KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

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KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 1.
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 – \(\frac{1}{2}\)) (3a – \(\frac{1}{2}\))
(v) (1.1m – 0.4) (1.1m – 0.4)
(vi) (a2 + b2) (-a2 + b2)
(vii) (6x – 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix) \(\left(\frac{x}{2}+\frac{3 y}{4}\right)\left(\frac{x}{2}+\frac{3 y}{4}\right)\)
(x) (7a – 9b) (7a – 9b)
Solution:
(i) (x + 3) (x + 3) = (x + 3)2
Using Identity (a + b)2 = a2 + 2ab + b2
(x + 3)2
= x2 + 2(x)(3) + (3)2
= x2 + 6x + 9

(ii) (2y + 5) (2y + 5) = (2y + 5)2
Using Identity (a + b)2 = a2 + 2ab + b2
(2y + 5)2
= (2y)2 + 2(2y)(5) + (5)2
= 4y2 + 20y + 25

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(iii) (2a – 7) (2a – 7) = (2a – 7)2
Using Identity (a + b)2 = a2 + 2ab + b2
(2a + 7)2
= (2a)2 – 2(2a)(7) + (7)2
= 4a2 – 28a + 49

(iv) (3a – \(\frac{1}{2}\)) (3a – \(\frac{1}{2}\)) = \(\left(3 a-\frac{1}{2}\right)^{2}\)
Using Identity (a – b)2 = a2 – 2ab + b2
\(\left(3 a-\frac{1}{2}\right)^{2}\) = (3a)2 – 2(3a) (\(\frac{1}{2}\)) + \(\left(\frac{1}{2}\right)^{2}\)
= 9a2 – 3a + \(\frac{1}{4}\)

(v) (1.1m – 0.4) (1.1m – 0.4)
Using Identity (a – b) (a + b) = (a2 – b2)
(1.1m – 0.4) (1.1m – 0.4)
= (1.1m)2 – (0.4)2
= 1.21m2 – 0.16

(vi) (a2 + b2) (-a2 + b2)
Using Identity (a – b) (a + b) = (a2 – b2)
(b2 + a2) (b2 – a2)
= (b2)2 – (a2)2
= b4 – a4

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(vii) (6x – 7) (6x + 7)
Using Identity (a – b) (a + b) = (a2 – b2)
(6x – 7) (6x + 7)
= (6x)2 – (7)2
= 36x2 – 49

(viii) (-a+ c) (-a+ c) = (c – a) (c – a) = (c – a)2
Using Identity (a – b)2 = a2 – 2ab + b2
(c – a)2
= (c)2 – 2(c)(a) + (a)2
= c2 – 2ca + a2

(ix) \(\left(\frac{x}{2}+\frac{3 y}{4}\right)\left(\frac{x}{2}+\frac{3 y}{4}\right)=\left(\frac{x}{2}+\frac{3 y}{4}\right)^{2}\)
Using Identity (a + b)2 = a2 – 2ab + b2
\(\left(\frac{x}{2}+\frac{3 y}{4}\right)^{2}=\left(\frac{x}{2}\right)^{2}+2\left(\frac{x}{2}\right)\left(\frac{3 y}{4}\right)+\left(\frac{3 y}{4}\right)^{2}\)
= \(\frac{x^{2}}{2}+\frac{3}{4} x y+\frac{9}{16} y^{2}\)

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(x) (7a – 9b) (7a – 9b) = (7a – 9b)2
Using Identity (a – b)2 = a2 – 2ab + b2
(7a – 9b)2
= (7a)2 – 2(7a) (9b) + (9b)2
= 49a2 – 126ab + 81b2

Question 2.
Use the identity (x + a) (x + b) = x2 – (a + b) x + ab to find the following products.
(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) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)
Solution:
(i) (x + 3) (x + 7)
Using Identity (x + a) (x + b) = x2 – (a + b)x + ab
(x + 3) (x + 7)
= x2 + (3 + 7) x + (7)
= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)
Using Identity (x + a) (x + b) = x2 – (a + b) x + ab
(4x + 5) (4x + 1)
= (4x)2 + (5 + 1)4x + 5 × 1
= 16x2 + 24x + 5

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(iii) (4x – 5) (4x – 1)
Using Identity (x + a) (x + b) = x2 + (a + b) x + ab
(4x – 5) (4x – 1)
= (4x)2 + (5 + 1) (4x) + (-5) (-1)
= 16x2 + 24x + 5

(iv) (4x + 5) (4x – 1) = (4x + 5) [4x + (-1)]
Using Identity (x + a) (x + b) = x2 + (a + b)x + ab
(4x + 5) [4x + (-1)]
= (4x)2 – (5 + 1) (4x) + (-5) (-1)
= 16x2 – 24x – 5

(v) (2x + 5y) (2x + 3y)
Using Identity (x + a) (x + b) = x2 + (a + b)x + ab
(2x + 5y) (2x + 3y)
= (2x)2 + (5y + 3y) (2x) + (5y) (3y)
= 4x2 + 16xy + 15y2

(vi) (2a2 + 9) (2a2 + 5)
Using Identity (x + a) (x + b) = x2 + (a + b)x + ab
(2a2 + 9) (2a2 + 5)
= (2a)2 + (9 + 5) (2a2) + 9 × 5
= 4a4 + 28a2 + 45

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(vii) (xyz – 4) (xyz – 2)
Using Identity (x + a) (x + b) = x2 + (a + b) x + ab
(xyz – 4) (xyz – 2)
= (xyz)2 + (-4 – 2)xyz + (-4) (-2)
= x2y2z2 – 6xyz + 8

Question 3.
Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy – 3z)2
(iii) (6x2 – 5y)2
(iv) \(\left(\frac{2}{3} m+\frac{3}{2} n\right)^{2}\)
(v) (0.4p – 0.5q)2
Solution:
(i) (b – 7)2
Using Identity (a – b)2 = a2 – 2ab + b2
(b – 7)2
= b2 – 2(b)(7) + (7)2
= b2 – 14b + 49

(ii) (xy – 3z)2
Using Identity (a – b)2 = a2 – 2ab + b2
(xy – 3z)2
= (xy)2 + 2(xy) (3z) + (3z)2
= x2y2 + 6xyz + 9z2

(iii) (6x2 – 5y)2
Using Identity (a – b)2 = a2 – 2ab + b2
(6x2 – by)2
= (6x2)2 – 2(6x2) (5y) + (5y)2
= 36x4 – 60x2y + 25y2

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(iv) \(\left(\frac{2}{3} m+\frac{3}{2} n\right)^{2}\)
Using Identity (a – b)2 = a2 – 2ab + b2
\(\left(\frac{2}{3} m+\frac{3}{2} n\right)^{2}=\left(\frac{2}{3} m\right)^{2}+2\left(\frac{2}{3} m\right)\left(\frac{3}{2} n\right)+\left(\frac{3}{2} n\right)^{2}\)
= \(\frac{4}{9} m^{2}+2 m n+\frac{9}{4} n^{2}\)

(v) (0.4p – 0.5q)2
Using Identity (a – b)2 = a2 – 2ab + b2
(0.4p – 0.5q)2
= (0.4p)2 – 2(0.4p) (0.5q) + (0.5q)2
= 0.16p2 – 0.40pq + 0.25q2

(vi) (2xy – 5y)2
Using Identity (a + b)2 = a2 + 2(a)(b) + b2
(2xy + 5y)2
= (2xy)2 + 2(2xy) (5y) + (5y)2
= 4x2y2 + 20xy2 + 25y2

Question 4.
Prove the following.
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2 + 180pq = (9p + 5q)2
(iii) \(\left(\frac{4}{3} m-\frac{3}{4} n\right)^{2}+2 m n=\frac{16}{9} m^{2}+\frac{9}{16} n^{2}\)
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
(i) (3x + 7)2 – 84x = (3x – 7)2
L.H.S. = (3x + 7)2 – 84x
= (3x)2 + (7)2 + 2(3x) (7) – 84x
= 9x2 + 49 + 42x – 84x
= 9x2 – 42x + 49
= (3x)2 – 2(3x) (7) + (7)2
= (3x – 7)2
= R.H.S.

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(ii) (9p – 5q)2 + 180pq = (9p + 5q)2
L.H.S. = (9p – 5q)2 + 180pq
= (9p)2 + (5q)2 – 2(9p) (5q) + 180pq
= 81p2 + 25q2 – 90pq + 180pq
= (9p)2 + (5q)2 + 2(9p) (5q)
= (9p + 5q)2
= R.H.S.

(iii) \(\left(\frac{4}{3} m-\frac{3}{4} n\right)^{2}+2 m n=\frac{16}{9} m^{2}+\frac{9}{16} n^{2}\)
L.H.S. = \(\left(\frac{4}{3} m\right)^{2}+\left(\frac{3}{4} n\right)^{2}-2 \times \frac{4}{3} m \times \frac{3}{4} n+2 m n\)
= \(\frac{16}{9} m^{2}+\frac{9}{16} n^{2}-2 \times \frac{4}{3} \times \frac{3}{4} m n+2 m n\)
= \(\frac{16}{9} m^{2}+\frac{9}{16} n^{2}-2 m n+2 m n\)
= \(\left(\frac{16}{9} m^{2}+\frac{9}{16} n^{2}\right)\)
= R.H.S.

(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
L.H.S. = (4pq + 3q)2 – (4pq – 3q)2
= (4pq)2 + (3q)2 + 2(4pq) (3q) – [(4pq)2 + (3q)2 – 2(4pq) (3q)]
= (16p2q2 + 9q2 + 24pq2) – (16p2q2 + 9q2 – 24pq2)
= 16p2q2 + 9q2 + 24pq2 – 16p2q2 – 9q2 + 24pq2
= 24pq2 + 24pq2
= 48pq2
= R.H.S.

(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
L.H.S. = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a2 – b2 + b2 – c2 + c2 – a2
= 0
= R.H.S.

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 5.
Solve
(i) 712
(ii) 992
(iii) 1022
(iv) 9982
(v) 5.22
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.92
(ix) 1.05 × 9.5
Solution:
(i) (71)2 = (70 + 1)2
(a + b)2 = a2 + 2ab + b2
= (70)2 + 2(70) (1) + (1)2
= 4900 + 140 + 1
= 5041

(ii) (99)2 = (100 – 1)2
(a – b)2 = a2 – 2ab + b2
= (100)2 – 2(100) (1) + (1)2
= 10000 – 200 + 1
= 9800 + 1
= 9801

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(iii) (102)2 = (100 + 2)2
(a + b)2 = a2 + 2ab + b2
= (100)2 + 2(100) (2) + (2)2
= 10000 + 400 + 4
= 10404

(iv) 9982 = (1000 – 2)2
(a – b)2 = a2 – 2ab + b2
= (1000)2 – 2(1000) (2) + 22
= 1000000 – 4000 + 4
= 996000 + 4
= 996004

(v) (5.22) = (5.0 + 0.2)2
(a + b)2 = a2 + 2ab + b2
= (5.0)2 + 2(5.0) (0.2) + (0.2)2
= 25 + 2.0 + 0.04
= 27.04

(vi) 297 × 303 = (300 – 3) × (300 + 3)
(a – b) (a + b) = (a2 – b2)
= (300)2 – (3)2
= 90000 – 9
= 89991

(vii) 78 × 82 = (80 – 2) (80 + 2)
(a – b) (a + b) = (a2 – b2)
= (80)2 – (2)2
= 6500 – 4
= 6396

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(viii) 8.02 = (10 – 1.1)2
(a – b)2 = a2 – 2ab + b2
= (10)2 – 2(10) (1.1) + (1.1)2
= 100 – 22.0 + 1.21
= 78.00 + 1.21
= 79.21

(ix) 1.05 × 9.5 = (1 + 0.5) (1 – 0.5)
= 1 × 10 – 1 × 0.5 + 0.5 × 10 – 0.5 × 0.5
= 10 – 0.5 + 0.5 – 0.25
= 10 – 0.0 – 0.25
= 10 – 0.75
= 9.925

Question 7.
Solve the following:
(i) 512 – 492
(ii) (1.02)2 – (0.98)2
(iii) 1532 – 1472
(iv) 12.12 – 7.92
Solution:
(i) 512 – 492
(a2 – b2) = (a – b) (a + b)
= (51 – 49)(51 + 49)
= (2)(100)
= 200

(ii) (1.02)2 – (0.98)2
(a2 – b2) = (a – b) (a + b)
= (1.02 – 0.98) (1.02 + 0.98)
= (0.04) (2.00)
= 0.08

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(iii) 1532 – 1472
(a2 – b2) = (a – b) (a + b)
= (153 – 147) (153 + 147)
= (6) (300)
= 1800

(iv) 12.12 – 7.92
(a2 – b2) = (a – b) (a + b)
= (12.1 – 7.9) (12.1 + 7.9)
= (4.2) (20)
= 84.0

Question 8.
Solve the following using the distributive property.
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
Solution:
(i) 103 × 104 = (100 + 3) (100 + 4)
(x + a) (x + b) = x2 + (a + b)x + ab
= (100)2 + (3 + 4) (100) + 3 × 4
= 10000 + 700 + 12
= 10712

(ii) 5.1 × 5.2
= (5 + 0.1) (5 + 0.2)
(x + a) (x + b) = x2 + (a + b) x + ab
= (5)2 + (0.1 + 0.2) (5) + 0.1 × 0.2
= 25 + (0.3) × 5 + 0.02
= 25 + 1.5 + 0.02
= 26.52

KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

(iii) 103 × 98
= (100 + 3) (100 – 2)
(x + a) (x – b) = x2 + (a – b) x – ab
= (100)2 + (3 – 2) (100) – 3 × 2
= 10000 + 100 – 6
= 10100 – 6
= 10094

(iv) 9.7 × 9.8
= (9.7 – 0.7) (9.0 + 0.8)
(x + a) (x + b) = x2 + (a + b) x + ab
= (9.0)2 + (0.7 + 0.8) 9.0 + 0.7 × 0.8
= 81.0 + 13.5 + 0.56
= 94.50 + 0.56
= 95.06