Vedanta Excel in Mathematcs Book -10 Final (2078)

H.C.F. and L.C.M.

Example 2: Find the H.C.F. of x5y + 4x3y + 16xy, x4y – 8xy and 2x3y + 4x2y + 8xy.
Solution:
Here, the 1st expression = x5y + 4x3y + 16xy

= xy (x4 + 4x2 + 16)
= xy [(x2)2 + 42 + 4x2]
= xy [(x2 + 4)2 – 2.x2.4 + 4x2]
= xy [(x2 + 4)2 – (2x)2]
= xy (x2 + 2x + 4) (x2 – 2x + 4)

The 2nd expression = x4y – 8xy = xy (x3 – 8)
= xy (x3 – 23)
= xy (x – 2) (x2 + 2x + 4)

The 3rd expression = 2x3y + 4x2y + 8xy
= 2xy (x2 + 2x + 4)

? The H.C.F. = xy (x2 + 2x + 4)

Example 3: Find the H.C.F. of x2 – 12x – 28 + 16y – y2, x2 + 2x – y2 + 2y and
x2 – y2 + 4y – 4.

Solution:
Here, the 1st expression = x2 – 12x – 28 + 16y – y2

= x2 – 2.x.6 + 62 – 62 – 28 + 16y – y2
= (x – 6)2 – (64 – 16y + y2)
= (x – 6)2 – (82 – 2.8.y + y2)
= (x – 6)2 – (8 – y)2
= (x – 6 + 8 – y) (x – 6 – 8 + y)
= (x – y + 2) (x + y – 14)

The 2nd expression = x2 + 2x – y2 + 2y
= x2 – y2 + 2(x + y)
= (x + y) (x – y) + 2(x + y)
= (x + y) (x – y + 2)

The 3rd expression = x2 – y2 + 4y – 4
= x2 – (y2 – 2. y. 2 + 22)
= x2 – (y – 2)2
= (x + y – 2) (x – y + 2)

? The H.C.F. = (x – y + 2)

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H.C.F. and L.C.M.

EXERCISE 8.1

General section

1. Find the H.C.F. of the following expressions:

a) 2x2 (x + 2) (x – 2) and 4x(x + 2) (x + 3)

b) 4xy2 (x – 1) (x + 2) and 6x2y (x – 1) (x – 4)

c) 6a2b2(a – b) (2a + 3b) and 9a3b3(2a + 3b) (a + b)

d) 12a2b3 (a – 3b) (a + b – 2) and 16a3b2 (a + 3b) (a + b – 2)

e) (p + q) (p – q), (p – q) (p2 + pq + q2) and (p – q) (p + q) (p2 + q2)

2. Find the H.C.F. of the following expressions:

a) a2 – b2 and a3 + b3 b) a2 – 2a and a4 – 8a

c) x2 – 9 and 3x + 9 d) x2 – y2 and x2 + 2xy + y2

e) 4x2 – 100 and 4x + 20 f) 4x3 + 8x2 and 5x3 – 20x

g) a2 – b2 – 2a + 1 and a2 – ab – a h) x3 – 8y3 and x2 + 2xy + 4y2

i) p4 – 16 and p2 – p – 6 j) x4 + 4y4 and 2x3y + 4xy3 + 4x2y2

k) a3 – 1 and a4 + a2 + 1 l) 16x4 – 4x2 – 4x – 1, 8x3 – 1

Creative Section

3. Find the H.C.F. of the following expressions:

a) a2 – 4, a3 + 8, a2 + 5a + 6
b) x2 – 9, x3 – 27, x2 + x – 12
c) 4x2 – 9, 2x2 + x – 3, 8x3 + 27
d) 3x2 – 8x + 4, 2x2 – 5x + 2, x4 – 8x
e) 5a3 – 20a, a3 – 3a2 – 10a, a3 – a2 – 2a + 8
f) m3 – m2 – m + 1, 2m4 – 2m, 3m2 + 3m – 6
g) (a + b)2 – 4ab, a3 – b3, a2 + ab – 2b2
h) x3 – 64y3, x2 – 6xy + 8y2, x2 – 16y2
i) 4x4 + 16x3 – 20x2, 3x3 + 14x2 – 5x, x4 + 125x
j) x3 – 1, x4 + x2 + 1, x3 + 1 + 2x2 + 2x
k) 8x3 + 27y3, 16x4 + 36x2y2 + 81y4, 4x3 – 6x2y + 9xy2
l) x3y + y4, x4 + x2y2 + y4, 2ax3 – 2ax2y + 2axy2
m) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2, c2 + 2ca + a2 – b2
n) 9x2 – 4y2 – 8yz – 4z2, 4z2 – 4y2 – 9x2 – 12xy, 9x2 + 12xz + 4z2 – 4y2
o) x2 – 18x – 19 + 20y – y2, x2 + x – y2 + y, x2 – y2 + 2y – 1

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H.C.F. and L.C.M.

8.2 L.C.M. of algebraic expressions

Let's take any two monomial expressions x2 and x3.
A few multiplies of x2 are x2, x3, x4, x5, x6,.....
A few multiplies of x3 are x3, x4, x5, x6, x7, ....
Here, the common multiplies are x3, x4, x5, x6, ...
Among these common multiplies, x3 is the lowest one.
So, the Lowest Common Multiple (L.C.M.) of x2 and x3 is x3.
Thus, to find the L.C.M. of the monomial expressions, at first we should find the
L.C.M. of the numerical coefficients. Then, the common variable with the highest
power is taken as the L.C.M. of the expressions. For Example,

24 is the L.C.M. of 6 and 8.

In 6x4y2 and 8x3y3 x4 is the L.C.M. of x4 and x3.

y3 is the L.C.M. of y2 and y3.

So, the L.C.M. of 6x4y2 and 8x3y3 is 24x4y3.
In case of polynomial expressions, their L.C.M. is obtained by the process of
factorisation. By this process the product of common factors and the factors which or
not common is taken as the L.C.M. of the polynomials

Worked-out examples

Example 1: Find the L.C.M. of 6x6 + 6x4 + 6x2 and 4x6 – 4x3.

Solution:

Here, the 1st expression = 6x6 + 6x4 + 6x2

= 6x2 (x4 + x2 + 1)

= 6x2 [(x2)2 + 12 + x2]

= 6x2 [(x2 + 1)2 – 2x2 + x2]

= 6x2 [(x2 + 1)2 – x2]

= 6x2 (x2 + x + 1) (x2 – x + 1)

The 2nd expression = 4x6 – 4x3

= 4x3 (x3 – 1)

= 4x3 (x – 1) (x2 + x + 1)

? The L.C.M.= 12x3 (x2 + x + 1) (x – 1) (x2 – x + 1)

= 12x3 (x3 – 1) (x2 – x + 1)

Example 2: Find the L.C.M. of x2 + 2xy + y2 – z2, y2 + 2yz + z2 – x2 and z2 + 2xz + x2 – y2.

Solution:

Here, the 1st expression = x2 + 2xy + y2 – z2

= (x + y)2 – z2

= (x + y + z) (x + y – z)

The 2nd expression = y2 + 2yz + z2 – x2

= (y + z)2 – x2

= (x + y + z) (y + z – x)

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H.C.F. and L.C.M.

The 3rd expression = z2 + 2zx + x2 – y2
= (z + x)2 – y2 = (x + y + z) (x – y + z)

? The L.C.M.=(x + y + z) (x + y – z) (y + z – x) (x – y + z)

Example 3: Find the L.C.M. of a3 – 2a2b + 2ab2 – b3, a3 + b3, a4 – b4.

Solution:
Here, the 1st expression = a3 – 2a2b + 2ab2 – b3

= a3 – b3 – 2ab (a – b)
= (a – b) (a2 + ab + b2) – 2ab (a – b)

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

The 2nd expression = a3 + b3 = (a + b) (a2 – ab + b2)

The 3rd expression = a4 – b4

= (a2)2 – (b2)2

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

? The L.C.M. = (a – b) (a + b) (a2 + b2) (a2 – ab + b2)

= (a2 – b2) (a2 + b2) (a2 – ab + b2) = (a4 – b4) (a2 – ab + b2)

EXERCISE 8.2

General section
1. Find the L.C.M. of the following expressions:

a) 3x (x +1) (x – 1) and 2x2 (x – 1) (x + 3)
b) 4x3 (x – 3) (x + 2) and 6x2 (x + 2) (x + 3)
c) 8a2b(a – b) (a2 + ab + b2) and 12ab2 (a – b) (a + b)
d) (x + 2) (x + 3), (x + 3) (x – 2), (x – 2)(x – 3)
e) (a – 3) (a – 4), (a – 4),(a – 5), (a – 5) (a – 3)

2. Find the L.C.M. of the following expressions:

a) 3x2 + 6x, 2x3 + 4x2 b) ax2 + ax, ax2 – a c) 4x5y4 + 2x4y5, 10x4y3 + 5x3y4
f) a3 – b3, a2 + ab + b2
d) x2 – xy, x3y – xy3 e) 4x2 – 2x, 8x3 – 2x i) a4b – ab4, a4b2 – a2b4
l) 6x2 – x – 1, 54x4 + 2x
g) x2 + 5x + 6, x2 – 4 h) x2 – 9, 3x3 + 81

j) x4 + x2y2 + y4, x3 – y3 k) a4 + a2b2 + b4, a3 + b3

Creative Section

3. Find the L.C.M. of the following expressions:

a) a2 – 4, a3 – 8, (a + 2)2 b) (a – 3)2, a2 – 9, a3 + 27

c) a3 – 2a2 + a, a3 + a2 – 2a, a3 – 4a d) x4 – y4, x2 – y2, x3 – y3

e) 4x3 – 10x2 + 4x, 3x4 – 8x3 + 4x2, x4 – 8x f) x3 – 9x, x4 – 2x3 – 3x2, x3 – 27

g) a3 – 4a, a4 – a3 – 2a2, a3 – 8 h) a4 + a2 + 1, a3 – 1, a3 – a2 + a

i) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2, c2 + 2ca + a2 – b2

j) x3 – 2x2y + 2xy2 – y3, x4 – y4, x3 + y3

k) x2 + 3x + 2, x2 + 5x + 6, x2 + 4x + 3

l) 2x3 + 2x2 – 12x, 6x3 – 6x2 – 72x, 4x3 – 24x2 + 32x

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Unit Simplification of Rational Expressions

9

9.1 Rational Expressions - review

x , x2 , 4a2b , a+b , etc. are a few examples of the rational expressions. However,
2 3 5 a–b

a+b is not a rational expression. Rational expressions can be expressed in the
a–b

form p , where q z 0.
q

9.2 Simplification of rational expressions

In case of addition and subtraction of rational expressions with a common
denominator, we should simply add or subtract the numerators. Then, the sum is
reduced to its lowest terms.

For example,

x – y = x– y = x–y = 1
x2 – y2 x2 – y2 x2 – y2 (x + y)(x – y) x+y

In case of addition and subtraction of rational expressions with unlike denominators,
at first, we should find the L.C.M. of the denominators. Then, the L.C.M. is divided
by each denominator and the quotient is multiplied by the corresponding numerator
as like the simplification of unlike fractions in arithmetic.

Let's learn the process of simplification from the following illustrations.

Worked-out examples

Example 1: Simplify 1 – 1
(x – y)(x – z) (z – x) (y – z)

Solution:

1 – 1 = 1 – 1
(x – y)(x – z) (z – x)(y – z) (x – y)(x – z) – (x – z)(y – z)

= 1 + 1
(x – y)(x – z) (x – z)(y – z)

= y–z+x–y
(x – y)(x – z) (y – z)

= x–z
(z – y)(x – z) (y – z)

= 1 Answer
(x – y) (y – z)

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Simplification of Rational Expressions

Example 2: Simplify (a – 2 – 3) + 2 + 1
2) (a (a – 1) (3 – a) (1 – a) (2 – a)

Solution:

2 + 2 + 1
(a – 2) (a – 3) (a – 1) (3 – a) (1 – a) (2 – a)

= (a 2 – 3) – 2 + (a – 1 2)
– 2) (a (a – 1) (a – 3) 1) (a –

= 2(a – 1) – 2(a – 2) + a – 3
(a – 2) (a – 3) (a – 1)

= 2a – 2 – 2a + 4 + a – 3 = (a – 2) a – 1 (a – 1) = 1 Answer
(a – 2) (a – 3) (a – 1) (a – 3) (a – 2) (a – 3)

Example 3: Simplify 1 +1 + 1
x2 – 5x + 6 x2 – 3x + 2 x2 – 8x + 15

Solution:

1 + 1 + 2
x2 – 5x + 6 x2 – 3x + 2 x2 – 8x + 15

= 1 + 1 + 2
x2 – 3x – 2x + 6 x2 – 2x – x + 2 x2 – 5x – 3x + 15

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

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

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

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

= 2(x – 5) + 2(x – 1) = 2x – 10 + 2x – 2
(x – 3) (x– 1) (x – 5) (x – 3) (x– 1) (x – 5)

= 4(x – 3) = 4 Answer
(x – 3) (x– 1) (x – 5) (x– 1) (x – 5)

Example 4: Simplify x + y + 2xy
Solution: (x – y) (x + y) y2 – x2

x + y + 2xy
(x – y) (x + y) y2 – x2
x y 2xy
= (x – y) + (x + y) – x2 – y2

= x(x + y) + y(x – y) – 2xy
(x – y) (x + y) x2 – y2

= x2+xy + xy – y2 – 2xy = x2+2xy – y2 – 2xy = x2 – y2 = 1 Answer
x2 – y2 x2 – y2 x2 – y2 x2 – y2

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Simplification of Rational Expressions

Example 5 : Simplify 1 – 1 – 2p
1 – p + p2 1 + p + p2 1 – p2 + p4

Solution:

1 – 1 p2 – 1 + 1 p2 – 2p = 1 + p2 + p – (1 + p2 – p) – 2p
p+ p+ 1 – p2 + p4 (1 + p2 – p) (1 + p2 + p) 1 + p4 – p2

= 1 + p2 + p – 1 – p2 + p – 2p
(1 + p2)2 – p2 1 + p4 – p2

= 2p – 2p
1 + 2p2 + p4 – p2 1 + p4 – p2

= 1 + 2p p2 – 2p
p4 + 1 + p4 – p2

= 2p(1 + p4 – p2) – 2p(1 + p4 + p2)
(1 + p4 + p2) (1 + p4 – p2)

= 2p + 2p5 – 2p3 – 2p – 2p5 – 2p3
(1 + p4)2 – (p2)2

= –4p3 = –4p3 Answer
1 + 2p4 + p8 – p4 1 + p4 + p8

Example 6 : Simplify y–2 + y2 y +2 4 – 16
y2 – 2y + 4 + 2y + y4 + 4y2 + 16

Solution:

y–2 + y2 y +2 4 – y4 + 16 + 16
y2 – 2y + 4 + 2y + 4y2

= (y – 2) (y2 + 2y + 4) + (y + 2) (y2 – 2y + 4) – 16
(y2 – 2y + 4) (y2 + 2y + 4) y4 + 4y2 + 16

= y3 – 23 + y3 + 23 – 16
[(y2 + 4) – 2y] [(y2 + 4) + 2y] y4 + 4y2 + 16

= (y2 + 2y3 (2y)2 – 16
4)2 – y4 + 4y2 + 16

= 2y3 – 16
y4 + 8y2 + 16 – 4y2 y4 + 4y2 + 16

= 2y3 – 16
y4 + 4y2 + 16 y4 + 4y2 + 16

2y3 – 16 2(y3 – 8)
= y4 + 4y2 +16 = (y2 – 2y +4) (y2 + 2y + 4)

= 2(y – 2) (y2 + 2y + 4) = 2(y – 2) Answer
(y2 – 2y +4) (y2 + 2y + 4) y2 – 2y +4

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Simplification of Rational Expressions

Example 7 : Simplify x2 – (y – z)2 + y2 – (x – z)2 + z2 – (x – y)2
(x + z)2 – y2 (x + y)2 – z2 (y + z)2 – x2

Solution:

x2 – (y – z)2 + y2 – (x – z)2 + z2 – (x – y)2
(x + z)2 – y2 (x + y)2 – z2 (y + z)2 – x2

= (x + y – z) (x – y + z) + (y + x – z) (y – x + z) + (z + x – y) (z – x + y)
(x + y + z) (x – y + z) (x + y + z) (x + y – z) (x + y + z) (y + z – x)

= x+y–z + y–x+z + z+x–y
x+y+z x+y+z x+y+z

= x+y–z+y–x+z+z+x–y
x+y+z

= x+y+z = 1 Answer
x+y+z

Example 8 : Simplify a4 + a4 – 1 1 – 1
a2 + 1 a2 – 1 a2 + a2 – 1

Solution:

a4 + a4 – 1 – 1 = a4 – 1 + a4 – 1
a2 + 1 a2 – 1 a2 + 1 a2 – 1 a2 + 1 a2 + 1 a2 – 1 a2 – 1

= a4 – 1 + a4 – 1
a2 + 1 a2 – 1

= (a2 + 1) (a2 – 1) + (a2 + 1) (a2 – 1)
a2 + 1 a2 – 1

= a2 – 1 + a2 + 1 = 2a2 Answer

Example 9 : Simplify 2 + 1 1 + 3a + 1 a
1+a a– 1 – a2 + a3

Solution:

2 + 1 + 3a + a = 2 – 1 + 3a + 1 a
1+a a–1 1 – a2 1 + a3 1+a 1–a 1 – a2 + a3

= 2 – 2a – 1 – a + 3a + (1 + a) a a + a2)
(1 + a) (1 – a) (1 –

= 1 + a
(1 + a) (1 – a) (1 + a) (1 – a + a2)

= 1 – a + a2 + a – a2
(1 + a) (1 – a) (1 – a + a2)

= (1 – 1 a3) Answer
a) (1+

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Simplification of Rational Expressions

Example 10: Simplify 1 – 2 1 + 1 – 2
x–1 2x – x+1 2x + 1

Solution:

1 – 2 1 + 1 – 1 = 1 + 1 – 2 – 2 1
x–1 2x – x+1 x+1 x–1 x+1 2x – 1 2x +

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

= 2x – 4x + 2 + 4x – 2
x2 – 1 4x2 – 1

2x – 8x
= x2 – 1 4x2 – 1

= 2x(4x2 – 1) – 8x(x2 – 1)
(x2 – 1) (4x2 – 1)

= 8x3 – 2x – 8x3 + 8x = 6x Answer
(x2 – 1) (4x2 – 1) (x2 – 1) (4x2 – 1)

Example 11: Simplify x–1 – x+1 – 2x + 16x
x–2 x+2 x2 + 4 x4 + 16

Solution:

x–1 – x+1 – 2x 4 + 16x
x–2 x+2 x2 + x4 + 16

= (x – 1) (x + 2) – (x + 1) (x – 2) – 2x + 16x
(x – 2) (x + 2) x2 + 4 x4 + 16

= x2 + x – 2 – x2 + x + 2 – 2x + 16x
x2 – 4 x2 + 4 x4 + 16

= 2x – 2x + 16x
x4 – 4 x4 + 4 x4 + 16

= 2x(x2 + 4) – 2x(x2 – 4) + 16x
(x2 – 4) (x2 + 4) x4 + 16

= 2x3 + 8x – 2x3 + 8x + 16x
x4 – 16 x4 + 16

= 16x 16x
x4 – 16 + x4 + 16

16x(x4 + 16) + 16x(x4 – 16)
= (x4 – 16) (x4 + 16)

= 16x5 + 256x + 16x5 – 256x = 32x5 Answer
x8 – 256 x8 – 256

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Simplification of Rational Expressions

EXERCISE 9.1

General section

1. Simplify:

a) a+ 1 + a2 – 1 b) p+ 2 – p– 2
a–1 a+1 p–2 p+2

c) 4x – x+ 1 d) x– 4 – x– 6
x2 – 1 x–1 x+6 x+4

e) a2 + ab + b2 + a2 – ab + b2 f) x–2 – x+1
a+ b a–b x2 – 1 x2 – 2x + 1

g) a+2 2 + 3 h) x + y
a2 + a – a2 – 1 x2 + 2xy + y2 x2 – y2

i) 1 + 1 j) 1 + 1
(x – y) (x – z) (x – z) (y – z) (a – b) (b – c) (c – b) (a – c)

Creative section - A

2. Simplify:

a) (x – 1 – 3) + (x – 1 – 1) – (x 2 1)
2) (x 2) (x – 3) (x –

b) (x 1 + (x + 3 – x) + (x 2 4)
– 3) (x + 2) 2) (4 – 3) (x –

c) 2 (a – 3) + a–1 + a–2
(a – 4) (a – 5) (3 – a) (a – 4) (5 – a) (a – 3)

d) (2x – x–1 + 2) + (x + 3 – 1) – 1
1) (x 2) (x (1 – x) (1 – 2x)

3. Simplify:

a) x2 – 1 + 6 – x2 – 2 + 3 – 1
5x 4x x2 –3x + 2

b) 1 – 2 – 1
a2 – 5a + 6 a2 – 4a + 3 a2 – 3a + 2

c) x2 x–1 2 + x2 x–2 6 + x–5
– 3x + – 5x + x2 – 8x + 15

d) a2 2a – 6 20 – a2 a– 1 12 – a2 a– 2 15
– 9a + – 7a + – 8a +

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Simplification of Rational Expressions

4. Simplify:

a) x + x – 4x b) 2 – 2 b + 4a
x+2 x–2 x2 – 4 a+b a– a2 – b2

c) x+y – x–y + 4xy d) a+b – a–b – 2ab
x–y x+y x2 + y2 a–b a+b a2 – b2

e) x+1 + x–1 – 2x2 f) x+y + x–y – 2 (x2 – y2)
x–1 x+1 x2 – 1 x–y x+y x2 + y2

5. Simplify:

a) 1 + 1 + x2 – 1 – 1 x2 – 2x
x x+ 1 + x2 + x4

b) x+3 + x2 x–3 9 – 54
x2 + 3x + 9 – 3x + x4 + 9x2 + 81

c) x2 x– y y2 + x2 x+y y2 – 2y3
– xy + + xy + x4 + x2y2 + y4

d) a+2 – a–2 – 2a2
1 + a + a2 1 – a + a2 1+ a2 + a4

e) 4a2 2a + b b2 + 2a – b – 2b3
+ 2ab + 4a2 – 2ab + b2 16a4 + 4a2b2 + b4

3x – 1 3x + 1 54x3
f) 9x2 – 3x + 1 – 9x2 + 3x + 1 + 81x4 + 9x2 + 1

6. Simplify:

a) a2 – (b – c)2 + b2 – (a – c)2 + c2 – (a – b)2
(a + c)2 – b2 (a + b)2 – c2 (b + c)2 – a2

b) (a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2

7. Simplify:

a) 1 x) – (1 1 x) + (1 x
(1 – + – x)

b) 2 y+ 3 – 5 x– y
x+ x– x–y
y

c) 1 –1 a) + 8 2 a
8 (1 – a ) 8 (1 + (1 – a)

d) 1 – 1) + 8 ( 1 + 2x
8( x x + 1) 8 (x – 1)

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Simplification of Rational Expressions

Creative section - B

8. Simplify:

a) 1 – 1 + 1 – 1 b) 1 – 1 + 1 – 1
x–5 x–3 x+5 x+3 a–3 a–1 a+3 a+1

c) 1 – 2 + 1 – 2 d) a3 + a3 – 1 + a 1 1
x–a 2x + a x+a 2x – a a–1 a+1 a–1 +

e) x3 – x2 – x x 1 + x 1 1 f) 1 – 3 + 3 – 1
x–1 x+1 – + a–1 a a+1 a+2

9. Simplify:

a) x 3 1 – x – x 2 1 – 5 1
– x2 + 1 + x2 –

b) 2 – a – 1 – 3
a–1 a2 + 1 a+1 a2 – 1

c) 4 – 2x – 2 + 6
x–1 x2 + 1 x+1 1 – x2

d) x4 + x4 – 1 – 1
x2 + 1 x2 – 1 x2 + 1 x2 – 1

10. Simplify:

a) 1 + 1 1 a + 2 + 4 b) 1 – 1 – 2 – 4
1–a + 1 + a2 1 + a4 a–1 a+1 a2 + 1 a4 + 1

c) 1 + 1 – 2x – 4x d) b + a b b + 2ab + 4a3b
x+1 x–1 x2 + 1 x4 + 1 a–b + a2 + b2 a4 + b4

e) 1 + 1 + 2x + 4x3 f) b – b + 6b2 – 8b4
x–y x+y x2 + y2 x4 + y4 a+b a–b a2 – b2 a4 – b4

g) a + a + 2a + 16a3 h) 1–x – 1+x – 4x – 8x
1 – 2a 1 + 2a 1 + 4a2 16a4 – 1 1+x 1–x 1 + x2 1 + x4

i) 1+x – 1–x + 4x + 8x3 j) 1+a + 1–a – 1 + a2 – 1 – a2
1–x 1+x 1 + x2 1 – x4 1–a 1+a 1 – a2 1 + a2

k) x+y – x–y + 4xy – 8x3y l) 1 – a + a2 + a4
x–y x+y x2 + y2 x4 + y4 a+1 a2 – 1 a4 – 1 a8 – 1

m) 1 + 1 + 2x 1 – x + 4x3 n) 1 + a b b + 2ab – a + 4a3b
x–1 x2 + x+1 x4 + 1 – a2 + b2 a+b a4 + b4

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Unit Indices

10

10.1 Indices - review

Let's multiply an algebraic term x four times by itself. It is written as x × x × x × x and
the product is shown in the abbreviated form which is x4. In this case, x is called the
base and 4 is called the index or exponent.

Thus, the index refers to the power to which a number is raised. In the example x4,
the base x is raised to the power 4. Indices is the plural form of index.

10.2 Laws of indices

There are certain verified rules which are used in the operations of indices of
algebraic terms. Product rule, quotient rule, power rule, etc. are the examples of
such rules. The verifications of these rules have been already discussed in our earlier
grades. The table given below shows the summary of these rules.

Laws of indices at a glance

(i) Product law am u an = am + n, where 'm' and 'n' are positive
integers

(ii) Quotient law am ÷ an = am – n when m > n
(iii) Power law of indices am ÷ an = 1 when m < n

an – m

(am)n = am u n, (ab)m = ambm, am = am
b bm

(iv) Law of negative index a– m = 1 or am = 1
am a–m

(v) Law of zero index aq = 1, bq = 1, xq = 1 and so on
(vi) Root law of indices
n am m

= an

Worked-out examples

Example 1: Evaluate 8 – 1 81 – 1 32 – 1
3 4 5
27 × 16 ÷
243

Solution:

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Indices

11 1
27 3 16 4 243 5
= 8 × 81 ÷ 32

= 3 3 × 1 × 2 4 × 1 ÷ 3 5 × 1
2 3 3 4 5

2

= 3 × 2 × 2 = 2 Answer
2 3 3 3

Example 2: Prove that 11x+2 + 11x = 1
122 × 11x
Solution:

L.H.S. = 11x+2 + 11x
122 × 11x
11x × 112 + 11x
= 122 × 11x

= 11x (121 + 1)
122 × 11x

= 11x × 122 = 1 R.H.S. Proved
122 × 11x

Example 3: Simplify 7a+ 1 + 9 × 7a
7a + 2 – 45 × 7a
Solution:

Here, 7a+ 1 + 9 × 7a = 7a × 7 + 9 × 7a
7a + 2 – 45 × 7a 7a × 72 – 45 × 7a

= 7a (7 + 9) = 16 = 4 Answer
7a (49 – 45) 4

Example 4: Simplify a a2 + ab + b2 b b2 + bc + c2 xc c2 + ca + a2
Solution: xa
xx ×b xx ×c

a a2 + ab + b2 b b2 + bc + c2 xc c2 + ca + a2
xa
xx ×b xx ×c

= x × x × x(a – b) (a2 + ab + b2) (c – a) (c2 + ca + a2)
(b – c) (b2 + bc + c2)

= xa3 – b3 × xb3 – c3 × xc3 – a3

= xa3 – b3 + b3 – c3 + c3 – a3 = x° = 1 Answer

Example 5: Simplify x x+y–z y y+z–x az z + x – y
Solution: ax
aa ×y aa ×z

x x+y–z y y+z–x az z + x – y y2 + yz – xy z2 + zx – yz
ax yz + z2 – zx zx + x2 – xy
aa ×y aa ×z x2 + xy – zx

= aa ×× aa ×× aaxy + y2 – yz

x2 + y2 + z2
x2 + y2 +z2
x2 + xy – zx + y2 + yz – xy + z2 + zx – yz

= aa = aa = 1 Answerxy + y2 – yz + yz + z2 – zx + zx + x2 – xy

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Indices

bc a

Example 6: Simplify xc × xa × xb
Solution: bc c ca a ab b

xb xc xa

b c a b c a

bc xc × ca xa × ab xb = xc ÷ bc × xa ÷ ca × xb ÷ ab
÷ bc ÷ ca ÷ ab
c a b c a b

xb xc xa xb xc xa

b × 1 × c × 1 a × 1 11 1
bc × ca ab x c2 × xa2 × x b2
= xc xa ×xb = =1 Answer
× 1 × 1 × 1 1 1 1
c bc a ca b ab x b2 × x c2 × xa2

xb xc ×xa

11 1

Example 7: Simplify m + (mn2)3 + (m2n)3 × 1 – n3
Solution: 1
m–n
m3

11 1 12 21 11

m + (mn2)3 + (m2n)3 × 1– n3 = m + m3 n3 + m3 n3 m3 – n3
1 ×1
m–n m–n m3
m3
12 2 11
11

m3 m3 + n3 + m3n3 m3 – n3
= m–n × 1
m3

11 12 11 12

m3 – n3 m3 + m3n3 + n3
=
m–n

13 13
m3 – n3
= m–n = m – n = 1 Answer
m – n

x+ 1 a 1 –x b
y y
Example 8: Simplify ×
Solution:
y+ 1 a 1 –y b
x x
×

x+ 1 a 1 –x b xy + 1 a 1 – xy b
y y y y
× ×
1 – xy b
1 a 1 b = xy + 1 a x
x x x
y+ × –y ×

= (xy + 1)a (1 – xy)b xa + b xa + b x a+b
× (xy + 1)a (1 – xy)b = ya + b = y
ya + b Answer

1 – a2 a 1 –a b–a a a+b
b2 b b
Example 9: Show that =
Solution:
1 – b2 b 1 +b a–b
a2 a

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1 a 1 –a b–a 1 a 1 a 1 – a b–a
b2 b +b a–b b b b
– a2 1 +a –a
a
L.H.S. = 1 b = 1 b 1 b 1 a–b
a2 a a a
– b2 +b –b +b

1 a 1 b–a+a 1 + ab a 1 – ab b
b b b b
1 +a –a
a 1 + ab a 1 – ab b
= b+a–b 1 b = a a
a
+b –b

= (1 + ab)a (1 – ab)b × aa + b
ba + b ab)a (1
(1 + – ab)b

= a a+b Proved.
b

Example 10: 111
Simplify 1 + ax – y + az – y + 1 + ay – z + ax – z + 1 + az – x + ay – x
Solution:

111
1 + ax – y + az – y + 1 + ay – z + ax – z + 1 + az – x + ay – x

= 1 az + 1 ax + 1 ay
ax ay ay az az ax
1+ ay + 1 + az + 1+ ax +

= 1 + 1 + 1
ay + ax + az az + ay + ax ax + az + ay
ay az ax

ay az ax ax + ay + az
= ax + ay + az + ax + ay + az + ax + ay + az = ax + ay + az = 1 Answer

Example 11: If a + b + c = 0, show that

Solution: 111
1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1

Here, a + b + c = 0, ? b + c = – a
111

1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a

xb xc 1
= xb (1 + xa + x–b) + xc (1 + xb + x–c) + 1 + xc + xb + c

xb xc 1
= xb + xa + b + 1 + xc + xb + c + 1 + 1 + xc + xb + c

xb.xc xc 1
= xc (xb + xa + b + 1) + xc + xb + c + 1 + 1 + xc + xb + c

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Indices

xb + c xc 1
= xb + c + xa + b + c +xc + xb + c + xc+ 1 + xb + c + xc + 1

xb + c xc 1
= xb + c + x0+ xc + xb + c + xc+1 + xb + c + xc + 1

xb + c xc 1 xb + c + xc + 1
= xb + c + xc+1 + xb + c + xc+1 + xb + c + xc + 1 = xb + c + xc + 1 = 1 Proved.

22

Example 12: If x2 + 2 = 33 + 3 3, show that 3x(x2 + 3) = 8.

Solution:

22

Here, x2 + 2 = 33 + 3 3

12 12

or, x2 = 33 + 3 3 – 2

12 12 1 1
or, x2 = 33 + 3 3 – 2.33.3 3

1 12

or, x2 = 33 – 3 3

11

or, x = 33 – 3 3

1 13

or, x3 = 33 – 3 3

13 13 11 1 1

or, x3 = 33 – 3 3 – 3.33.3 3 33 – 3 3

or, x3 = 3– 1 – 3x
8
3

or, x3 + 3x = 3
or, 3x3 + 9x = 8

or, 3x (x2 + 3) = 8 Proved.

EXERCISE 10.1

General section

1. Evaluate:

8 – 1 4 – 1 125 – 1 2 25 – 1
27 3 2 3 2
a) ÷ b) ÷
9 64 4

– 1 – 1 – 1 –1
3 4 5
c) 8 × 81 ÷ 32 d) 729
27 16
243 3
64

– 1 2 1 1 1
3 2 3
125 25 – 125 3 8 –
64 ÷
e) 64 f) 16 27

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Indices

2. Simplify:

a) 2x × 3 – 2x b) 5m + 2 – 5m c) 4m + 4m + 1
2x+2 – 2x 5m+1 + 5m 4m+2 – 4m
+1

d) 6n + 2 – 6n e) 2x +4 – 2x f) 5n + 2 – 2.5n
6n + 1 + 6n 5.2x 23.5n

g) 5n + 2 – 10 × 5n h) 33a + 2 – 33a + 1 i) 132x +1 + 5 × 169x
3 × 5n 6 × 27a 9 × 169x

j) 112x + 1 – 6 × 121x k) 7a + 1 + 9 × 7a l) 9n + 2 + 10 × 9n
5× 121x 7a + 2 – 45 × 7a 9n + 1 × 11 – 8 × 9n

m) 11n + 2 – 55.11n – 1 n) 273n – 1 (243)– 4n 2n
11n × 116 5
o) (243) 5 . 32n + 1
9n + 1 33n – 5 9n + 1 × 32(n – 2)

3. Simplify: b) 3n a3 × an n – 1 2
e) a6b–2c4 ÷ 4 a4b–4c8
2 h) 3 3x7y11z–1 × 3 72x–1yz4 c) (125a3 ÷ 27b–3) 3

a) (64x3 ÷ 27a–3) 3 f) 3 27a12b9 ÷ 4 16a16b2

3 2

d) (81a4 ÷ 16b–4) 4 i) 3 (x + y)–8 .(x + y)3

g) 3 9x–2y7 × 3 3x5y–1

1

j) 3 (a + b)–7 .(a + b)3

4. Simplify: b) ax + y x – y × ay + z y – z × az – x z + x
a) xb – c × xc – a × xa – b
ya b zb a xa b
c) xa + b × xa – b × xc – 3a d) za . xb . ya
xc – a
f) 1 + 1
e) 1 + 1 1 – ax – y 1 – ay – x
1 + xa – b 1 + xb – a
h) a2 + b2 + c2
g) x–1y–1 + y–1z–1 + z–1x–1 a–2b–2 + b–2c–2 + c–2a–2
x+y+z

Creative section - A

5. Simplify:

xa + b a – b xc + a c – a xb + c b – c xa a – b xb b – c xc c + a
a) xc × xb × xa b) x–b × x–c × xa

xl l2 + lm + m2 xm m2 + mn + n2 xn n2 + nl + l2
c) xm × xn × xl

d) (xa ÷ xb)a2 + ab + b2 × (xb ÷ xc)b2 + bc + c2 × (xc ÷ xa)c2 + ca + a2

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Indices

(aax + y)2 (ay+z)2 (az + x)2 xa2 + b2 a–b xb2 +bc b–c xc2 +ac c –
(ax.ay.az)4 x–ab x–c2 x–a2
e) f) × ×

xb b + c – a xc c + a – b xa a + b – c xb + c a – b xa + b b – c xa + b c – a
g) xc × xa × xb h) xc – a × xa – c × xb – c

xl + m n – l xm + n l – m xn + l m – n xa a – b xb b – c xc c – a
i) xl – m × × xn – l j) x.x–b × x.x–c × x.x–a
xm – n

111 b 1 c 1 a 1
xb xc xa ab bc ca ab
xc bc xa ac xb xc xa xb
k) l) c × a × b
× ×

xb xc xa

m) 1 1 1 1 . 1 1 111
y–x z–y
ax–y x–z . ay–z az–x n) xm2 – n2 m+ n × xn2 – p2 n+ p × xp2 – m2 p+ m

11 1

o) ax .ay x – y az y+z × az z+x p) p + (pq2)3 + (p2q)3 × 1– q3
ay ax p–q 1

p3

6. Simplify: b) xy ax × yz ay × zx az
a) xa +b a2 – b2 × b +c xb2 – c2 × c +a xc2 – a2 ay az ax

yzx 1 111
ab 1 1
c) yz az × zx ax × xy ay d) xa × bc xb × ca xc
z x y 1 1 1

ay az ax xb xc xa

Creative section - B

7. Simplify:

a+ 1 x 1 –a y 1 –x a x+ 1 a
b b y y
a) × b) ×

b+ 1 x 1 –b y y+ 1 a 1 –y a
a a x x
× ×

a2 – 1 a a– 1 b–a 1+ x x 1– y y
b2 b y x x–y
× x–y ×

c) 1 b 1 a–b d) y x x y
a2 a x x–y y x–y
b2 – × b+ +1 × –1

y+ 1 x+y x– 1 x+y
x y
e) ×

x2 – 1 x y2 – 1 y
y2 x2
×

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Indices

8. Simplify:

a) 1 + 1 + xc – b +1 + 1 + xa – c + 1 + 1 + xb – a
xa – b xb – c xc – a

b) 1 + ax 1 + ax – z + 1 + ay 1 + ay – z + 1 + az 1 + az – y

–y –x –x

y y2 – yz + z2 z z2 – zx + x2
a a ax x2 – xy + y2
9. a) If x3 + y3 + z3 = 1, show that, a–y a–z a–x = a2

y y–z
a a ax x – y z z–x

b) If x2 + y2 + z2 = xy + yz + zx, show that ay . az . ax =1

c) If pqr = 1, show that 1 + 1 q–1 + 1 + 1 r–1 + 1 + 1 p–1 = 1.
p+ q+ r+

d) If abc + 1 = 0, prove that 1 + 1 + 1 = 1
1 – a – b–1 1 – b – c–1 1 – c – a–1

22

e) If x2 + 2 = 23 + 2 3, show that 2x (x2 + 3) = 3

12

f) If x = 33 + 33 , prove that x3 – 9x – 12 = 0
111

g) If a + b + c = 0, show that 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1

10.3 Exponential equation

Let's take an equation 2x = 8. In this equation, the unknown variable 'x' is the base
and 2 is its coefficient.

Now, take another equation 2x = 8. In this case, the unknown variable 'x' is the
exponent of the base 2. Such an equation in which the unknown variable appears as
an exponent of a base is known as exponential equation. The following axioms are
useful while solving the exponential equations:

(i) If ax = ab, then x = b (ii) If ax = 1, then ax = a° and x = 0

Thus, while solving an exponential equation, we should simplify the equation till
the equation will be obtained in the forms ax = ab or ax = 1.

Worked-out examples

Example 1: Solve 2x + 2 + 2x + 3 = 1
2
Solution:
2x + 2 + 2x + 3 = 1
Here, 2 Check

or, 2x × 2 2 + 2x × 23 × 2–1 = 1 2x + 2 + 2x + 3 = 1
2
or, 2x (22 + 22) = 1 or, 2–3 + 2 + 2– 3 + 3 = 1
or, 2x = 1
8 or, 1 2 1 =1
+
or, 2x = 1 22
23
or, 1 = 1
or, 2x = 2–3

? x =–3

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Indices

Example 2: Solve 3x + 2 × 5x – 1 = 405
Solution:

Here, 3x + 2 × 5x – 1 = 405

or, 3x × 3 2 + 5x × 5–1 = 405

or, 3x × 5x × 9 = 405
5

or, (3 × 5)x = 225

or, 15x = 152

? x =2

Example 3: Solve 2x + 3 + 1 – 9 = 0
2x
Solution:
2x + 3 + 1 – 9 = 0
Here, 2x

or, 2x × 23 + 1 – 9 = 0
2x

Let 2x = a 8a + 1 – 9 = 0

Then, a

or, 8a2 + 1 – 9a = 0
a

or, 8a2 – 9a + 1 = 0

or, 8a2 – 8a – a + 1 = 0

or, 8a (a – 1) – 1 (a – 1) = 0

or, (a – 1) (8a – 1) = 0

Either, a – 1 = 0 or, 8a – 1 = 0

i.e. a = 1 i.e. a = 1 = 1 = 2–3
8 23
i.e. 2x = 2° i.e.
? x =0 ? 2x = 2–3
So, x = 0, – 3 x = –3

Example 4: If a = bc, b = ca and c = ab, prove that abc = 1.

Solution:

Here, a = bc, b = ca and c = ab Alternative process
1 Here, a = bc,

Since c = ab, i.e. a = cb a = (ca)c [ b = ca]
a = cac
Now, a = bc [Given]

1 1

or, cb = bc [Putting a = cb]

1 a = (ab)ac [ c = ab]

or, cb = (ca)c [Putting b = ca] a = aabc
? abc = 1 Proved.
1

or, cb = cac

or, 1 = ac
b

or, abc = 1 proved.

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Indices

Example 5: If ap.aq = (ap) , prove that p(q – 2) + q (p – 2) = 0.
Solution:

Here, ap.aq = (ap)

or, ap + q = apq

or, p + q = pq

or, 2(p + q) = 2pq (multiplying both sides by 2)

or, pq + pq – 2p – 2q = 0

or, pq – 2p + pq – 2q = 0

or, p (q – 2) + q (p – 2) = 0 Proved.

Example 6: If xa = yb = zc and y2 = xz, show that 2 = 1 + 1 .
bac
Solution:

Here, xa = yb, b Alternative process
Also, zc = xa, Let, xa = yb = zc = k
Now, i.e. x = ya
or,
or, a b ab 11 1
or,
or, i.e. z = x c = ya c = yc

y2 = xz Then, x = ka, y = kb , z = k c

bb Now, y2 = xz

y2 = ya . y c 1 11

bb or, k2 b = ka . k c

y2 = ya c or, 2 = 1 + 1
c
kb ka
b b
2 = a + c or, 2 = 1 + 1 Proved.
b a c
2 =b 1 + 1
a c

or, 2 = 1 + 1 Proved.
b a c

Example 7: If p x = q y = r z and xyz = 1, prove that p + q + r = 0.
Solution:
Here, p x = q y = r z = k (suppose)

Then, p x = k

1

or, xP = k

or, x = kp

Similarly, y = kq and z = kr

Now, xyz = 1

or, kp.kq.kr. = 1

or, kp + q + r = k0

or, p + q + r = 0 Proved

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EXERCISE 10.2

General section b) 9x – 1 = 3x + 1 c) 42x – 1 = 2x + 1
1. Solve. 1 f) 103y – 3 =0.0101
0.04 i) 3x + 1 – 3x = 54
a) 2x – 4 = 4x – 6
l) 4x – 1 = ( 2)x
d) 32x + 1 = 92x –1 e) 25x + 3 =
x
g) 3 × 81x = 9x + 4 h) 2x + 1 – 2x = 8
j) 3x + 2 + 3x + 1 = 113 o) (0.5)2 = 0.25
k) 4x + 1 = 1
8x r) 5x + 5x + 1 + 5x + 2 = 155

m) ( 2)3x – 1 = ( 4)x – 2 n) ( 9)x – 3 = ( 3)x + 2

p) 3x – 2 + 3x = 10 q) 32x + 3 – 2.9x + 1 = 1
9 3

2. Solve.

a) 2x + 3 × 3x + 4 = 18 b) 2x – 3 × 3x – 4 = 3–1 c) 23x – 5 ax – 2 = 2x – 2 a1 – x
d) 5x – 3 × 32x – 8 = 225 e) 2x – 5 × 5x – 4 = 5 7
Creative section f) 72x + 1 × 52x – 1 = 5

3. Solve.

a) 22x + 3.2x – 4 = 0 b) 22x – 6.2x + 1 + 32 = 0 c) 4.3x + 1 – 9x = 27

d) 2a – 2 + 23 – a = 3 e) 3.2x + 1 – 4x = 8 f) 5.4x + 1 – 16x = 64
26 h) 5x + 5–x = 25215
g) 51 – x + 5x – 1 = 5 i) 5x + 1 + 52 – x = 126

j) 2x + 16 = 10 k) 3x +3 + 1 – 28 = 0 l) 7x + 343 = 56
2x 3x 7x
1 1 1 = 414 1 1
m) 4x + 4x = 16 16 n) 2x + 2x o) 3x + 3x = 9 9

p) 3x + 2 + 1 = 30
3x – 2

4. a) If ax = b, by = c and cz = a, prove that xyz = 1.

b) If x = yz, y = zx and z = xy, prove that xyz = 1.

c) If ax = by and b = a2, show that x – 2y = 0.

d) If a = 10x, b = 10y and aybx = 100, show that xy = 1.

e) If xa.xb = (xa)b, prove that a + b = ab – 2.
b a
3 1 1
f) If xa = yb = zc and y3 = xz, show that b = a + c

g) If ap = bq = cr and b2 = ac, prove that q = 2pr
p+r
1 1 2x.
h) If 2x = 3y = 12z, show that z = y +

i) If x a = y b = z c and abc = 1, prove that x + y + z = 0.

5. a) In how many years does Rs 2,000 amount to Rs 2,420 at 10% p.a. compound
interest?

b) In how many years does Rs 8,000 amount to Rs 9,261 at 5% p.a. compound interest?

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Unit Surds

11

11.1 Surds - Review

0, 1, 2, –3, –5, 2 , 4 , etc. are the examples of rational numbers. A rational number
can be 3 9
p
expressed in q form where q z 0. When a rational number is expressed in

decimal, whether the decimal part may be terminating or none-terminating recurring

decimal. For example,

1 = 0.5, 2 = 0.4, 3 = 0.75, 5 = 0.625 , etc. are terminating decimals.
2 5 4 8

1 = 0.333…, 5 = 0.833…, 4 = 0.5714285714..., etc. are none-terminating recurring
3 6 7

decimals.

On the other hand, there is another set of numbers which are not rational and cannot
p
be expressed in q form. These numbers are called irrational numbers. When an

irrational number is expressed in decimal, the decimal part is neither terminating

nor none–terminating recurring. 2, 3 , 3 6 , 4 8 , S, etc. are the example of irrational
numbers. Such irrational numbers are called surds.

In n a , ‘n’ is called the order of the surd and ‘a’ is called the radicand.
Here, n a is called the nth order surd, where n is the natural number and a is the
rational number greater than zero (0).

11.2 Laws of surds

There are some verified rules which are used in the operations of surds. Such rules
are called the laws of surds. The table given below shows some of the laws of surds.

Laws of surds Examples

(i) n a = 1 So, n a n 1 35 33

n, =a 3 2 = 23, = 53 = 5

(ii) n ab = n a . n b 3×5= 3× 5

(iii) n a= na 4 4= 44
b nb 5 45

(iv) n a + n a = 2 n a 3 5 + 3 5 =23 5

m na n ma mn a 3 64 = 4 = 2 = 2×3 64 = 6 64

(v) = =

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Surds

11.3 Like and unlike surds

2 3 , 5 3 , 3 , etc. are the like surds. They have the same order and the equal
radicand.
3 5 , 3 3 5 , 7 3 5 , etc. also have the same order and the equal radicand. So, they are
also the like surds.

The surds with the same order and the equal radicands are known as like surds.
On the other hand, 2 3 , 2 , 3 2 , 7 4 6 , etc. are unlike surds. Unlike surds do not
have the same order and the equal radicands.

11.4 Simplification of surds

(i) Addition and Subtraction of Surds

We can add or subtract the rational factors of the surds only in the case of like
surds. The process is exactly the same as in the case of addition and subtraction
of the algebraic terms with the same base and powers.

For example,

5 2 + 4 2 = (5 + 4) 2 = 9 2 As like 5x + 4x = 9x
7 3 4 – 3 3 4 = (7 – 3) 3 4 = 4 3 4 As like 7x – 3x = 4x

(ii) Multiplication and division of surds

We can multiply or divide the rational factors as well as the radicands only in the
case of the surds with the same order. For example,

3 6×5 7 =3×5 6×7 = 15 42
435 × 234 = 4 × 235 × 4 = 8 3 20

8 15 y 2 5 = (8 y 2) 15 ÷ 5 = 4 3

Worked-out examples

Example 1: Simplify (i) 50 + 18 – 8 2

(ii) 3 128 + 3 3 54 – 2 3 250

Solution:

(i) 50 + 18 – 8 2 = + 9 × 2 – 8 2

=5 2+3 2–8 2

=8 2–8 2 =0

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Surds

(ii) 3 128 + 3 3 54 – 2 3 250 = 3 64 × 2 + 3 3 27 × 2 – 2 3 125 × 2

= 4 3 2 + 3.3 3 2 – 2.5 3 2

= 13 3 2 – 10 3 2

= 332

Example 2: Simplify (i) 5 × 15 × 3 (ii) 3 128 – 3 16
Solution: 23 2

(i) 5 × 15 × 3 = 5 × 15 × 3

= 5×5×3×3

= 5 × 3 = 15

(ii) 3 128 – 3 16 = 3 64 × 2 – 3 8 × 2
23 2 23 2

= 43 2 – 23 2
23 2

= 23 2 = 1
23 2

Example 3: Simplify (3 5 – 4 2 ) (2 5 + 3 2 )
Solution:

Here, (3 5 – 4 2 ) (2 5 + 3 2 ) = 3 × 2 5. 5 + 3.3 5. 2 – 4 × 2 2. 5 – 4 × 3 2. 2
= 6 × 5 + 9 10 – 8 10 – 12 × 2
= 30 + 10 – 24
= 6 + 10

Example 4: Simplify 5x – 9 .
Solution: 3 + 5x

Here, 5x – 9 = ( 5x)2 – 32 = ( 5x + 3) ( 5x – 3) = 5x – 3
3 + 5x 3 + 5x
3 + 5x

Example 5: Simplify 7 3 16 – 2 3 54 .
5 3 32 + 2 3 108

Solution:

Here, 7 3 16 – 2 3 54 = 7 3 8 × 2 – 2 3 27 × 2
5 3 32 + 2 3 108 5 3 8 × 4 + 2 3 27 × 4

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Surds

= 7 × 232 – 2 × 332
5 ×2 3 4 + 2 × 3 3 4

= 14 3 2 – 6 3 2
10 3 4 + 6 3 4

= 832 = 1 3 2 = 1 31
16 3 4 2 4 2 2

EXERCISE 11.1

General section

1. Simplify: b) 5 3 + 6 3 – 3 c) 5 – 4 5 + 6 5
a) 2 + 5 2 + 2 2 e) 3 7 + 4 3 7 – 9 3 7 f) 2 4 6 – 4 6 – 3 4 6

d) 8 3 4 – 3 4 – 3 3 4

2. Simplify: b) 6 × 3 × 2 2 c) 3 9 × 3 3 × 3 3 2
a) 2 × 3 × 5

d) 5 8 ÷ 2 2 e) 5 3 108 ÷ 3 3 2 f) 4 360 ÷ 3 20

3. Simplify:

a) 32 + 8 – 72 b) 27 + 75 – 8 3

c) 4 45 – 3 20 +8 5 d) 12 – 75 + 48
e) 3 16 + 3 54 – 3 250 f) 5 3 81 – 2 3 24 + 3 375
g) 4 4 405 – 3 4 80 – 2 4 5 h) 3 2 + 4 2500 – 4 64 + 6 8

4. Simplify:

a) ( 3 + 2) ( 3 – 2) b) ( 5 – 3) ( 5 + 3)

c) (2 5 + 3 2) (2 5 – 3 2) d) ( 2 + 3)2

e) ( 5 – 3)2 f) ( x + a – x – a)2

g) (2 2 – 3) (3 2 + 3) h) (3 5 – 4 2) (2 5 + 2 3)

5. Simplify: b) x2 – 9 c) 3 25 – x2
x–3 3x + 5
a) a2 – b2
a–b e) 3x – 16 f) 49 – 5x
4 + 3x 7 – 5x
d) x – 4
x +2

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Surds

6. Simplify. b) 4 3 54 – 2 3 250 c) 2 75 + 4 108 – 3 48
6 3 128 33 3
a) 24 + 54
10 6 e) 50 + 18 f) 5 3 81 – 2 3 24
7 8 – 128 2 3 48 + 3 3 162
d) 3 3 81 – 3 3 24 + 2 3 375
13 3 192

11.5 Rationalisation

Let’s take a surd 3 .

3 is an irrational number. When it is multiplied by 3 , the product is 3 and it is
a rational number. The process of changing a surd into a rational number is called
rationalisation. Look at the following examples:

2× 2 = ( 2)2 = 2 o 2 is the rationalising factor of 2.

2 3× 3 = 2( 3)2 = 6 o 3 is the rationalising factor of 2 3 .

( x + a) ( x + a) = ( x + a)2= x + a o x + a is the rationalising factor of x + a.

Thus, if the product of two surds is a rational number, each of them is called the
rationalising factor of the other.

11.6 Conjugate

Let’s take a binomial surd, 5 + 3 .
The rationalising factor of 5 + 3 is 5 – 3 .
Here, 5 – 3 is called the conjugate of 5 + 3 or vice versa.
Thus, a binomial surd can be rationalised multiplying by its conjugate.

Worked-out examples

Example 1: Rationalise the denominators of: (i) 2 (ii) 3 6
3 25
Solution:

(i) Multiplying the numerator and denominator by 3 ,

2 =2 × 3 = 23
33 3 3

(ii) Multiplying the numerator and denominator by 5,

3 6 =3 6 × 5 = 3 30 = 3 30
2 5 2 5 5 2 ×5 10

Example 2: Rationalise the denominator and simplify: 3 5+ 3 .
5– 3
Solution:

Multiplying the numerator and denominator by 5 + 3 .

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Surds

3 5+ 3 = 3 5+ 3 × 5+ 3
5– 3 5– 3 5+ 3

= 3 × 5 + 3 15 + 15 + 3
( 5)2 – ( 3)2

= 15 + 4 15 + 3 = 18 +4 15 = 2 (9 +2 15) =9+2 15
5–3 2 2

Example 3: Rationalise the denominator of: x+2– x–2 .
x+2+ x–2
Solution:

Here, x + 2 – x – 2 = x + 2 – x – 2 × x + 2 – x – 2
x+2+ x–2 x+2+ x–2 x+2– x–2

= ( x + 2 – x – 2 )2
( x + 2)2 – ( x – 2 )2

=( x + 2)2 – 2 x + 2. x – 2 ) + ( x – 2 )2
x+2–x+2

= x + 2 – 2 x2 – 4 + x – 2
4

= 2x – 2 x2 – 4
4

= 2(x – x2 – 4) = x – x2 – 4
42

Example 4: Rationalise the denominator of: 3 5.
3+ 2–
Solution:

Here, 3 = 3 × ( 3 + 2) + 5

3+ 2– 5 ( 3 + 2) – 5 ( 3 + 2) + 5

= 3 + 6 + 15
( 3 + 2)2 – ( 5)2

= 3 + 6 + 15
( 3)2 + 2. 3. 2 + ( 2)2 – 5

= 3 + 6 + 15
3 + 2 6+ 2 – 5

= 3+ 6+ 15 × 6
26 6

=3 6+6+ 90
2×6

= 6+3 6+ 9 × 10
12

= 6+3 6+3 10 = 3(2 + 6+ 10) = 2 + 6+ 10
12 12 4

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Surds

Example 5: Simplify 5– 3 – 5+ 3 .
5+ 3 5– 3
Solution:
( 5 – 3)2 – ( 5+ 3)2
Here, 5– 3 – 5+ 3 = ( 5 + 3) ( 5– 3)
5+ 3 5– 3

= ( 5)2 – 2. 5. 3+( 3)2 – ( 5)2 – 2. 5. 3–( 3)2
( 5)2 – ( 3)2

= –4 15
5–3

= –4 15 = –2 15
2

Note: In the above example, the denominator of each term is conjugate to each other. In
such case, instead of rationalising the denominators, we can take their L.C.M. and
then simplify them.

Example 6: Simplify 32 6 + 6 3– 43 2 .
3+ 2+ 6+
Solution:

Here, 3 2 + 6 – 43
3+ 6 2+ 3 6+ 2

= ( 3 2 ( 3 – 6) 6) +( 6 ( 2 – 3) 3) –( 4 3 ( 6 – 2)
3 + 6) ( 3 – 2 + 3)( 2 – 6 + 2) ( 6 – 2)

= 3 2 ( 3 – 6) + 6 ( 2 – 3) – 4 3 ( 6 – 2)
( 3)2 – ( 6)2 ( 2)2– ( 3)2 ( 6)2 – ( 2)2

= 3 2 ( 3 – 6) + 6 ( 2 – 3) – 4 3 ( 6 – 2)
–3 –1 4

=– 2× 3+ 2× 6– 6× 2+ 6× 3– 3× 6+ 3× 2

= – 6 + 12 – 12 + 18 – 18 + 6 = 0

Example 7: Simplify a2 + 1 + a2 – 1 + a2 + 1 – a2 – 1
a2 + 1 – a2 – 1 a2 + 1 + a2 – 1
Solution:

Here, a2 + 1 + a2 – 1 + a2 + 1 – a2 – 1
a2 + 1 – a2 – 1 a2 + 1 + a2 – 1

= ( a2 + 1 + a2 – 1 )2 + ( a2 + 1 – a2 – 1 )2
( a2 + 1 – a2 – 1 ) ( a2 + 1 + a2 – 1 )

= ( a2 + 1)2 +2 a2 + 1. a2 – 1 ) + ( a2 – 1 )2 + ( a2 + 1)2 – 2 a2 + 1. a2 – 1 ) + ( a2 – 1 )2
( a2 + 1)2 – ( a2 – 1 )2

= a2 + 1 + 2 a4 – 1 + a2 – 1 + a2 + 1 – 2 a4 – 1 + a2 – 1 = 4a2 = 2a2
a2 + 1 – a2 + 1 2

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Surds

EXERCISE 11.2

General section

1. Rationalise the denominators and simplify:

a) 1 b) 1 c) 3 d) 4
2+1 3–2 5– 2 2 3– 2

e) 5 f) 2 2 g) 5 3 h) 4 5
3 7+2 3 6+ 3 2 3– 2 2 3+ 5

2. Rationalise the denominators and simplify:

a) 2 + 1 b) 3 – 1 c) 3 – 2 d) 5 + 3
2–1 3+1 3+ 2 5– 3

e) 2 3 – 3 2 f) 3 5 – 3 g) a + b – a – b h) x + 1 – x – 1
2 3+3 2 5– 3
a+b+ a–b x+1+ x–1

i) a + 2 – a – 2 j) a + 5 + a – 5 k) 2 l) 2
3+ 2+1
a+2+ a–2 a+5– a–5 2+ 3– 5

3. Simplify:

a) 3 5 – 1 b) 7 + 2 3 c) 3 + 5
5 3 5 2

d) 7 – 3 e) 2 + 3 f) ) 7 + 5
3 4 5 2 3 27

Creative Section - A b) 5 + 3 5 – 5 – 3 5

4. Simplify: 5+2 5–2
a) 3 + 1 + 3 – 1
3–1 3+1

c) 3 + 2 + 3 – 2 d) x + a – x – a
x– a x+ a
3– 2 3+ 2

e) x + x2 – 1 – x – x2 – 1 f) a – a2 – 1 + a + a2 – 1
x – x2 – 1 x + x2 – 1 a + a2 – 1 a – a2 – 1

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5. Simplify: b) 72 – 48 – 45 + 2 98
a) 3 20 + 4 + 5 + 3 50 128
5 5–3
d) 3 2 – 4 3 + 6
c) 2 10 – 2 5 – 10
3+1 6+2 2+1 6+ 3 6+ 2 3+ 2

e) 7 3 + 25 + 32 f) 5 2 – 8 5 + 3 10
10 + 3 6+ 5 15 + 3 3 5( 2 + 1) 10 + 2 2+ 5

Creative Section - B

6. Simplify:

a) x + 1 + x – 1 + x + 1 – x – 1 b) a + b + a – b + a + b – a – b

x+ 1 – x– 1 x+ 1 + x– 1 a+ b – a– b a+ b + a– b

c) 2x + 3 + 2x – 3 + 2x + 3 – 2x – 3 d) x2 + 2 + x2 – 2 + x2 + 2 – x2 – 2

2x + 3 – 2x – 3 2x + 3 + 2x – 3 x2 + 2 – x2 – 2 x2 + 2 + x2 – 2

11.7 Simple surd equations

Let's consider an equation x = 5.
Here, the unknown variable is a surd. Such an equation is known as the surd
equation.

To solve a surd equation, we should remove the radical from the variable. For this, we
should give nth power to both sides of the equation to remove nth order of radical.

∴ If n x = a, then

nx n

= an

n

i.e. xn = an
i.e. x = an.

Worked-out examples

Example 1: Solve x + 5 = 3.
Solution:

Here, x + 5 = 3
Squaring both sides of the equation, we get,

( x + 5)2 = 32
or, x + 5 = 9
or, x = 9 – 5
or, x = 4
Now, substituting x = 4 in the original equation, we get,

4 + 5 =3
or, 9 =3
or, 3 = 3 which is true.
So, the required value of x is 4.

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Example 2: Solve x + 4 = x – 2.
Solution:

Here, x + 4 = x – 2

Squaring both sides of the equation, we get,

or, ( x + 4)2 = (x – 2)2
or,
or, x+4 = x2 – 4x + 4
Either, x2 – 5x =0
or, x (x – 5) =0
=0
x = 0, i.e. x = 5
x–5

Substituting x = 0 in the original equation, we get,

0+4 =0–2

or, 4 = – 2
or, 2 = – 2 which is false.

Substituting x = 5 in the given equation, we get,

5+4 =5–2

or, 9 = 3
or, 3 = 3 which is true.
So, the required value of x is 5.

Example 3: Solve x – 9 – x = 3

Solution:

Here, x – 9 – x = 3
or, x – 9 = 3 + x

Squaring both sides of the equation, we get,

( x – 9 )2 = (3 + x )2
or, x – 9 = 32 + 2.3. x + ( x )2
or, x – 9 = 9 + 6 x + x
or, 6 x = – 18

x =–3

Again, squaring both side,

( x )2 = (– 3)2
or, x = 9

Now, substituting x = 9 in the original equation, we get,

9–9 – 9 =3
or, 0 – 3 = 3
or, – 3 = 3 which is false.
So, the equation does not have any solution.

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Example 4: Solve 3x + 1 – x – 1 = 2
Solution:
Here, 3x + 1 – x – 1 = 2
or, 3x + 1 = 2 + x – 1
Squaring both sides of the equation, we get,

( 3x + 1)2 = (2 + x – 1)2
or, 3x + 1 = 22 + 2.2. x – 1 + ( x – 1)2
or, 3x + 1 = 4 + 4 x – 1 + x – 1
or, 2x – 2 = 4 x – 1

x–1 =2 x–1
Again, squaring both sides,

(x – 1)2 = (2 x – 1 )2

or, x2 – 2x + 1 = 4 (x – 1)

or, x2 – 2x + 1 – 4x + 4 = 0
or, x2 – 6x + 5 = 0
or, x2 – 5x – x + 5 = 0
or, x(x – 5) – 1(x – 5 = 0
or, (x – 5) (x – 1) = 0
Either, x – 5 = 0, i.e., x = 5

or, x – 1 = 0, i.e., x = 1

When, x = 5, 3 × 5 + 1 – 5 – 1 = 2, i.e. 2 = 2 which is true.
When, x = 1, 3 × 1 + 1 – 1 – 1 = 2, i.e. 2 = 2 which is true.
So, the requested values of x are 5 and 1.

Example 5: Solve 3 x – 4 = 15 + 3 x
x +2 x + 40
Solution:

Here (3 x – 4) ( x + 40) = ( x + 2) (15 + 3 x )

or, 3x +120 x – 4 x – 160 = 15 x + 3x + 30 + 6 x

or, 95 x = 190

or, x = 2
or, ( x )2 = 22

x =4

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Surds

Now, substituting x = 4 in the given equation, we get

3 4 – 4 = 15 + 3 4
4 +2 4 + 40

or, 6–4 = 21
4 42

or, 1 = 1 which is true
2 2

So, the required value of x is 4.

Example 6: Solve x–1 =4+ x –1
Solution: x +1 2

Here, x–1 =4+ x –1
x +1 2

or, ( x )2 – 12 = 8+ x –1
x +1 2

or, ( x + 1) ( x – 1) 7+ x
x +1 =2

or, 2 x – 2 = 7 + x

or, 2 x – x = 7 + 2

or, x = 9

or, ( x )2 = 92

or, x = 81

Now, substituting x = 81 in the given equation, we get,

81 – 1 =4+ 81 – 1
81 + 1 2

or, 80 =4+4
10

or, 8 = 8 which is true

Hence, the required value of x is 81.

Example 7: Solve x + 5 + 6 = 3
x+5– 6

Solution:

Here, x+5+ 6 =3
x+5– 6

By using the rule of componendo and dividendo, we get

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Surds

x+5+ 6+ x+5– 6 3+1 If a = c then
x+5+ 6– x+5+ 6 = 3–1 b d

or, 2 x + 5 = 4 a+b = c+d
26 2 a–b c–d

is called the rule of

x+5 componendo and
or, = 2
dividendo.
6
Squaring both sides, we get,

x+5 2
6
= 22

or, x + 5 = 4
6

or, x = 19

Now, substituting x = 19 in the given equation, we get
19 + 5 + 6 = 3
19 + 5 – 6

or, 24 + 6 = 3
24 – 6

or, 2 6 + 6 = 3
2 6– 6

or, 3 6 = 3
6

or, 3 = 3 which is true.

Hence, the required value of x is 19.

Example 8: Solve x+ 2 + x– 2 =6
x– 2 x+ 2

Solution:

Here, x+ 2 + x– 2 =6
x– 2 x+ 2

or, ( x + 2)2 + ( x – 2)2 = 6
( x – 2) ( x + 2)

or, ( x )2 +2. x . 2 + ( 2)2 + ( x )2 – 2. x . 2 + ( 2)2 = 6
( x )2 – ( 2)2

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or, x + 2 + x + 2 = 6
x–2

or, 2x + 4 = 6x – 12

or, 4x = 16
or, x = 16
Now, Substituting x = 4 in the given equation, we get,

4+ 2 + 4– 2 =6
4– 2 4+ 2

or, 2 + 2 + 2 – 2 = 6
2– 2 2+ 2

or, 22 + 2 . 2 . 2 + ( 2)2 + 22 – 2. 2. 2 + ( 2)2 = 6
(2 – 2) (2 + 2)

or, 4 + 2 + 4 + 2 = 6
4–2

or, 12 = 6
2

or, 6 = 6 which is true.
So, the required value of x is 4.

EXERCISE 11.3

General section

1. Solve:

a) x + 1 = 5 b) 2x – 5 = 7 c) 4x – 3 = 3
f) 3 x + 5 – 1 = 2
d) 2x + 3 – 3 = 0 e) 3x – 2 – 2 = 3 i) 4 a – 3 = 5a + 7
g) 4 2x – 1 – 2 = 1 h) x + 5 = 2x – 3 l) 4 6x + 5 = 4 x – 20
j) 3 4x + 1 = 3 x + 13 k) 3 7x + 11 = 3 3x + 5
c) x + 8 – 2 = x
2. Solve: f) 2x – 4x2 – 15 = 3
i) x2 + 5 – 1 = x
a) x – 7 = x – 1 b) x + 3 = x + 1
d) x – 4 + x = 2 e) x = 6 – x – 24 c) x – 25 = 9
x –5
g) x + x2 – 20 = 10 h) 9x2 – 20 = 3x – 2

3. Solve: b) x –9 = 1
x–1 x +3

a) x + 1 = 1

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d) 5x – 4 5x + 2 e) 5y – 4 = 2 – 5y – 3 f) x–3 = 2
5x + 2 = 2 – 2 5y – 2 2 x 5

g) x–4 = 3 h) y + 5 = 3 i) x + 7 = 3
x 7 y– 5 x– 7

Creative Section - A b) 2x +9 = 13 – x
4. Solve: d) 3x – 7x +2 = 2
f) 2x2 + x – 3 = x – 1
a) 2x + 7 = x + 2 h) 4 – x + x + 9 = 5
c) 3x + 4 + x = 12 j) x + 5 + x + 12 = 2x + 41
e) 2x + 1 = 4x2 + 3x + 6

g) 3x + 1 – x – 1 = 2
i) 4x – 3 + 2x + 3 = 6

5. Solve: 105
b) x + x – 15 = x – 15
a) x+ 5+x = 15
5+x

c) 2 x – 4x – 3 = 1 d) 5 x –3 = 3+5 x
4x – 3 x +2 x +5

e) x + x + 13 = 91 f) x + x – 1 – x = 1
Creative Section - B x + 13

6. Solve:

a) x–1 x –1 b) x – 4 = 2 + x –2
x +1 =4+ 2 2+ x 2

y – 25 =4+ y –5 d) 5y – 4 = 2 + 5y – 3
c) 5 + y 5 5y + 2 2

e) 3x – 4 – 3x – 2 = 2 f) 7x – 36 =9–5 7x – 11
2 + 3x 2 6 + 7x 3

7. Solve:

a) x + 4 + 2 = 2 b) x + 6 + 3 = 3
x+4 – 2 x+6 – 3

c) x+2 – x–2 = 1 d) x + 4 + x – 4 = 2
x+2 + x–2 2 x+4 – x–4

e) x + 5 + x – 5 = 4 f) x + a + x – a = 6
x– 5 x+ 5 x– a x+ a

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Unit Geometry - Area of Triangles
and Quadrilaterals
13

13.1 Relation between the area of triangles and quadrilaterals

In this chapter, we will discuss the relation of area of triangles and quadrilaterals
which are formed between parallel lines. The relation of area of triangles and
quadrilateral standing on the same base or on the equal base and between the same
parallel lines are verified by using the following various theorems.

Theorem 1

Diagonal of a parallelogram bisect the parallelogram.

or

The area of each of the triangles formed by drawing a diagonal of a parallelogram is half

of the area of the parallelogram. DC

Given: ABCD is a parallelogram in which AC is its diagonal.

To prove: ' ABC = 1 ABCD A
Proof 2
B

Statements Reasons
1. In ∆ ABC and ∆ ACD
1.

(i) AB = DC (S) (i) Sides of the parallelogram

(ii) BC = AD (S) (ii) Same as (i)

(iii) AC = AC (S) (iii) Common side

(iv) ? ∆ ABC # ∆ ACD (iv) S.S.S. axiom
2. Area of ∆ ABC = Area of ∆ ACD 2. Area of congruent triangles

3. ∆ ABC + ∆ ACD = ABCD 3. Whole part axiom

4. 2 ∆ ABC = ABCD 4. From statements (2) and (3)

5. Area of ∆ ABC = 1 area of ABCD 5. From statement (4)
2

Proved

Theorem 2
Parallelograms on the same base and between the same parallels are equal in area.

Experimental verification

Step 1: Three parallelograms ABCD with different sizes of bases and heights are drawn.

Another parallelogram BCEF on the same base BC and between the same parallels

BC and AE is drawn in each figure.

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Geometry– Area of Triangles and Quadrilateral

Step 2: DM A BC is drawn in each figure, where DM is the height (altitude) of the

parallelograms. A D F E A FD E

A FD E

B MC BM C B MC
(i) (ii) (iii)

Step 3: The base BC and the height DM are measured and the areas of ABCD and BCEF

are calculated. Result

Fig. Base Height Area of ABCD Area of BCEF ABCD = BCEF
(BC × DM)
BC DM (BC × DM)
(i)

(ii) ABCD = BCEF

(iii) ABCD = BCEF

Conclusion: Parallelograms on the same base and between the same parallels are equal in

area.

Theoretical proof A FD E

Given: Parallelograms ABCD and BCEF are on the same base BC
and between the same parallels BC and AE.

To prove: ABCD = BCEF BC

Proof

Statements Reasons
1. In ∆s ABF and DCE
1.

(i) ‘ AFB = ‘ DEC (A) (i) BF // CE and corresponding angles

(ii) ‘ BAF = ‘ CDE (A) (ii) BA // CD and corresponding angles

(iii) AB = DC (S) (iii) Opposite sides of a parallelogram

(iv) ∆ ABF # ∆ DCE (iv) A. A. S. axiom
2. ∆ ABF = ∆ DCE 2. Area of congruent triangles
3. ∆ ABF + trapezium BCDF 3. Adding the same trapezium BCDF to

= ∆ DCE + trapezium BCDF both sides
4. ABCD = BCEF
4. Whole part axiom

Corollary A DE Proved

Parallelograms on the equal bases and between the same H
parallels are equal in area.

In the figure, AH // BG and BC = FG B C FG
So, area of ABCD = area of EFGH

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Geometry – Area of Triangles and Quadrilateral

Theorem 3

The area of a triangle is equal to half of the area of a parallelogram standing on the same
base and between the same parallels.

Experimental verification
Step 1: Three different figures with a triangle BCD and a parallelogram ABCE on the same

base BC and between the same parallels AD and BC are drawn.

Step 2: AM A BC is drawn in each figure. AM is the height of ∆ BCD and ABCE.

A ED DA E AD E

MB C BM C BM C
(i) (ii)
(iii)

Step 3: The base BC and the height AM in each figure are measured. The areas of ∆ BCD and
ABCE are calculated.

Fig. Base BC Height AM Area of ∆ BCD Area of ABCE Result
(i) 1
(ii) ∆ BCD = 2 ABCE
(iii)
∆ BCD = 1 ABCE
2

∆ BCD = 1 ABCE
2

Conclusion: The area of triangle is equal to half of the area of a parallelogram on the same

base and between the same parallels. A FE D
Theoretical proof

Given: ∆ BCD and parallelogram ABCE are on the base BC

and between the same parallels AD and BC.

To prove: ∆ BCD = 1 ABCE B C
2
Construction: BF // CD is drawn. BF meets AD at F.

Proof

Statements Reasons

1. BCDF is a parallelogram 1. BC // FD and BF // CD

2. ∆ BCD = 1 BCDF 2. Diagonal bisects a parallelogram.
2

3. BCDF = ABCE 3. They are on the same base and between the same

parallels.

4. ∆ BCD = 1 ABCE 4. From statements (2) and (3)
2

Proved

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Geometry– Area of Triangles and Quadrilateral

Alternative process A ED
Construction: EF A BC is drawn.

EF is the height of the parallelogram ABCE and ∆ BCD.

Proof B FC

Statements Reasons
Area of a parallelogram is base × height
1. ABCE = BC × EF 1.
2.
2. ∆ BCD = 1 BC × EF Area of a triangle is 1 base × height
2 2

3. ∆ BCD = 1 BCEA 3. From statements (1) and (2)
2

Proved

Corollary

If a triangle and a parallelogram stand on equal bases and A DG

between the same parallels, the area of the triangle is equal to

one half of the area of the parallelogram. In the figure, AG // BF

and BC = EF

So, area of ' ABC = 1 area of DEFG. B CE F
2
Theorem 4

Triangles on the same base and between the same parallels are equal in area.

Experimental verification

Step 1: Three pairs of triangles ABC and BCD are drawn on the same base BC and between
the same parallels AD and BC.

Step 2: AM A BC and DN A BC are drawn.

AD A D AD

BM NC MB CN B CN
M
(i) (ii) (iii)

Step 3: The heights AM and DN and the base BC of each figure are measured. The areas of
∆ ABC and ∆ BCD are calculated.

Fig. Base Height Area of Base Height Area of Result
BC DN ∆ BCD
BC AM ∆ ABC ∆ ABC = ∆ BCD
(i) ∆ ABC = ∆ BCD
(ii) ∆ ABC = ∆ BCD
(iii)

Conclusion: Triangles on the same base and between the same parallels are equal in area.

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Geometry – Area of Triangles and Quadrilateral

Theoretical proof A E D

Given: ∆ ABC and ∆ BCD are on the same base BC and
between the same parallels BC and AD.

To prove: ∆ ABC = ∆ BCD

Construction: BE // CD is drawn. BE meets AD at E. BC

Proof

Statements Reasons

1. BCDE is a parallelogram 1. Opposite sides are parallel

2. ∆ ABC = 1 BCDE 2. They are on the same base and between the same
2 parallels.

3. ∆ BCD = 1 BCDE 3. Diagonal bisects a parallelogram.
2

4. ∆ ABC = ∆ BCD 4. From statements (2) and (3)

Alternative process A D Proved

Construction: AM A BC is drawn.

AM is the height of ∆ ABC and ∆ BCD.

MB C

Statements Reasons

1. Area of ∆ ABC = 1 BC × AM 1. Area of a triangle is 1 (base × height)
2 2

2. Area of ∆ BCD = 1 BC × AM 2. Area of a triangle is 1 (base × height)
2 2

3. ∆ ABC = ∆ BCD 3. From statements (1) and (2)

Corollary A Proved

Triangles on equal bases and between the same parallels are D
equal in area.
F
In the figure, AD // BF and BC = EF

So, area of ' ABC = area of ' DEF B CE

Worked-out examples

S R
T

Example 1: In the figure, the area of ' PQT is 28 cm2. Find the
area of ' PRS.
Solution:
(i) Area of P Q

PQRS = 2 area of ' PQT [' PQT and PQRS are standing on
= 2 × 28 cm2 the same base and between the same
= 56 cm2 parallels.]

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Geometry– Area of Triangles and Quadrilateral

(ii) Area of ' PRS = 1 Area of PQRS [Diagonal PR bisects PQRS]
2
1
= 2 × 56 cm2 = 28 cm2

So, the required area of ' PRS is 28 cm2. D E C
Example 2: In the given figure, E is the mid point of DC. If

the area of ABCD is 24 cm2, find the area of the

quadrilateral ABCE.

Solution: = 1 Area of ABCD = 1 × 24 cm2 = 12 cm2 A B
(i) Area of ' ABE 2 2

(ii) Area of ' BCE = Area of ' AED [Both the triangles are standing on the equal base

1 and between the same parallels.]
2
? Area of ' AED = × 12 cm2 = 6 cm2 [Area of (' BCE + ' AED) = 12 cm2]

(iii) Area of quadrilateral ABCE = Area of (' ABE + ' BCE)
= 12 cm2 + 6 cm2 = 18 cm2

So, the required area of the quadrilateral ABCE is 18 cm2.

Example 3: In the figure, ABCD is a rhombus in A DE
which the diagonals AC = 20 cm and
BD = 28 cm. If AD is produced to E, find the
area of ' BCE.

Solution: B C
S
(i) Area of rhombus ABCD = 1 × AC × BD T
2
1
= 2 × 20 × 28 cm2 = 280 cm2

(ii) Area of ' BCE = 1 area of ABCD
2
1
= 2 × 280 cm2 = 140 cm2

So, the required area of ' BCE is 140 cm2. P

Example 4: In the given figure, PQRS is a square with its Q
Solution: diagonal PR = 4 3 cm. If PS is produced to T, find
the area of ' TQR.

R

(i) Area of the square PQRS = 1 × PR2
2

= 1 × (4 3 cm)2 = 24 cm2
2 =
(ii) Area of ' TQR = 1 area of PQRS 1 × 24 cm2 = 12 cm2
2 2

So, the required area of ' TQR is 12cm2.

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Geometry – Area of Triangles and Quadrilateral

Example 5: In the given ∆ PQR, S and T are the mid–points of P
PQ and PR respectively. Prove that ∆ RPS = ∆ QPT.

Solution: In ∆ PQR, S is the mid–point of PQ and T is the S T
Given: R
mid–point of PR. O
To prove: ∆ RPS = ∆ QPT Q

Construction: S and T are joined.

Proof

Statements Reasons
ST joins the mid–points of two
1. ST // QR 1. sides of ∆ PQR
2. They are on the same base QR
2. ∆ QRT = ∆ QRS 3. and between QR // ST.
The same ∆ QRO is subtracted
3. ∆ QRT – ∆ QRO = ∆ QRS – ∆ QRO from both the sides of the
statement (2)
4. ∆ ORT = ∆ OQS 4. Remaining part of whole
The same quadrilateral PSOT is
5. ∆ ORT + quad. PSOT = ∆ OQS + quad.PSOT 5. added to both the sides of the
statement (4)
6. ∆ RPS = ∆ QPT 6. Whole part axiom

Example 6: In the given figure, O be any point on the diagonal D F Proved
BD of ABCD. If EF // AD and GH // CD, EF and GH
C
pass through O, prove that AEOH = CGOF. HO G

Solution: B

Given: ABCD is a parallelogram. O is any point on the diagonal A E
BD, EF // AD, GH // CD.

To prove: AEOH = CGOF.

Proof

Statements Reasons
1. EBGO is a parallelogram 1. Opposite sides are parallel.
2. ∆ EBO = ∆ BGO 2. Diagonal BO bisects EBGO.
3. DFOH is a parallelogram 3. Opposite sides are parallel.
4. ∆ HOD = ∆ OFD 4. Diagonal OD bisects DFOH.
5. ∆ ABD = ∆ BCD 5. Diagonal BD bisects ABCD.
6. ∆ ABD – ∆ EBO – ∆ HOD 6. The same triangles EBO and HOD are

= ∆ BCD – ∆ EBO – ∆ HOD subtracted from both the sides of (5).
7. ∆ ABD – ∆ EBO – ∆ HOD
7. From statements (2), (4) and (6)
= ∆ BCD – ∆ BGO – ∆ OFD
8. AEOH = CGOF 8. Remaining part of a whole

Proved

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Geometry– Area of Triangles and Quadrilateral

Example 7: In the given figure, PARM is a parallelogram. B is M R

any point on diagonal AM. Prove that 'PMB and X

' RMB are equal in area. B

Solution : PA

Given: PARM is a parallelogram. AM is its diagonal and B is any point on the diagonal.

To prove: Area of 'PMB = Area of 'RMB

Construction: P and R are joined. Diagonals AM and PR bisect each other at X.

Proof

Statements Reasons

1 X is the mid-point of diagonals AM and PR 1. From construction and given

2. 'BPX = 'BRX 2. BX is the median of ' BPR

3. 'PMX = 'RMX 3. MX is the median of ' PMR
4. 'BPX + 'PMX = 'BRX + 'RMX 4. Adding the statements (2) and (3)

5. 'PMB = 'RMB 5. Whole parts axiom
6. ? 'PMB and 'RMB are equal in area 6. From statement (5)

Example 8: In the given figure, D is the mid-point of side A Proved

BC of ' ABC, E is the mid-point of AD, F is the C
mid-point of AB and G is any point of BD. Prove that F E C

'ABC = 8'EFG. B GD
Solution :

Given: D, E and Fare the mid-points of BC, AD and AB respectively. A

G is any point of BD. FE

To prove: 'ABC = 8 'EFG

Construction: D and F are joined. B GD
Proof

Statements Reasons

1 EF//BD 1. E and F are the mid-points of the sides

AD and AB of ' ABD.

2. 'ABD = 1 'ABC 2. Median AD bisects ' ABC.
2

3. 'EFD = 'EFG 3. Both of them are standing on the same
base and between the same parallels.

4. 'ADF = 1 'ABD 4. Median DF bisects ' ABD.
2 5. Median FE bisects 'ADF

5. 'EFD = 1 'ADF
2

6. 'EFG = 1 × 1 'ABD 6. From statements (3), (4) and (5)
2 2

7. ' EFG = 21×12×12 ' ABC 7. From statements (1) and (6)

' EFG = 1 ' ABC. So, ' ABC = 8 ' EFG
8

Proved

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Geometry – Area of Triangles and Quadrilateral

EXERCISE 13.1

General section

1. Find the area of the shaded regions.

a) D E C b) S R c) 7.4 cm
D

12.5cm
10 cm

12 cm

A B PT Q A B
16.5 cm R T
18 cm 9.6 cm
S

2. a) In the figure, parallelogram PQSR and ' PQT are standing
on the same base and between, the same parallel lines. The
area of ' PQT is 50 cm2. Find the area of ' PRS.

P Q

AD

b) In the figure, the area of trapezium ABCD is 100 sq.cm

and the area of ' ADC is 40 sq.cm. Find the area of

' DEC. B EC

P W X
Z Y
c) In the figure, WXYZ is a rhombus in which XW
is produced to the point P. If WY = 10 cm and
XZ = 9 cm, find the area of 'PYZ.

P ST

d) In the given figure, PQRS is a square whose each side is
4 cm. Find the area of ' QRT.

Q R
D
A
E
e) In the figure, AE // BC, square ABCD and ' EBC are
standing on the same base BC and between the same C
parallels. If AC = 6 cm, find the area of ' EBC. B

B F

A

f) In the given figure, ABCD is a square whose perimeter is
40 cm and AB is produced to the point F. If E is the mid
point of DC, find the area of ' EFC.

DE C

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Geometry– Area of Triangles and Quadrilateral

D C

g) In the adjoining figure, AF // DC, ED // FC and ABCD is a EB
square. If AC = 5 2 cm, find the area of the parallelogram E
DEFC.

A F
D
A

h) In the given figure, if ABCE is a rectangle, find the area 8 cm
of ' ADE.
C
B 10 cm

P Q
M
i) In the adjoining figure, PS = 5 cm and SM = 8 cm.
Calculate the area of ' PQN. SR

N

Q

j) PQRS is a quadrilateral in which PR = 10 cm, perpendiculars P

from S and Q PR are 3.4 cm and 4.6cm respectively. Calculate MN
the area of the quadrilateral.

RS
Q

k) Find the area of the quadrilateral PQRS R A P
given in the adjoining figure in which B
3RB = 2PA = QS = 12 cm.

l) Calculate the area of the following trapeziums. S

(i) A 18 cm B (ii) 17 cm Q (iii) E 12 cm B

P A

13 cm 26 cm 13 cm

E D 23 cm C S 7 cm R D C
28 cm

m) In the given figure, ABCD is a trapezium. If AB = 10 cm, 10 cm A 16 cm D
BC = 22 cm, AD = 16 cm, AD // BC and DC A BC, calculate the B 22 cm C
area of ' ADC.

n) The areas of two parallelograms are equal and their altitudes are 6 cm and 9
cm. If the base of the first parallelogram is 12 cm, find the base of the second
parallelogram.

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Geometry – Area of Triangles and Quadrilateral
o) The areas of two parallelograms are equal. The altitude of one parallelogram is

4 cm and the base of the other is 6 cm. If the base of the first parallelogram is 9 cm,
find the height of the second parallelogram.

Creative section A DF E

3. a) In the adjoining figure, ABCD and C
BCEF are parallelograms, AE // BC. S
Prove that ABCD = BCEF.

B
P

b) In the figure alongside, QR // PS.
Prove that ' PQR = ' QRS.

c) In the given figure, DX // AB. Prove that QR
D CX

area of ' ABX = 1 area of ABCD.
2

d) In the given figure, ABCD is a parallelogram. AB
X is any point within it. Prove that the sum of DC
the areas of ' XCD and ' XAB is equal to half of
the area of ABCD. X
AB

M N

e) In the adjoining figure, triangles AMB and ANB O B
are standing on the same base AB and between A
the same parallel lines AB and MN. Prove that
area of ' AOM = area of 'BON.

M AN

f) In the given figure, 'ABC and parallelogram MBCN are on
the same base BC and between the same parallels MN and
BC. Prove that,

area of ' ABC = area of rectangle APCN. B P C
X C
D

g) In the given parallelogram ABCD, X and Y are B
any points on CD and AD respectively. Prove that Y
the area of ' AXB = the area of ' BYC.

A

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Geometry – Area of Triangles and Quadrilateral

A D
C
h) In the adjoining parallelogram ABCD, A is joined to any
point E on BC. AE and DC produced meet at F. Prove that
the area of 'BEF = the area of ' CDE

B E

F

P AS B

i) In the given figure, rectangle PQRS and
parallelogram AQRB are on the same base QR and
between the same parallels PB and QR. Prove that

(i) ' PQA # ' SRB Q R
(ii) Area of rectangle PQRS = Area of AQRB.

j) In the adjoining parallelogram ABCD, PQ // AB and AP D
RS // BC. Prove that the area of ROPA = the area of
QCSO. R S
O C

BQ

D MC
B
k) In the given figure, ABCD is a parallelogram. If M and N are N
any points on CD and DA respectively,
prove that ' AMB = ' CDN + 'ANB.

A

A PB
Q
l) In the given diagram, ABCD and PQRD are two
C
parallelograms. Prove that ABCD = PQRD. D R

AB

m) In the given figure, if AB // DC // EF, AD // BE and D HC
AF // DE, prove that parm. DEFH = parm . ABCD. G

EF

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