 45°
OA R OP YX

‘AOB is an Obtuse angle
d) N f) O 300° G
e) A C
125° D
O T

M



In the figures given below, OX be the fixed line segment. It is also called the initial
line segment. OP be the revolving line segment that turns about a fixed point O in
anti-clockwise direction.

‘XOP = 90° is formed ‘XOP = 180° is formed ‘XOP = 270° is formed ‘XOP = 360° is formed

due a quarter turn due a quarter turn due a quarter turn due a quarter turn

Thus, when a fixed line segment is rotated about a fixed point, various angles are
formed at the fixed point.
Here, ‘XOP is the name of angle. O is called the vertex of angle. XO and OP are the
arms of the angle.

219 Vedanta Excel in Mathematics - Book 7

Geometry: Angles

13.2 Different pairs of angles – Review

There are various pairs of angles according to their structures and properties. For
example, adjacent angles, vertically opposite angles, complementary angles, and
supplementary angles. Let's review these various pairs of angles.

(i) Adjacent angles CC B
In the given figure, ‘AOB and ‘BOC have B A
common vertex O and a common arm OB.
They are called adjacent angles. Thus, a AO
pair of angles having a common vertex and O
a common arm are called adjacent angels.

(ii) Linear pair B

In the given figure, ‘AOB and ‘BOC are a pair of adjacent

angles. Their sum is a straight angle (180q), 180°

i.e. ‘AOB + ‘BOC = 180q CO A
‘AOB and ‘BOC are called a linear pair.

Thus, the adjacent angles in a straight line are known as a linear pair.

(iii) Vertically opposite angles

In the given figure, ‘AOC and ‘BOD are formed by C
intersected line segments and they lie to the opposite A

side of the common vertex. They are called vertically

opposite angles. ‘AOD and ‘BOC are another pair of D O B
vertically opposite angles.

Vertically opposite angles are always equal.

? ‘AOC = ‘BOD and ‘AOD = ‘BOC.

(iv) Complementary angles

In the given figure, the sum of ‘AOB and ‘BOC is a right angle C B
(90q), i.e., ‘AOB + ‘BOC = 90q A
90°
‘AOB and ‘BOC are called complementary angles. O

Thus, a pair of angles whose sum is 90q are called complementary
angles. The complementary angles may or may not be adjacent.

Here,complement of ‘AOB = 90q – ‘BOC.

complement of ‘BOC = 90q – ‘AOB

(v) Supplementary angles B

In the given figure, the sum of ‘AOB and ‘BOC is two 180°
right angles (180q),

i.e., ‘AOB + ‘BOC = 180q. CO A
‘AOB and ‘BOC are called supplementary angles.

Thus, a pair of angles whose sum is 180q are called supplementary angles. The

supplementary angles may or may not be adjacent.

Here,supplement of ‘AOB = 180q – ‘BOC

Supplement of ‘BOC = 180q – ‘AOB.

Vedanta Excel in Mathematics - Book 7 220

Geometry: Angles

13.3 Verification of properties of angles

Activity - 1

Let's verify experimentally that the vertically opposite angles formed due to the
intersection of two line segments are equal.

Verification
(i) Let's draw three different sets of two line segments AB and CD intersecting at O.

A D D
B
O D B
C O
AO B
fig (i)
C A fig (iii) C
fig (ii)

(ii) Let's measure each pair of vertically opposite angles: ‘AOD and ‘BOC, ‘AOC
and ‘BOD with a protractor and write the measurements in the table.

Fig. Vertically opposite angles Result

‘AOD and ‘BOC ‘AOC and ‘BOD ‘AOD = ‘BOC
‘AOC = ‘BOD
(i)

(ii)

(iii)

Conclusion: From the above table, we conclude that the vertically opposite
angles formed due to the intersection of two line segments are equal.

Activity - 2
Let's verify experimentally that the sum of adjacent angles in linear pair is 180°.
Verification
(i) Let's draw three different sets of adjacent angles ‘AOC and ‘BOC in linear pair.

C C A OB

AO B AO B C
fig (i) fig (iii)
fig (ii)

(ii) Let's measure ‘AOC and ‘BOC with a protractor and write the measurements
in the table.

221 Vedanta Excel in Mathematics - Book 7

Geometry: Angles

Fig. No. ‘AOC ‘BOC ‘AOC + ‘BOC Result
(i)
(ii) ‘AOC + ‘BOC = 180°
(iii)

Conclusion: From the above table we conclude that the sum of adjacent angles
in linear pair is 180°.

Activity - 3

Let's verify experimentally that the angle formed by a revolving line in a complete
rotation at a point is 360°.

Verification
(i) Let's draw three different sets of angles ‘AOB, ‘BOC and ‘COA formed by a

revolving line OA in a complete rotation at O.

AC

B

OA OC OB

C B A fig (iii)
fig (i) fig (ii)

(ii) Let's measure ‘AOB, ‘BOC, and ‘COA with a protractor and write the

measurements in the table.

Fig. ‘AOB ‘BOC ‘COA ‘AOB + ‘BOC + ‘COA Result
No.

(i)
(ii) ‘AOB + ‘BOC + ‘COA = 360°
(iii)

Conclusion: The angle formed by a revolving line in a complete rotation at a
point O is 360°.

Worked-out examples

Example 1: If 2xq and (x + 30)q are a pair of complementary angles, find them.

Solution:

Here,2xq + (x + 30)q = 90q [The sum of a pair of complementary angles]

or, 3xq= 90q – 30q

or, xq = 60° = 20q
3

? xq = 2 u 20q = 40q and (x + 30)q = (20q + 30q = 50q

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Geometry: Angles

Example 2: A pair of supplementary angles are in the ratio 7 : 3, find them.

Solution:

Let the required supplementary angles be 7xq and 3xq.

Now,

7xq + 3xq = 180q [The sum of a pair of supplementary angles]

or, 10xq = 180q

or, xq = 180° = 18q
10

? 7xq= 7 u 18q = 126q and 3xq = 3 u 18q = 54q

Example 3: In the adjoining figure, find the sizes of unknown angles.

Solution: 5x°
4x°
Here, 4xq + 5xq + 6xq = 180q [Being the sum a straight angle] 6x°
or, r° p°
15xq = 180q
or, xq = 11850°= 12q q°

? 4xq = 4 u 12 = 48°, 5xq = 5 u 12q = 60q and 6xq = 6 u 12q = 72q.

Again, pq = 4xq = 48q, q = 5xq = 60q and rq = 6xq = 72q [Each pair is vertically opposite
angles]

EXERCISE 13.1
General Section - Classwork
1. Let's fill in the blanks as quickly as possible.

a) If two angles are complementary, then the sum of their measures is .................

b) Two angles are supplementary if the sum of their measures is .................

c) If a° and b° are in linear pair, then a° + b° = .................

d) The complement of 40° is .................

e) The supplement of 100° is .................

f) If p° and 80° are vertically opposite angles, then p° = .................

g) If two lines intersect at a point and one pair of vertically opposite angles are
acute angles, the other pair of vertically opposite angles are .................

Creative Section - A
2. a) Define vertex and arms of an angle with an example.

b) Define acute, obtuse, and reflex angles with examples.
c) Define adjacent angles, vertically opposite angles, complementary angles

and supplementary angles with examples.
d) Define a linear pair with an example.
e) What is the relation between a right angle and a straight angle?

223 Vedanta Excel in Mathematics - Book 7

Geometry: Angles

3. Let's state with reason whether the angles a and b are adjacent or not.
a) b) c) d)

b a ba
a b ba

4. Let's find the complements of :

a) 60° b) 80° c) 15° d) 28° e) 51°

5. Let's find the supplements of :

a) 110° b) 135° c) 150° d) 70° e) 45°

6. a) If 4x° and 5x° form a linear pair, find them.

b) If (2x – 10)° and (3x + 20)° are a linear pair, find them.

c) A pair of complementary angles are in the ratio 2 : 3, find them.

d) The difference of complementary angles is 10°. Find the measures of the
angles.

e) Find the size of an angle which is five times its supplement.

f) Find the supplementary angles in which one angle is 40° more than the
other angle

g) If the supplement of an angle is four times its complement, find the angle.

7. Let's find the sizes of unknown angles.

a) b) c) d)

110° x° 75° y° x° y°
120° 40°

e) f) g) h) (x 1)°

x° (x+30)° 3y° 2y° 2x° 3x° (2x+1)°
x° (x+20)°

i) j) k) 20° l) 3a°

(x+5)° x° z° x°
x
2x°x° y° a°
(x+25)° 4x°3x° y°
30°
m) n) 88° o) p)

p° (p+10)° y° q° x°
z° 128° 3q° (x+25)°
(p+20)° y° 140° x° 2x° 3x°
z° x° b° a° p° (x+35)°

2b°

Vedanta Excel in Mathematics - Book 7 224

Creative Section - B Geometry: Angles

8. a) In the figure given alongside, if x = 1 y, y° x°
3
X
show that y = 135° Y

b) In the adjoining figure, ‘XOY = 2‘YOZ, and OZ

‘WOX = 3‘YOZ. Show that ‘WOX = 90q. W

c) In the figure alongside, if a + b + c = 180°, a x
show that x = a + b.
bc
d) In the given figure, if ‘a = ‘b, prove that ‘x = ‘y.
a
x
by

e) In the adjoining figure, ‘m = ‘n , show that‘p = ‘q. p
m
nq

9. a) An angle is 21° more than twice its complement. Find it.

b) An angle exceeds three times its complement by 10°, find it.

c) The supplement of an angle is 6° less than three times its complement. Find
the angle.

d) A pair of supplementary angles are in the ratio 4 : 5. Find the complement of
smaller angle.

It's your time - Project Work!

10. a) Draw the diagrams of four clocks. Then show the angles made by hour-hand
(shorter) and minute-hand (longer) at

(i) 3 o'clock (ii) 6 o'clock (iii) 9 o'clock (iv) 12 o'clock

Also, mention the angles formed due the quarter turn, half turn, three-quarter
turn, and complete turn. At what time are these angles formed?

b) Explore experimentally the relationship between vertically opposite angles
formed due to the intersection of two line segments.

c) Experimentally verify that the sum of adjacent angles in linear pair is 180°.

225 Vedanta Excel in Mathematics - Book 7

Geometry: Angles

13.4 Pairs of angles made by a transversal with parallel lines.

In the given figure, straight line PQ intersects two parallel lines A P B
AB and CD at M and N respectively. Here, the straight line PQ is C D
called a transversal . The transversal PQ makes various pairs of M
angles at M and N between the parallel lines AB and CD.
N
(i) Exterior and alternate exterior angles Q

In the given figure, ‘a, ‘b, ‘c, and ‘d are lying outside the parallel lines.

They are called exterior angles. ‘a and ‘c are lying to opposite side of the

transversal. They are called alternate exterior angles. P

Similarly ‘b and ‘d are another pair of alternate exterior A ab B
angles. M

Thus, alternate exterior angles are the pair of non-adjacent N D
exterior angles which lie to the opposite side of transversal. C dc

The alternate exterior angles made by transversal with Q

parallel lines are always equal.

? ‘a = ‘c and ‘b = ‘d

(ii) Interior and alternate interior angles P

In the given figure, ‘a, ‘b, ‘c, and ‘d are lying A M B
inside the parallel lines. They are called interior C ab D
angles. ‘a and ‘c are lying to the opposite side of the
transversal. They are called alternate interior (or simply dc
alternate) angles. Similarly, ‘b and ‘d are another pair of N
alternate angles.
Q

Thus, alternate angles are the pair of non-adjacent interior angles which lie to

the opposite sides of transversal.

Alternate angles always lie in the ' ' - shaped of parallel lines.

The alternate (interior) angles made by a transversal with parallel lines are

always equal.

? ‘a = ‘c and ‘b = ‘d P P

(iii) Corresponding angles A ab BA M B
‘a is an exterior and ‘c is an M cd

interior angles lying to the same cd DC N D
side of the transversal and they C N ab

are not adjacent to each other. Q Q

They are called corresponding

angles. ‘b and ‘d are another pair of corresponding angles.

Thus, a pair of non-adjacent interior and exterior angles lying on the same side

of transversal are said to be corresponding angles.

Corresponding angles always lie in the ' ' - -shaped of parallel lines.

A pair of corresponding angles made by a transversal with parallel lines are
always equal.

? ‘a = ‘c and ‘b = ‘d

Vedanta Excel in Mathematics - Book 7 226

Geometry: Angles

(iv) Co-interior angles P B
M D
In the figure given alongside, ‘a and ‘d are a pair of ab
interior angles lying to same side of the transversal. A dc
They are called co-interior (consecutive interior) angles.
Similarly, ‘b and ‘c are another pair of co-interior angles. C N
Q
Thus, a pair of interior angles lying on the same
side of transversal are said to be co-interior angles.
Co-interior angles always lie in the ' '-shaped of parallel lines.

The sum of a pair of co-interior angles made by a transversal with parallel
lines is always 180q.

Pairs of angles between parallel lines at a glance Properties
Parts of angles Diagrams

Alternate ‘a = ‘b
angles

Corresponding ‘a = ‘b
angles

Co-interior ‘a+‘b = 180°
angles

Worked-out examples

Example 1: Find the sizes of unknown angles in the following figures.

a) b)

wx 40°
100° y
z x
25°

Solution:

a) (i) w = 100q [Being vertically opposite angles]

(ii) x = w = 100q [Being corresponding angles]

(iii) y = x = 100q [Being vertically opposite angles]

y + z = 180q [Being the sum of a pair of co-interior angles]

or, 100q + z = 180q

or, z = 180q – 100q = 80q

So, w = x = y = 100q and z = 80q

227 Vedanta Excel in Mathematics - Book 7

Geometry: Angles

b) Through the point E, PQ parallel to the given parallel lines AB and CD is drawn.

(i) ‘a = 40q [Being alternate angles] A 40° B
(ii) ‘b = 25q [Being alternate angles]

‘a + ‘b + ‘x =360q [Being part of a complete turn] P ax Q
or, 40q+25q + ‘x = 360 C bE D
or, ‘x = 360q 65q 25°

? ‘x = 295q

EXERCISE 13.2
General Section - Classwork
1. From the given figure, let's tell and write the answers in the blanks spaces.

a) ‘d and ........................... are alternate angles.

b) ‘a and ............................ are corresponding angles. ab
c) ‘c and ............................ are co – interior angles. dc
d) ‘q and ‘d are ...................................... angles.
e) ‘c and ‘m are ...................................... angles. mn
f) ‘d and ‘m are ...................................... angles. qp

2. In each of the following questions, there are four options. Out of which one
option is correct. Let's tick (— ) the correct option.
a) Which of the following statement is not true?
(i) A pair of alternate angles between parallel lines are equal.
(ii) A pair of corresponding angles between parallel lines are equal.
(iii)A pair of co- interior angles between parallel lines are complementary.
(iv) A pair of co- interior angles between parallel lines are supplementary.

b) In the figure alongside, a pair of corresponding angles is ab
(i) ‘a and ‘c dc
(ii) ‘b and ‘r pq
(iii)‘d and ‘q sr
(iv) ‘p and ‘a

c) In the given figure, which of the following is true? 12
(i) ‘1 = ‘8 43
(ii) ‘4 = ‘5
(iii)‘2 = ‘7 56
(iv) ‘8 = ‘2 87

Vedanta Excel in Mathematics - Book 7 228

Geometry: Angles

d) In the adjoining figure, which of the following is not true? ab
(i) ‘m + ‘x =180° nm
(ii) ‘a + ‘y = 180° pq
(iii) ‘n + ‘p = 180° yx
(iv) ‘a + ‘q = 180°

3. Each of the pair of these angles are formed between parallel lines. Let's tell
and write the answers as quickly as possible.

a) If x° and 45° are a pair of alternate angles, then x° = ......................

b) If a° and 120° are a pair of corresponding angles, then a° = ......................

c) If p° and q° are a pair of co-interior angles, then p° + q° = ......................

d) If x° and 40° are a pair of co-interior angles, then x° = ......................

e) If m° and 150° are a pair of alternate exterior angles, then m° = ......................

Creative Section - A

4. a) What is a transversal? Write with a diagram.

b) Define alternate angles with the help of a diagram.

c) Define corresponding angles with the help of a diagram.
d) Define co-interior angles with the help of a diagram.
e) Write the properties of alternate angles, corresponding angles and co-interior

angles made by a transversal between parallel lines.

5. Let's name the pairs of alternate angles, corresponding angles and co-interior
angles in the following figures.

a) A b) P c) d) C
B
PQ Z YD
BC W
X Y XT A
Q R

6. Let's find the sizes of unknown angles.

a) b) c) p° d) c°

y° b° 60° q° b° d°
x° c° r° a°
40° a°
45° 65°
e°

229 Vedanta Excel in Mathematics - Book 7

Geometry: Angles

e) f) f° g) w° h)

110° w° 105° b° c° 85° y°
z° y° x°
e° 110° x° y° x° 130°
70° a° d° z°

i) c° j) k) l)

145° z° y° c° b° 55°
x° 115°
a° b° a° 150° a°
65° b°

m) n) 100q o) p) w°
w°
y° z° z° a° x° z° 55°
x° 80° x° y° 35°
y° x° y° 65°
25° b°

7. From the given figure, let's find the sizes of unknown angles .

a) b) c) d)

2x° (2x 25)q (3a 5)q (2x 69)q (x 96)q
(2a 5)q
y° x° (x 55)q

Creative Section - B

8. In the given figures, find the measurements of unknown angles.
a) b) c) d)

35° 150° 140° 140°
100°
x° x° y° x°
25° 20° 130°

9. a) In the figure alongside, show that w° x° a° b°
(i) ‘w = ‘c (ii) ‘x = ‘s (iii) ‘y = ‘g z° y° d° c°
(iv) ‘a = ‘r (v) ‘d = ‘q (vi) ‘p = ‘c p° q° e° f°
s° r° h° g°

b) In the adjoining figure, show that x° y°
‘a + ‘b + ‘c = 2 right angles. a°

b° c°

c) In the given figure, show that ‘a + ‘b + ‘c + ‘d = 360q. y° c° b°

d° a° x°

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Geometry: Angles

d) In the given figure, show that y = 120°. y°
x°
60°

e) In the given figure, show that ‘a + ‘b = ‘c + ‘d b° a°
d°
f) In the figure alongside, find ‘x and ‘y, and
show that BF and CE are parallel to each other. B c°

10. a) In the given figure, AB // DE, ‘BAC = 120° and F E
‘CDE = 150°, find the value of x° A 100°
[Hint: Produce ED to intersect AC at F.]
x° y°
b) From the figure given alongside, find the value of x. 45° C 55°D
[Hint: Through C, draw CZ // DY // BX]
A B
E
120°
FD

150°

CA X
B 3x°

C 90° Y

D 2x°
E

m°

c) In the figure, if m : n = 4 : 5, find the measure of ‘p. p°
n°

It's your time - Project work!

11. a) Let's cut three sets of 3 paper strips from a chart paper. From each set of 3
paper strips make the models of alternate angles ( -shaped), corresponding
angles ( -shaped), and co-interior angles ( -shaped) between two parallel
strips intersected by another strip using glue stick.

b) With the help of a ruler and set squares, let's draw three pairs of parallel lines
separately. Intersect each pair of parallel lines by a transversal. Measure all
angles made by the transversal between each pair of parallel lines with the
help of a protractor. Then, explore the relationships between the following
pairs of angles.

(i) Relationship between alternate angles

(ii) Relationship between corresponding angles

(iii) Relationship between co-interior angles

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Unit Congruency and Similarity

15

15.1 Congruent figures – Introduction
Let's observe the following Pair of figures

a) Do the line segments AB and PQ b) Do ‘MON and ‘XOY seem to be
seem to be equal in length? equal?

NY

A BP Q MO
O
X

c) Are 'ABC and 'DEF exactly d) Are these two rectangles identical

the same? to each other?

AD A DP

2.5 cm S

2.5 cm
4 cm
4 cm

1cm
1cm
B 2 cm C Q
B 3 cm C E 3 cm F 2 cm R

The above pair of figures are seemed to be the same in shape and size. They are
identical or twins. Such figures or object are said to be congruent if they have
exactly the same shape and size.

The symbol # is used to denote ' is congruent to'. Thus, from above examples
(i) AB # PQ (ii) ‘MON # ‘XOY (iii) 'ABC # 'DEF (iv) ABCD # PQRS

15.2 Congruent triangles C D D' C
Let's take a rectangular sheet of paper. Fold it DD'
A B B'
along one of its diagonals and cut the folding edge C D'

into two right angled triangular pieces. Place the BB' A D
triangular pieces one above another as shown in A

the figure. Now, answer these questions.

Are the triangular pieces exactly fitted to one above another ?

Which are the angles exactly fitted ? B' B
Which are the arms exactly fitted ?

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Congruency and Similarity

Could you investigate some ideas about congruent triangles ?

The figures which are exactly the same in shape and size are called congruent
figures.

In the adjoining figures, 'ABC and 'PQR A P

are congruent triangles because their shapes 2.5 cm
and sizes are exactly the same. In this case, 2.5 cm
each part of 'ABC can exactly be fitted over 3 cm 85° 3 cm 85°
45° 50° 45° 50°
B 4 cm C Q 4 cm R

the corresponding parts of 'PQR.

'ABC is congruent to ' PQR is written as ' ABC # ' PQR.

15.3 Conditions of congruency of triangles
Two triangles are congruent under the following conditions.

Condition 1: Side – Side – Side (S.S.S.) axiom

If three sides of one triangle are respectively equal to three corresponding sides of
another triangle, the triangles are said to be congruent.

In 'ABC and 'PQR, AP
(i) AB = PQ (S)

(ii) BC = QR (S) B CQ R
(iii)CA = RP (S)

(iv) ? 'ABC # 'PQR (S.S.S. axiom)

(v) ? ‘A = ‘P, ‘B = ‘Q and ‘C = ‘R [Corresponding angles of congruent triangles]

The angles of triangles lying to the opposite of equal sides are called
corresponding angles. For example: sides BC = QR, the angle opposite to
BC is ‘A and the angle opposite to QR is ‘P. So, ‘A and ‘P are corresponding
angles.

Condition 2: Side-Angle-Side (S.A.S.) axiom

If two sides of one triangle are respectively equal to two sides of another triangle
and the angle made by them are also equal, the triangle are said to be congruent.

In 'ABC and 'PQR, AP
(i) AB = PQ (S) B CQ R
(ii) ‘B = ‘Q (A)
(iii)BC = QR (S)
(iv) ? 'ABC # 'PQR (S.A.S. axiom)

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(v) ‘C = ‘R and ‘A = ‘P [Corresponding angles of congruent triangles]
(vi) AC = PR [Corresponding sides of congruent triangles]

The sides opposite to the equal angles of congruent triangles are called
corresponding sides. For example: ‘B = ‘Q, the side opposite to ‘B is AC
and the side opposite to ‘Q is PR. So, AC and PR are corresponding sides.

Condition 3: Angle – Side – Angle (A.S.A) axiom

If two angles of one triangle are respectively equal to two angles of another triangle
and the adjacent sides of the angles are also equal, the triangles are said to be
congruent.

In 'ABC and 'PQR, AP

(i) ‘B = ‘Q (A)

(ii) BC = QR (S) B CQ R
(iii)‘C = ‘R (A)

(iv) ? 'ABC # 'PQR (A.S.A. axiom)

(v) AC = PR and AB = PQ [Corresponding sides of congruent triangles]

(vi) ‘A = ‘P [Corresponding angles of congruent triangles]

Condition 4: Right angle-Hypotenuse-Side (R.H.S.) axiom

If hypotenuse and one of two other sides of a right angled triangle are respectively
equal to the hypotenuse and a side of other right angled triangle, the triangles are
said to be congruent.

In rt. ‘ed 'sABC and PQR, AP

(i) ‘B = ‘Q (R)

(ii) AC = PR (H) B CQ R
(iii) AB = PQ (S)

(iv) ? 'ABC # 'PQR (R.H.S. axiom)

(v) BC = QR [Corresponding sides of congruent triangles]

(vi) ‘C = ‘R and ‘A = ‘P [Corresponding angles of congruent triangles]

15.4 Similar triangles

The figures which are exactly the same in shape P
75°
but their sizes may be different are known as A 60°

similar figures. In the given figure, 'ABC and 75°
60° 45°
'PQR are similar triangles because they have equal B CQ 45° R
angles and hence the same shape.

Thus, if three angles of one triangle are respectively equal to three angles of
another triangle, they are said to be similar.

'ABC is similar to 'PQR is written as 'ABC ~ 'PQR. The symbol ~ is used to
denote ‘is similar to’.

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Congruency and Similarity

Here, AB and PQ, BC and QR, CA and RP are the corresponding sides of the similar
triangles.

The corresponding sides of similar triangles are always proportional, i.e., the ratios
of the corresponding sides are equal.

? AB = BC = CA .
PQ QR RP

Worked-out examples

Example 1: Use necessary axiom and show that 'ABC # 'PQR. Also, write their

corresponding sides and angles.

Solution: AP

In 'ABC and 'PQR,

(i) ‘B = ‘Q = 60q (A)

(ii) BC = QR = 3.5 cm (S) 60° 75° 60° 75°
(iii)‘C = ‘R = 75q (A) Q R
(iv) ?'ABC # 'PQR [A.S.A. axiom] B 3.5 cm C 3.5 cm

(v) AB = PQ and AC = PR [Corresponding sides of congruent triangles]

(vi) ‘A = ‘P [Corresponding angles of congruent triangles]

Example 2: In the figure, 'DEF ~ 'XYZ. Find the length of YZ.
Solution:

Here, 'DEF ~ 'XYZ D X
60° 60°
? DE = EF 8 cm Y 40° 80° Z
XY YZ 6 cm
8 4
or, 6 = YZ 40° 80°
4 cm
or, 8YZ = 24 E F

or, YZ = 24 = 3 cm
8

EXERCISE 15.1
General Section - Classwork

1. Which of the following pairs of figures are congruent or similar? Tell and
write "congruent or similar" in the blank space.

A B These line segments are
D .........................
a)

C

A

P These triangle are
.........................
b)
RC
Q
B

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P SW Z These squares are
RX Y .......................
c)
These circles are
Q .........................

d)

e) These figures are
.........................

f) These squares are
.........................

2. The following pairs of triangles are congruent. Let's tell and write the necessary
axioms of congruency.
a) b)

....................... axiom ....................... axiom

c) d)

....................... axiom ....................... axiom

3. From these pairs of congruent triangles, tell and write the corresponding sides
and corresponding angles as quickly as possible.

A P Corresponding angle of ‘A is .........................
Corresponding angle of ‘R is .........................
a)

Corresponding angle of ‘B is .........................

B CQ R

b) F X Corresponding side of FD is .........................
Corresponding side of YZ is .........................
D EZ Y Corresponding angle of ‘F is .........................

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4. Let's tell and write the equal ratios of corresponding sides of these pairs of
similar triangles.
X

a) A b) F

P

B CQ R D EY Z
................................................
................................................

Creative Section - A

5. a) Define the meaning of congruent figures and similar figures.

b) Are all congruent figures similar?

c) Are all similar figures congruent?

d) Draw the sketch of two congruent triangles and name them. Write the name
of corresponding sides and corresponding angles.

e) Write the necessary conditions (or axioms) to be two triangles congruent.

6. Lets use the necessary axioms and show that the following pair of triangles are

congruent. Also write their corresponding sides and angles.

a) A D b) P X

B 50°4 cm C E 50° F Q 110° 35° R Y 110° 35° Z
3.5 cm 4 cm3.5 cm 2 cm 2 cm
3.6 cm
c) K 3.6 cmRd)EO
5 cm
5 cm 3.2 cm
3.2 cm

L 4 cm M S 4 cm T F 4.5 cm G P 4.5 cm Q

7. Let's use the necessary axioms and show that the following pairs of triangles

are congruent. Also, write their corresponding sides and angles.

a) A P b) X ZD F

B CQ R Y E
R C
c) K d) Q

L MS TO PB A

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8. In the following pairs of similar triangles, find the unknown sizes of the sides.

a) A P b) F Z
30°
50° 50° 30°
8 cm
6 cm 6 cm
a cm
B 55° 75° C Q 55° 75° R 115° 115°
4 cm x cm 35°
D 6 cm E X 35° Y
3 cm G
c) d)
12 cm S x cm K E
R 50° 45° U 50° 45° V 80° y cm 80° 6 cm

85° 10 cm 85° L 60° 40° M 60° 40°
T 5 cm 15 cm 9 cm
F
W

Creative Section - B A C
9. a) In the figure alongside, use A.S.A. axiom and prove that
B D
'ABO # 'CDO. S R
P
b) In the adjoining figure, use S.A.S. axiom and prove that Q
'PQR # 'PRS. A

c) In the figure alongside, AD A BC and AB = AC,
prove that ∆ABD = ∆ ACD by using RHS axiom.

d) In the adjoining figure, ABCD is a square. B DC
DC
Prove that 'ABC # 'ADC by using
AB
(i) S.A.S. axiom (ii) R.H.S. axiom

(iii) S.S.S. axiom (iv) A.S.A. axiom separately.

It's your time - Project work!

10. a) Let's take a square sheet of paper and fold it through one of its diagonals.
(i) Does the diagonal divide a square into two congruent triangles?
Again, fold the same square paper through another diagonal and cut out
all four triangles so formed.
(ii) Do two diagonals of a square divide it into four congruent triangles?

b) Take a rectangular sheet of paper and fold it through one of its diagonals.
(i) Does the diagonals divide a rectangle into two congruent triangles?
Again fold the same rectangular paper through another diagonal and cut
out all 4 triangles so formed.
(ii) Do two diagonals of a rectangle divide it into four congruent triangles?

c) Write a short report about your finds on the activities a) and b). Then, present
in the class.

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Unit Statistics

21

21.1 Statistics – Review

The modern age is the age of information and communication. In the various field
of the society, we need information in the form of numerical figures to accomplish
various economical, commercial, technical activities. Such numerical figures are
called data.

Statistics is a branch of mathematics that deals about the collection, tabulation, and
presentation of data.

21.2 Types of data and frequency table

The numerical information given below are the amounts of pocket money (in Rs)
brought by 20 students of class VII on a day.

20, 15, 20, 25, 20, 25, 10, 30, 20, 30,
25, 20, 15, 20, 30, 25, 40, 20 25, 40

The above data are not presented in the proper order. Such data are called raw data.

Now, let's arrange these data in the ascending order of the values.
10, 15, 15, 20, 20, 20, 20, 20, 20, 20,

25, 25, 25, 25, 25, 30, 30, 30, 40, 40

The above data which are presented in the proper order (either ascending or
descending order) are called arrayed data.

From an arrayed data, it is easier to observe how many times a particular figure is
repeated

Rs 10 is repeated only one time. o Frequency of Rs 10 is 1.

Rs 15 is repeated two times. o Frequency of Rs 15 is 2.

Rs 20 is repeated seven times. o Frequency of Rs 20 is 7.

Rs 25 is repeated five times. o Frequency of Rs 25 is 5.

Rs 30 is repeated three times. o Frequency of Rs 30 is 3.

Rs 40 is repeated two times. o Frequency of Rs 40 is 2.

In this way, a frequency is the number of times a figure (or value) occurs.

We can present the data and their frequencies in a table. Such table is called
frequency distribution table (or simply frequency table).

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Frequency table Frequency

Pocket money (in Rs) Tally marks 1
2
10 | 7
15 || 5
20 || 3
25 2
30 |||
40 || 20

Total

Tallying is a system of showing frequencies using diagonal lines grouped in fives.
Each time five is reached, a horizontal line is drawn through the tally marks to make
a group of five. The next line starts a new group. For example,
1 is |, 2 is ||, 5 is , 7 is ||, and so on.

21.3 Cumulative frequency table of ungrouped data

The word 'cumulative' is related to the word 'accumulated', meaning to 'pile up'.
The table given below shows the pocket money of 20 students.

Wages (in Rs.) Frequency (f) Cumulative frequency (c.f.)
10 1 1
15 2 1+2=3
20 7 3 + 7 = 10
25 5 10 + 5 = 15
30 3 15 + 3 = 18
40 2 18 + 2 = 20
20
Total

21.4 Grouped and continuous data

Let's consider the following marks obtained by 20 students in a Mock Test of
Mathematics.

25, 32, 45, 15, 8, 42, 26, 19, 28, 45,

36, 48, 22, 29, 18, 34, 6, 38, 41, 24

The above mentioned data are called individual data. Another way of organizing

data is to present them in a grouped form. For grouping the given data, we should
first see the smallest and the largest values. Then, we have to divide the data into
appropriate class intervals.

In the above data, smallest value is 6 and the largest value is 48. So, the class
intervals can be 0 – 10, 10 – 20,…, 40 – 50. The data arranged in this way are called
grouped data.

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The frequency table of the grouped data are given below.

Marks Tally marks No of students (f)
0 – 10 || 2

10 – 20 ||| 3

20 – 30 | 6
30 – 40 |||| 4
40 – 50 5
Total 20

Now, the cumulative frequency of above data can be constructed as follows.

Marks No. of Cumulative frequency (c.f)
students (f)

0 – 10 22 o 2 students obtained less

10 – 20 3 2+3 = 5 o 5 students obtained less than 20 marks

20 – 30 6 5+6 = 11 o 11 students obtained less than 30 marks

30 – 40 4 11+4 = 15 o15 students obtained less than 40 marks

40 – 50 5 15+5 = 20 o 20 students obtained less than 50 marks

Total 20

Thus, cumulative frequency corresponding to a class – interval is the sum of all
frequencies up to and including that interval.

21.5 Bar graph

We have already discussed to present data in frequency tables. Alternatively, we can
also present data graphically. Different types of diagrams are used for this purpose.
Here, we shall discuss about a bar graph (or bar diagram).

(i) Simple bar graph

A simple bar graph is drawn to present a single type of data. In a bar graph, data
are represented in a series of bars that are equally wide. Bars can touch each
other or be separated by gaps of equal width. The height of the bars represent
the frequency of the data.

We should follow the rules given below while drawing a bar graph:

Choose a suitable scale to represent the whole data. Mention the frequencies in
y-axis.

Decide how wide the bars are and how much space you leave between them.

Construct the bars of the same width and at equal distance.

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Worked-out examples

Example 1: The table given below shows the number of students in class 7 of
Solution: a school in five years. Represent the numbers in a bar graph.

Years (B.S.) 2073 2074 2075 2076 2077

No. of students 30 35 45 60 50

2073 2074 2075 2076 2077

(ii) Multiple bar graph
A multiple bar graph is drawn to show two or more related components of the
given data. For example, if we want to show the number of boys and girls of
a school separately, we should draw the multiple bar graph. The method of
drawing a multiple bar graph is exactly the same as a simple bar graph.

Example 2: The number of boys and girls in the primary level of a school are
given in the table below. Draw a multiple bar graph to show their
numbers.

Solution: Classes 1 2345
No. of boys 12 15 16 25 25
No. of girls 18 20 24 20 10

No. of boys and girls 35 20 24 25 25
30 15 16 20 10
5
25 23 4
Classes
20 18
15 12
10

5

0
1

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1. a) EXERCISE 21.1
b)
c) Define raw and arrayed data with examples.
d) Define frequency with examples.
Define cumulative frequency with examples.
2. a) Write a difference between a simple bar graph and a multiple bar graph.

The amount of pocket money (in Rs) brought by 20 students of class VII on a
day are given below. Let's construct a frequency distribution table with tally
marks.
30, 40, 25, 35, 40, 50, 40, 30, 25, 20,
40, 45, 35, 40, 30, 35, 40, 45, 30, 35

b) Daily wages (in Rs) of 30 workers in a factory are given below. Arrange the
data in proper order. Lets construct a frequency table to present the data.

850, 900, 600, 750, 800, 750, 600, 900, 800, 750,

700, 800, 750, 600, 750, 800, 900, 850, 750, 700,

800, 750, 700, 800, 850, 750, 800, 750, 700, 750.
c) In an interview of 20 married couples about their desired number of children,

the responses were as follows:

1, 2, 3, 2, 3, 1, 4, 2, 2, 1

4, 3, 1, 2, 2, 2, 1, 3, 1, 2

Let's present the above data in a frequency table using tally marks and answer
the following questions.

(i) How many couples desired a single child?

(ii) How many couples desired two children?

(iii) How many couples desired three children?

(iv) How many couples desired more than three children?

(v) What is the desirable number of children for the maximum number of
couples?

3. Let's construct the cumulative frequency table to represent the following
ungrouped data,

a) Age (in years) 6 8 10 12 14 16

No. of students 5 3 7 2 6 2

b) Weight (in Kg) 20 30 40 50 60 70
No. of persons 9 6 15 10 6 4

4. Let's construct the cumulative frequency table to represent the following
grouped data.

a) Marks obtained 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of Students 2 4 7 10 9 4

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b) Age (in years) 0 – 15 15 – 30 30 – 45 45 – 60 60 – 75 75 – 90
No. of patients
20 10 6 9 15 4

c) Daily sales 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
(in Rs. 1000) 35 18 25 12 10

No. of Shops

5. a) The number of students in the primary level (class 1 to 5) are given below.
Draw a simple bar graph to show the number.

Classes 12345
No. of students 45 30 50 25 15

b) The table given below shows the number of students who secured 'A' grade in
S.E.E. from your school in the last five years. Let's draw a simple bar diagram to
show the data.

Years (B.S.) 2073 2074 2075 2076 2077

No. of students 25 40 30 45 40

6. a) The table given below shows the number of boys and girls of a school from
class 6 to 10. Illustrate the numbers by drawing a multiple bar graph.

Classes 6 7 8 9 10

No. of boys 10 20 15 25 20

No. of girls 15 10 20 15 10

b) Class 7 students conducted a survey and recorded the following number
of male and female who had positive results of rapid diagnostic test for
conronavirus in the last five days. Draw a multiple bar graph to show the
numbers.

Days Sunday Monday Tuesday Wednesday Thursday

No. of male patients 7 10 12 10 7

No. of female patients 3 5 4 3 5

c) The table given below shows the S.E.E. result of a school for the last 4 years
Let's draw a multiple bar graph to show the numbers.

Year (in B.S.) 2074 2075 2076 2077
'A+' grade 5 10 15 12
'A' grade 10 15 20 27
'B+' grade 15 5 15 11

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7. The multiple bar graph given below shows the number of boys and girls who
secured 'A' grade in the final examination from different classes. Answer the
following questions.

No. of boys and girls boys
girls

6 78 9

Classes

a) How many students are represented by a room in vertical column?

b) How many boys and girls secured 'A' grade in the final exam from class 6?

c) If all the students of class 7 have grade 'A' in the exam, how many students were

there in class 7?

d) If 80 students appeared in the exam from class 9, find the percentage of students

who secured 'A' grade.

e) If 30 boys appeared in the exam from class 8, find the percentage of boys who

could not secure 'A' grade.

f) If 20 students could not secure 'A' grade from class 6, find the percentage of

students who secured 'A' grade.

21.6 Average (or Mean)
Suppose Ram has Rs 10 and Hari has Rs 8.

Then, we often say that in average Ram and Hari have Rs 9.

Let's think how we get an average of Rs 9 from Rs 10 and Rs 8.

Now, let's answer what the averages of (i) 2 and 4 (ii) 5 and 9 (iii) 2, 3 and 4 are.

Thus, an average is a single number that represents the central value of a set of data.
Of course, an average indicates the quality of the given data.

An average of the given data is calculated by adding them together and dividing the
sum by the number of data.

i.e., Average = Total sum of the data
Number of data

If x is used to represent the data, n is used to represent the number of data and x is
used to represent the average or mean of the data, then

Average (x) = 6nx
Here, the symbol’ 6' means summation of whole data.

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