NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities

NCERT
SOLUTIONS
CLASS - 8TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-9 Maths

Class : 8th
Subject : Maths
Chapter : 9
Chapter Name : Algebraic Expressions and Identities

Exercise 9.1

Q1 Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz − 3zy
2

(ii) 1 + x + x 2

(iii) 4x y − 4x y z + z
2 2 2 2 2 2

(iv) 3 − pq + qr − rp
x y
(v) + − xy
2 2

(vi) 0.3a − 0.6ab + 0.5b

Answer. The terms and the respective coefficients of the given expressions are as follows.

Page : 140 , Block Name : Exercise 9.1

Q2 Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not

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fit in any of these three categories?
2 3 4 2 2 3
x + y, 1000, x + x + x + x , 7 + y + 5x, 2y − 3y , 2y − 3y + 4y , 5x − 4y + 3xy

2 2 2
4z − 15z , ab + bc + cd + da, pqr, p q + pq , 2p + 2q

Answer. The given expressions are classified as
Monomials: 1000, pqr
Binomials: x + y, 2y − 3y , 4z − 15z , p q + pq , 2p + 2q
2 2 2 2

Trinomials: 7 + y + 5x, 2y − 3y + 4y , 5x − 4y + 3xy 2 3

Polynomials that do not fit in any of these categories are
2 3 4
x + x + x + x , ab + bc + cd + da

Page : 140 , Block Name : Exercise 9.1

Q3 Add the following
(i) ab − bc, bc − ca, ca − ab
(ii) a − b + ab, b − c + bc, c − a + ac
(iii) 2p q − 3pq + 4, 5 + 7pq − 3p q
2 2 2 2

2 2 2 2 2 2
l + m ,m + n ,n + l
(iv)
2lm + 2mn + 2nl

Answer. The given expressions written in separate rows, With like terms one below the other
and then the addition of these expressions are as follows.

Page : 140 , Block Name : Exercise 9.1

Q4 (a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3

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(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
2 2
4p q − 3pq + 5pq − 8p + 7q − 10 from
(c) Subtract 2 2
18 − 3p − 11q + 5pq − 2pq + 5p q

Answer. The given expressions in separate rows, with like terms one below the other and
then the subtraction of these expressions is as follows.

Page : 140 , Block Name : Exercise 9.1

Exercise 9.2

Q1 Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) −4p, 7p
(iii) −4p, 7pq
(iv) 4p , −3p
3

(v) 4p, 0

Answer. The product will be as follows.
(i) 4 × 7p = 4 × 7 × p = 28p
(ii) −4p × 7p = −4 × p × 7 × p = (−4 × 7) × (p × p) = −28p 2

(iii) −4p × 7pq = −4 × p × 7 × p × q = (−4 × 7) × (p × p × q) = −28p q 2

(iv) 4p × −3p = 4 × (−3) × p × p × p × p = −12p
3 4

(v) 4p × 0 = 4 × p × 0 = 0

Page : 143 , Block Name : Exercise 9.2

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Q2 Find the areas of rectangles with the following pairs of monomials as their lengths and breadths
respectively.
2 2 2
(p, q); (10m, 5n); (20x , 5y ) ; (4x, 3x ) ; (3mn, 4np)

Answer. We know that,
Area of rectangle = Length x Breadth
Area of 1 rectangle = p × q = pq
at

Area of 2 r rectangle = 10m × 5n = 10 × 5 × m × n = 50mn
nd

Area of 3 rectangle = 20x × 5y = 20 × 5 × x × y = 100x y
rd 2 2 2 2 2 2

Area of 4 rectangle = 4x × 3x = 4 × 3 × x × x = 12x
th 2 2 3

Area of 5 rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn p
th 2

Page : 143 , Block Name : Exercise 9.2

Q3 Complete the table of products.

Answer. The tabel can be completed as follows.

Page : 144 , Block Name : Exercise 9.2

Q4 Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a , 7a
2 4

(ii) 2p, 4q, 8r
(iii) xy, 2x y, 2xy
2 2

(iv) a, 2b, 3c

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Answer. We know that,
Volume = Length x Breadth X Height
(i) Volume = 5a × 3a × 7a = 5 × 3 × 7 × a × a × a = 105a
2 4 2 4 7

(ii) Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr
(iii) Volume = xy × 2x y × 2xy = 2 × 2 × xy × x y × xy = 4x y
2 2 2 2 4 4

(iv) Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc

Page : 144 , Block Name : Exercise 9.2

Q5 Obtain the product of
(i) xy, yz, zx
(ii) a, −a , a
2 3

(iii) 2, 4y, 8y , 16y
2 3

(iv) a, 2b, 3c, 6abc
(v) m, −mn, mmp

Answer. (i) xy × yz × zx = x y z 2 2 2

(ii) a × (−a ) × a = −a
2 3 6

(iii) 2 × 4y × 8y × 16y = 2 × 4 × 8 × 16 × y × y × y = 1024y
2 3 2 3 6

(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a b c 2 2 2

(v) m × (−mn) × mnp = −m n p 3 2

Page : 144 , Block Name : Exercise 9.2

Exercise 9.3

Q1 Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a − b
(iii) a + b, 7a b 2 2

(iv) a − 9, 4a
2

(v) pq + qr + rp, 0

Answer. (i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
(ii) (ab) × (a − b) = (ab × a) + [ab × (−b)] = a b − ab 2 2

(iii) (a + b) × (7a b ) = (a × 7a b ) + (b × 7a b ) = 7a b + 7a b
2 2 2 2 2 2 3 2 2 3

(iv) (a − 9) × (4a) = (a × 4a) + (−9) × (4a) = 4a − 36a
2 2 3

(v) (pq + qr + rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) = 0

Page : 146 , Block Name : Exercise 9.3

Q2 Complete the table.

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Answer. The table can be completed as follows.

Page : 146 , Block Name : Exercise 9.3

Q3 Find the product.
(i) (a ) × (2a ) × (4a
2 22 26
)

−9
(ii) ( 2

3
xy) × (
10
2
x y )
2

(iii) (−
10 3 6 3
pq ) × ( p q)
3 5

(iv) x × x × x × x 2 3 4

Answer. (i) (a ) × (2a 2 22
) × (4a
26
) = 2 × 4 × a
2
× a
22
× a
26
= 8a
50

2 −9 2 −9 −3
(ii) ( 3
xy) × (
10
2
x y ) = (
2

3
) × (
10
) × x × y × x
2
× y
2
=
5
3
x y
3

−10 6 −10 6
(iii) ( 3
3
pq ) × (
5
3
p q) = (
3
) × (
5
) × pq
3 3
× p q = −4p q
4 4

(iv) x × x × x × x 2 3 4
= x
10

Page : 146 , Block Name : Exercise 9.3

Q4 (a) Simplify 3x(4x − 5) + 3 and find its values for
(i)x = 3
1
(ii)x =
2

2
a (a + a + 1) + 5 and find its value for \)\((i)a = 0, \)\((ii)a = 1
(b) Simplify
(iii) a = −1

Answer. (a) 3x(4x − 5) + 3 = 12x − 15x + 3 2

(i) For x = 3, 12x − 15x + 3 = 12(3) − 15(3) + 3
2 2

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= 108 - 45 + 3
= 66
2

(ii) For x =
1 2 1 1
, 12x − 15x + 3 = 12( ) − 15 ( ) + 3
2 2 2

1 15
= 12 × − + 3
4 2
15 15
= 3 − + 3 = 6 −
2 2
12−15 −3
= =
2 2

(b) a (a + a + 1) + 5 = a + a + a + 5
2 3 2

(i) For a = 0, a + a + a + 5 = 0 + 0 + 0 + 5 = 5
3 2

(ii) For a = 1, a + a + a + 5 = (1) + (1) + 1 + 5
3 2 3 2

= 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a + a + a + 5 = (−1) + (−1) + (−1) + 5
3 2 3 2

= −1 + 1 − 1 + 5 = 4

Page : 146 , Block Name : Exercise 9.3

Q5 (a) Add: p(p − q), q(q − r) and r(r − p)
(b) Add: 2x(z − x − y) and 2y(z − y − x)
(c) Subtract: 3l(l − 4m + 5n) from 4l(10n − 3m + 2l)
(d) Subtract: 3a(a + b + c) − 2b(a − b + c) from 4c(−a + b + c)

Answer. (a) First expression = p(p − q) = p − pq 2

Second expression = q(q − r) = q − qr 2

Third expression = r(r − p) = r − pr 2

Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p + q + r − pq − qr − rp. 2 2 2

(b) First expression = 2x(z − x − y) = 2xz − 2x − 2xy 2

Second expression = 2y(z − y − x) = 2yz − 2y − 2yx 2

Adding the two expressions, we obtain

Therefore, the sum of the given expressions is −2x − 2y 2 2
− 4xy + 2yz + 2zx .
(c) 3l(l − 4m + 5n) = 3f − 12/m + 15 ln 2

2
4/(10n − 3m + 2l) = 40/n − 12lm + 8t

Subtracting these expressions , we obtain

Therefore, the result is 5f + 25/n 2

(d) 3a(a + b + c) − 2b(a − b + c) = 3a + 3ab + 3ac − 2ba + 2b − 2bc 2 2

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2 2
= 3a + 2b + ab + 3ac − 2bc
2
4c(−a + b + c) = −4ac + 4bc + 4c

Subtracting these expressions, we obtain

Therefore, the result is −3a − 2b + 4c − ab + 6bc − 7ac.
2 2 2

Page : 146 , Block Name : Exercise 9.3

Exercise 9.4

Q1 Multiply the binomials.
(i) (2x + 5) and (4x − 3)
(ii) (y − 8) and (3y − 4)
(iii) (2.5l − 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q ) and (3pq − 2q )
2 2

3
(vi) ( 2 2 2 2 2
a + 3b ) and 4 (a − b )
4 3

Answer. (i) (2x + 5) × (4x − 3) = 2x × (4x − 3) + 5 × (4x − 3)
2
= 8x − 6x + 20x − 15
2
= 8x + 14x − 15( By adding like terms)

(ii) (y − 8) × (3y − 4) = y × (3y − 4) − 8 × (3y − 4)
2
= 3y − 4y − 24y + 32
2
= 3y − 28y + 32( By adding like terms )

(iii) (2.51 − 0.5m) × (2.5/ + 0.5m) = 2.5/ × (2.5/ + 0.5m) − 0.5m(2.5/ + 0.5m)
2 2
= 6.25l + 1.25/m − 1.25/m − 0.25m
2 2
= 6.25l − 0.25m

(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q ) × (3pq − 2q ) = 2pq × (3pq − 2q ) + 3q
2 2 2 2
× (3pq − 2q )
2

2 2 3 3 4
= 6p q − 4pq + 9pq − 6q
2 2 3 4
= 6p q + 5pq − 6q

(vi) (
3 2 2 2 2 2 3 2 2 2 8 2
a + 3b ) × [4 (a − b )] = ( a + 3b ) × (4a − b )
4 3 4 3

3 2 2 8 2 2 2 8 2
= a × (4a − b ) + 3b × (4a − b )
4 3 3

4 2 2 2 2 4
= 3a − 2a b + 12b a − 8b
4 2 2 4
= 3a + 10a b − 8b

Page : 148 , Block Name : Exercise 9.4

Q2 Find the product.
(i) (5 − 2x)(3 + x)
(ii) (x + 7y)(7x − y)

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(iii) (a + b) (a + b )
2 2

(iv) (p − q ) (2p + q)
2 2

Answer. (i) (5 − 2x)(3 + x) = 5(3 + x) − 2x(3 + x)
2
= 15 + 5x − 6x − 2x
2
= 15 − x − 2x

(ii) (x + 7y)(7x − y) = x(7x − y) + 7y(7x − y)
2 2
= 7x − xy + 49xy − 7y
2 2
= 7x + 48xy − 7y

(iii) (a + b) (a + b ) = a (a + b ) + b (a + b )
2 2 2 2 2

3 2 2 3
= a + a b + ab + b

(iv) (p − q ) (2p + q) = p (2p + q) − q (2p + q)
2 2 2 2

3 2 2 3
= 2p + p q − 2pq − q

Page : 148 , Block Name : Exercise 9.4

Q3 Simplify:
(i) (x − 5) (x + 5) + 25
2

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

(iii) (t + s ) (t − s) 2 2

(iv) (a + b)(c − d) + (a − b)(c + d) + 2(ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x − y)
(vi) (x + y) (x − xy + y ) 2 2

(vii) (1.5x − 4y)(1.5x + 4y + 3) − 4.5x + 12y
(viii) (a + b + c)(a + b − c)

Answer. (i) (x − 5) (x + 5) + 25 2

2
= x (x + 5) − 5(x + 5) + 25

3 2
= x + 5x − 5x − 25 + 25

3 2
= x + 5x − 5x

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

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

2 3 2 3
= a b + 3a + 5b + 15 + 5

2 3 2 3
= a b + 3a + 5b + 20

(iii) (t + s ) (t − s) 2 2

2 2 2
= t (t − s) + s (t − s)

3 2 2 3
= t − st + s t − s

(iv) (a + b)(c − d) + (a − b)(c + d) + 2(ac + bd)
= a(c − d) + b(c − d) + a(c + d) − b(c + d) + 2(ac + bd)

= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd

= (ac + ac + 2ac) + (ad − ad) + (bc − bc) + (2bd − bd − bd)

= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x − y)
= x(2x + y) + y(2x + y) + x(x − y) + 2y(x − y)

2 2 2 2
= 2x + xy + 2xy + y + x − xy + 2xy − 2y

2 2 2 2
= (2x + x ) + (y − 2y ) + (xy + 2xy − xy + 2xy)

2 2
= 3x − y + 4xy

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(vi) (x + y) (x − xy + y ) 2 2

2 2 2 2
= x (x − xy + y ) + y (x − xy + y )

3 2 2 2 2 3
= x − x y + xy + x y − xy + y

3 3 2 2 2 2
= x + y + (xy − xy ) + (x y − x y)

3 3
= x + y

(vii) (1.5x − 4y)(1.5x + 4y + 3) − 4.5x + 12y
= 1.5x(1.5x + 4y + 3) − 4y(1.5x + 4y + 3) − 4.5x + 12y

2 2
= 2.25x + 6xy + 4.5x − 6xy − 16y − 12y − 4.5x + 12y

2 2
= 2.25x + (6xy − 6xy) + (4.5x − 4.5x) − 16y + (12y − 12y)

2 2
= 2.25x − 16y

(viii) (a + b + c)(a + b − c)
= a(a + b − c) + b(a + b − c) + c(a + b − c)

2 2 2
= a + ab − ac + ab + b − bc + ca + bc − c

2 2 2
= a + b − c + (ab + ab) + (bc − bc) + (ca − ca)

2 2 2
= a + b − c + 2ab

Page : 148 , Block Name : Exercise 9.4

Exercise 9.5

Q1 Use a suitable identity to get each of the following products.
(i) (x + 3)(x + 3)
(ii) (2y + 5)(2y + 5)
(iii) (2a − 7)(2a − 7)
(iv) (3a − 1

2
) (3a −
1

2
)

(v) (1.1m − 0.4)(1.1m + 0.4)
(vi) (a + b ) (−a + b )
2 2 2 2

(vii) (6x − 7)(6x + 7)
(viii) (−a + c)(−a + c)
3y 3y
(ix) (
x x
+ )( + )
2 4 2 4

(x) (7a − 9b)(7a − 9b)

Answer. The products will be as follows.
2
(i) (x + 3)(x + 3) = (x + 3)

2 2 2 2 2
= (x) + 2(x) + (3) [(a + b) = a + 2ab + b ]

2
= x + 6x + 9
2
( ii) (2y + 5)(2y + 5) = (2y + 5)

2 2 2 2 2
= (2y) + 2(2y)(5) + (5) [(a + b) = a + 2ab + b ]

2
= 4y + 20y + 25
2
( iii) (2a − 7)(2a − 7) = (2a − 7)

2 2 2 2 2
= (2a) − 2(2a)(7) + (7) [(a − b) = a − 2ab + b ]

2
= 4a − 28a + 49

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2

(iv) (3a −
1 1 1
) (3a − ) = (3a − )
2 2 2

2
2 1 1
= (3a) − 2(3a) ( ) + ( )
2 2 2 2 2
[(a−b) =a −2ab+b ]

2 1
= 9a − 3a +
4

(v) (1.1m − 0.4)(1.1m + 0.4)
2 2 2 2
= (1.1m) − (0.4) [(a + b)(a − b) = a − b ]

2
= 1.21m − 0.16

(vi) (a + b ) (−a + b ) = (b + a ) (b − a )
2 2 2 2 2 2 2 2

2 2
2 2 2 2
= (b ) − (a ) [(a + b)(a − b) = a − b ]

4 4
= b − a

(vii) (6x − 7)(6x + 7) = (6x) − (7) [(a + b)(a − b) = a − b ] 2 2 2 2

2
= 36x − 49

(viii) (−a + c)(−a + c) = (−a + c) 2

2 2 2 2 2
= (−a) + 2(−a)(c) + (c) [(a + b) = a + 2ab + b ]

2 2
= a − 2ac + c
2
3y 3y 3y
(ix) ( x

2
+
4
)(
x

2
+
4
) = (
x

2
+
4
)

2 2
x x 3y 3y
= ( ) + 2( )( ) + ( )
2 2 4 4 2 2 2
[(a+b) =a +2ab+b ]

2 2
x 3xy 9y
= + +
4 4 16

(x) (7a − 9b)(7a − 9b) = (7a − 9b) 2

2 2 2 2 2
= (7a) − 2(7a)(9b) + (9b) [(a − b) = a − 2ab + b ]

2 2
= 49a − 126ab + 81b

Page : 151 , Block Name : Exercise 9.5

Q2 Use the identity (x + a)(x + b) = x + (a + b)x + ab to find the following products. 2

(i) (x + 3)(x + 7)
(ii) (4x + 5)(4x + 1)
(iii) (4x − 5)(4x − 1)
(iv) (4x + 5)(4x − 1)
(v) (2x + 5y)(2x + 3y)
(vi) (2a + 9) (2a + 5)
2 2

(vii) (xyz − 4)(xyz − 2)

Answer. The products will be as follows.
(i) (x + 3)(x + 7) = x + (3 + 7)x + (3)(7) 2

2
= x + 10x + 21

(ii) (4x + 5)(4x + 1) = (4x) + (5 + 1)(4x) + (5)(1) 2

2
= 16x + 24x + 5

(iii) (4x − 5)(4x − 1) = (4x) + [(−5) + (−1)](4x) + (−5)(−1) 2

2
= 16x − 24x + 5

(iv) (4x + 5)(4x − 1) = (4x) + [(5) + (−1)](4x) + (5)(−1) 2

2
= 16x + 16x − 5

(v) (2x + 5y)(2x + 3y) = (2x) + (5y + 3y)(2x) + (5y)(3y) 2

2 2
= 4x + 16xy + 15y
2
(vi) (2a + 9) (2a + 5) = (2a ) + (9 + 5) (2a ) + (9)(5)
2 2 2 2

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4 2
= 4a + 28a + 45

(vii) (xyz − 4)(xyz − 2)
2
= (xyz) + [(−4) + (−2)](xyz) + (−4)(−2)

2 2 2
= x y z − 6xyz + 8

Page : 151 , Block Name : Exercise 9.5

Q3 Find the following squares by using the identities.
(i) (b − 7) 2

(ii) (xy + 3z) 2

2
(iii) (6x − 5y) 2

2

(iv) (
2 3
m + n)
3 2

(v) (0.4p − 0.5q) 2

(vi) (2xy + 5y) 2

Answer. (i) (b − 7) 2
= (b)
2
− 2(b)(7) + (7)
2
[(a − b)
2
= a
2
− 2ab + b ]
2

2
= b − 14b + 49

(ii) (xy + 3z) 2
= (xy)
2
+ 2(xy)(3z) + (3z)
2
[(a + b)
2
= a
2
+ 2ab + b ]
2

2 2 2
= x y + 6xyz + 9z
2 2
(iii) (6x − 5y) 2
= (6x )
2 2
− 2 (6x ) (5y) + (5y)
2
[(a − b)
2
= a
2
− 2ab + b ]
2

4 2 2
= 36x − 60x y + 25y
2 2 2
3 3 3
(iv) ( 2

3
m +
2
n) = (
2

3
m) + 2(
2

3
m) (
2
n) + (
2
n)
2 2 2
[(a+b) =a +2ab+b ]

4 2 9 2
= m + 2mn + n
9 4

(v) (0.4p − 0.5q) 2
= (0.4p)
2
− 2(0.4p)(0.5q) + (0.5q)
2

2 2 2
[(a − b) = a − 2ab + b ]
2 2
= 0.16p − 0.4pq + 0.25q

(vi) (2xy + 5y) 2
= (2xy)
2
+ 2(2xy)(5y) + (5y)
2

2 2 2
[(a + b) = a + 2ab + b ]

2 2 2 2
= 4x y + 20xy + 25y

Page : 151 , Block Name : Exercise 9.5

Q4 Simplify.
2
(i) (a − b )2 2

(ii) (2x + 5) − (2x − 5) 2 2

(iii) (7m − 8n) + (7m + 8n) 2 2

(iv) (4m + 5n) + (5m + 4n) 2 2

(v) (2.5p − 1.5q) − (1.5p − 2.5q) 2 2

(vi) (ab + bc) − 2ab c 2 2

2
(vii) (m − n m) + 2m n
2 2 3 2

2 2 2
Answer. (i) (a − b ) 2 2
= (a )
2 2 2
− 2 (a ) (b ) + (b )
2
[(a − b)
2
= a
2
− 2ab + b ]
2

4 2 2 4
= a − 2a b + b

(ii) (2x + 5) − (2x − 5) 2 2
= (2x)
2
+ 2(2x)(5) + (5)
2
− [(2x)
2
− 2(2x)(5) + (5) ]
2

Page 12 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-9 Maths
2 2 2
[(a − b) = a − 2ab + b ]

2 2 2
[(a + b) = a + 2ab + b ]

2 2
= 4x + 20x + 25 − [4x − 20x + 25]

2 2
= 4x + 20x + 25 − 4x + 20x − 25 = 40x

(iii) (7m − 8n) + (7m + 8n) 2 2

2 2 2 2
= (7m) − 2(7m)(8n) + (8n) + (7m) + 2(7m)(8n) + (8n)

2 2 2 2 2 2
[(a − b) = a − 2ab + b and (a + b) = a + 2ab + b ]

2 2 2 2
= 49m − 112mn + 64n + 49m + 112mn + 64n

2 2
= 98m + 128n

(iv) (4m + 5n) + (5m + 4n) 2 2

2 2 2 2
= (4m) + 2(4m)(5n) + (5n) + (5m) + 2(5m)(4n) + (4n)

2 2 2
[(a + b) = a + 2ab + b ]

2 2 2 2
= 16m + 40mn + 25n + 25m + 40mn + 16n

2 2
= 41m + 80mn + 41n

(v) (2.5p − 1.5q) − (1.5p − 2.5q) 2 2

2 2 2 2
= (2.5p) − 2(2.5p)(1.5q) + (1.5q) − [(1.5p) − 2(1.5p)(2.5q) + (2.5q) ]

2 2 2
[(a − b) = a − 2ab + b ]

2 2 2 2
= 6.25p − 7.5pq + 2.25q − [2.25p − 7.5pq + 6.25q ]

2 2 2 2
= 6.25p − 7.5pq + 2.25q − 2.25p + 7.5pq − 6.25q

2 2
= 4p − 4q

(vi) (ab + bc) − 2ab c 2 2

2 2 2 2 2 2
= (ab) + 2(ab)(bc) + (bc) − 2ab c [(a + b) = a + 2ab + b ]

2 2 2 2 2 2
= a b + 2ab c + b c − 2ab c

2 2 2 2
= a b + b c
2
(vii) (m − n m) + 2m n
2 2 3 2

2 2
2 2 2 2 3 2 2 2 2
= (m ) − 2 (m ) (n m) + (n m) + 2m n [(a − b) = a − 2ab + b ]

4 3 2 4 2 3 2
= m − 2m n + n m + 2m n

4 4 2
= m + n m

Page : 151 , Block Name : Exercise 9.5

Q5 Show that
(i) (3x + 7) − 84x = (3x − 7)2 2

(ii) (9p − 5q) + 180pq = (9p + 5q)2 2

2

(iii) ( 4

3
m −
3

4
n) + 2mn =
16

9
m
2
+
9

16
n
2

(iv) (4pq + 3q) − (4pq − 3q) = 48pq 2 2 2

(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)(c + a) = 0

Answer. (i) L.H.S = (3x + 7) − 84x 2

2 2
= (3x) + 2(3x)(7) + (7) − 84x

2
= 9x + 42x + 49 − 84x

2
= 9x − 42x + 49

R.H.S = (3x − 7) 2
= (3x)
2
− 2(3x)(7) + (7)
2

Page 13 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-9 Maths
2
= 9x − 42x + 49

L.H.S = R.H.S
(ii) L.H.S = (9p − 5q) + 180pq 2

2 2
= (9p) − 2(9p)(5q) + (5q) − 180pq

2 2
= 81p − 90pq + 25q + 180pq

2 2
= 81p + 90pq + 25q

2
R.H.S = (9p + 5q)
2 2
= (9p) + 2(9p)(5q) + (5q)

2 2
= 81p + 90pq + 25q

L.H.S = R.H.S
2
4 3
(iii) L.H.S = = ( 3
m −
4
n) + 2mn

2 2
4 4 3 3
= ( m) − 2( m) ( n) + ( n) + 2mn
3 3 4 4

16 2 9 2
= m − 2mn + n + 2mn
9 16

16 2 9 2
= m + n = R. H. S
9 16

(iv) L.H.S = (4pq + 3q) − (4pq − 3q) 2 2

2 2 2 2
= (4pq) + 2(4pq)(3q) + (3q) − [(4pq) − 2(4pq)(3q) + (3q) ]

2 2 2 2 2 2 2 2
= 16p q + 24pq + 9q − [16p q − 24pq + 9q ]

2 2 2 2 2 2 2 2
= 16p q + 24pq + 9q − 16p q + 24pq − 9q

2
= 48pq = R. H . S

(v) L.H.S = (a − b)(a + b) + (b − c)(b + c) + (c − a)(c + a)
2 2 2 2 2 2
= (a − b ) + (b − c ) + (c − a ) = 0 = R. H . S

Page : 151 , Block Name : Exercise 9.5

Q6 Using identities, evaluate.
(i) 71 2

(ii) 99 2

(iii) 102 2

(iv) 998 2

(v) 5.2 2

(vi) 297 × 303
(vii) 78 × 82
(viii) 8.9 2

(ix) 1.05 × 9.5

Answer.
2 2
(i)71 = (70 + 1)

2 2 2 2 2
= (70) + 2(70)(1) + (1) [(a + b) = a + 2ab + b ]

= 4900 + 140 + 1 = 5041
2 2
(ii)99 = (100 − 1)

2 2 2 2 2
= (100) − 2(100)(1) + (1) [(a − b) = a − 2ab + b ]

= 10000 − 200 + 1 = 9801
2 2
(iii)102 = (100 + 2)

2 2 2 2 2
= (100) + 2(100)(2) + (2) [(a + b) = a + 2ab + b ]

= 10000 + 400 + 4 = 10404

Page 14 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-9 Maths
2 2
(iv) 998 = (1000 − 2)

2 2 2 2 2
= (1000) − 2(1000)(2) + (2) [(a − b) = a − 2ab + b ]

= 1000000 − 4000 + 4 = 996004
2 2
(v)(5.2) = (5.0 + 0.2)

(v) = (5.0) + 2(5.0)(0.2) + (0.2) [(a + b)
2 2 2
= a
2 2
+ 2ab + b ]

= 25 + 2 + 0.04 = 27.04

vi)297 × 303 = (300 − 3) × (300 + 3)

2 2 2 2
= (300) − (3) [(a + b)(a − b) = a − b ]

= 90000 − 9 = 89991

(vii)78 × 82 = (80 − 2)(80 + 2)

2 2 2 2
= (80) − (2) [(a + b)(a − b) = a − b ]

= 6400 − 4 = 6396
2 2
(viii) 8.9 = (9.0 − 0.1)

2 2 2 2 2
= (9.0) − 2(9.0)(0.1) + (0.1) [(a − b) = a − 2ab + b ]

= 81 − 1.8 + 0.01 = 79.21

(ix)1.05 × 9.5 = 1.05 × 0.95 × 10

= (1 + 0.05)(1 − 0.05) × 10

2 2
= [(1) − (0.05) ] × 10

2 2
= [1 − 0.0025] × 10 [(a + b)(a − b) = a − b ]

= 0.9975 × 10 = 9.975

Page : 152 , Block Name : Exercise 9.5

Q7 Using a − b = (a + b)(a − b), find:
2 2

(i) 51 − 49
2 2

(ii) (1.02) − (0.98) 2 2

(iii) 153 − 147
2 2

(iv) 12.1 − 7.9
2 2

Answer.
2 2
( i) 51 − 49 = (51 + 49)(51 − 49)

= (100)(2) = 200
2 2
(ii) (1.02) − (0.98) = (1.02 + 0.98)(1.02 − 0.98)

= (2)(0.04) = 0.08
2 2
(iii) 153 − 147 = (153 + 147)(153 − 147)

= (300)(6) = 1800
2 2
( iv )12.1 − 7.9 = (12.1 + 7.9)(12.1 − 7.9)

= (20.0)(4.2) = 84

Page : 152 , Block Name : Exercise 9.5

Q8 Using (x + a)(x + b) = x + (a + b)x + ab, find 2

(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8

Page 15 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-9 Maths

Answer.
(i) 103 × 104 = (100 + 3)(100 + 4)

2
= (100) + (3 + 4)(100) + (3)(4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2 = (5 + 0.1)(5 + 0.2)

2
= (5) + (0.1 + 0.2)(5) + (0.1)(0.2)

= 25 + 1.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3)(100 − 2)

2
= (100) + [3 + (−2)](100) + (3)(−2)

= 10000 + 100 − 6

= 10094

( iv )9.7 × 9.8 = (10 − 0.3)(10 − 0.2)

2
= (10) + [(−0.3) + (−0.2)](10) + (−0.3)(−0.2)

= 100 + (−0.5)10 + 0.06 = 100.06 − 5 = 95.06

Page : 152 , Block Name : Exercise 9.5

Page 16 of 16 Aglasem Schools