In this page we have *class 8 maths exercise 8.5 solutions * for
Algebraic Expressions.This exercise has questions about Standard identities and how to apply the identities to solve various problems. Hope you like them and do not forget to like , social share
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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

(vii) (6x – 7) (6x + 7)

(viii) (– a + c) (– a + c)

(ix) [ (x/2)+(3y/4)] [ (x/2)+(3y/4)]

(x) (7a – 9b) (7a – 9b)

We will be using below identities in these question

(i) This question can be solved using the first identify

(x + 3) (x + 3)

=x

(ii) This question can be solved using the first identify

(2y + 5) (2y + 5)

=4y

(iii) This question can be solved using the second identify

(2a – 7) (2a – 7)

= 4a

(iv) This question can be solved using the second identify

(3a-1/2)(3a-1/2)= 9a

(v) This question can be solved using the third identity

(1.1m – 0.4) (1.1m + 0.4)

= 1.21m

(vi) This question can be solved using the third identity

(a

= (b

= a

(vii) This question can be solved using the third identity

(6x – 7) (6x + 7)

=36x

(viii) This question can be solved using the third identity

= c

(ix) This question can be solved using the first identity

[ (x/2)+(3y/4)] [ (x/2)+(3y/4)]

=(x

(x) This question can be solved using second identity

= 49a

Use the identity (x + a) (x + b) = x

(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

(vii) (xyz – 4) (xyz – 2)

(i) x

= x

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

= 16x

= 16x

(iii) (4x – 5) (4x – 1)

= 16x

= 16x

(iv) (4x + 5) (4x – 1)

= 16x

= 16x

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

= 4x

= 4x

(vi) (2a

= 4a

= 4a

(vii) (xyz – 4) (xyz – 2)

= x

= x

Find the following squares by using the identities.

(i) (b – 7)

(ii) (xy + 3z)

(iii) (6x

(iv) [(2m/3}) + (3n/2)]

(v) (0.4p – 0.5q)

(vi) (2xy + 5y)

- b
^{2}- 14b + 49 - x
^{2}y^{2}+ 6xyz + 9z^{2} - 36x
^{4}- 60x^{2}y + 25y^{2} - (4m
^{2}/9) +(9n^{2}/4) +2mn - 0.16p
^{2}- 0.4pq + 0.25q^{2} - 4x
^{2}y^{2}+ 20xy^{2}+ 25y^{2}

Simplify.

(i) (a

(ii) (2x + 5)

(iii) (7m – 8n)

(iv) (4m + 5n)

(v) (2.5p – 1.5q)

(vi) (ab + bc)

(vii) (m

(i) a

(ii) (2x + 5)

= 4x

= 4x

= 40

(iii) (7m – 8n)

= 49m

= 98m

(iv) (4m + 5n)

= 16m

= 41m

(v) (2.5p – 1.5q)

= 6.25p

= 4p

(vi) (ab + bc)

= a

= a

(vii) (m

= m

= m

Show that.

(i) (3x + 7)

(ii) (9p – 5q)

(iii)

(iv) (4pq + 3q)

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

(i) LHS = 9x

= 9x

RHS = 9x

LHS = RHS

(ii) LHS = 91p

= 91p

RHS = 91p

(iii) LHS = $\frac {16}{9}m^2 + \frac {9}{16}n^2 -2mn + 2mn=\frac {16}{9)m^2 + \frac {9}{16}n^2$ RHS=$\frac {16}{9}m^2 + \frac {9}{16}n^2$ (iv) LHS= 16p

= 48pq

(v) LHS= a

= 0

Using identities, evaluate.

(i) 71²

(ii)

(iii) 102

(iv) 998²

(v) 5.2²

(vi) 297 x 303

(vii) 78 x 82

(viii) 8.9

(ix) 10.5 x 9.5

We will be using below identities in these question

(i)71

= 70

= 4900 + 140 +1= 5041

(ii) 99²

(iii) 102

= 100

= 10000 + 400 + 4

= 10404

(iv) 998

= 1000

= 1000000 - 4000 + 4

= 996004

(v) 5.2

= 5

= 25 + 2 + 0.4

= 27.4

(vi) = (300 - 3 )(300 + 3)

= 300

= 90000 - 9

= 89991

(vii) = (80 - 2)(80 + 2)

= 80

= 6400 - 4

= 6396

(viii) 8.9

= 9

= 81 - 1.8 + 0.01

= 79.21

(ix) = (10 + 0.5)(10 - 0.5)

= 10

= 100 - 0.25

= 99.75

Using a

(i) 51

(ii) (1.02)

(iii) 153

(iv) 12.1

(i) = (51 + 49)(51 - 49)

= 100 x 2

= 200

(ii) = (1.02 + 0.98)(1.02 - 0.98)

= 2 x 0.04

= 0.08

(iii) = (153 + 147)(153 - 147)

= 300 x 6

= 1800

(iv) = (12.1 + 7.9)(12.1 - 7.9)

= 20 x 4.2

= 84

Using (x + a) (x + b) = x

(i) 103 x 104

(ii) 5.1 x 5.2

(iii) 103 x 98

(iv) 9.7 x 9.8

(i) = (100 + 3)(100 + 4)

= 100

= 10000 + 1200 + 12

= 11212

(ii) = (5 + 0.1)(5 + 0.2)

= 5

= 25 + 1.5 + 0.02

= 26.52

(iii) = (100 + 3)(100 - 2)

= 100

= 10000 + 100 - 6

= 10094

(iv) = (9 + 0.7 )(9 + 0.8)

= 9

= 81 + 13.5 + 0.63

= 95.13

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