NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- Class 8 Maths Algebraic Expressions and Identities Exercise 9.5
- Algebraic Expressions and Identities Class 8 Extra Questions
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5
Ex 9.5 Class 8 Maths 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) (a
2
+ b
2
) (-a
2
+ b
2
)
(vii) (6x – 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix) (\(\frac { x }{ 2 }\) + \(\frac { 3y }{ 4 }\)) (\(\frac { x }{ 2 }\) + \(\frac { 3y }{ 4 }\))
(x) (7a – 9b) (7a – 9b)
Solution:
Ex 9.5 Class 8 Maths Question 2.
Use the identity (x + a)(x + b) = x
2
+ (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) (2a
2
+ 9) (2a
2
+ 5)
(vii) (xyz – 4) (xyz – 2)
Solution:
Ex 9.5 Class 8 Maths Question 3.
Find the following squares by using the identities.
(i) (b – 7)
2
(ii) (xy + 3z)
2
(iii) (6x
2
– 5y)
2
(iv) (\(\frac { 2 }{ 3 }\) m + \(\frac { 3 }{ 2 }\) n)
2
(v) (0.4p – 0.5q)
2
(vi) (2xy + 5y)
2
Solution:
Ex 9.5 Class 8 Maths Question 4.
Simplify:
(i) (a
2
– b
2
)
2
(ii) (2x + 5)
2
– (2x – 5)
2
(iii) (7m – 8n)
2
+ (7m + 8n)
2
(iv) (4m + 5n)
2
+ (5m + 4n)
2
(v) (2.5p – 1.5q)
2
– (1.5p – 2.5q)
2
(vi) (ab + bc)
2
– 2ab
2
c
(vii) (m
2
– n
2
m)
2
+ 2m
3
n
2
Solution:
Ex 9.5 Class 8 Maths Question 5.
Show that:
(i) (3x + 7)
2
– 84x = (3x – 7)
2
(ii) (9p – 5q)
2
+ 180pq = (9p + 5q)
2
(iii) (\(\frac { 4 }{ 3 }\) m – \(\frac { 3 }{ 4 }\) n)
2
+ 2mn = \(\frac { 16 }{ 9 }\) m
2
+ \(\frac { 9 }{ 16 }\) n
2
(iv) (4pq + 3q)
2
– (4pq – 3q)
2
= 48pq
2
(v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
Ex 9.5 Class 8 Maths Question 6.
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
Solution:
Ex 9.5 Class 8 Maths Question 7.
Using a
2
– b
2
= (a + b) (a – b), find
(i) 51
2
– 49
2
(ii) (1.02)
2
– (0.98)
2
(iii) 153
2
– 147
2
(iv) 12.1
2
– 7.9
2
Solution:
(i) 51
2
– 49
2
= (51 + 49) (51 – 49) = 100 × 2 = 200
(ii) (1.02)
2
– (0.98)
2
= (1.02 + 0.98) (1.02 – 0.98) = 2.00 × 0.04 = 0.08
(iii) 153
2
– 147
2
= (153 + 147) (153 – 147) = 300 × 6 = 1800
(iv) 12.1
2
– 7.9
2
= (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84
Ex 9.5 Class 8 Maths Question 8.
Using (x + a) (x + b) = x
2
+ (a + b)x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
Solution:
(i) 103 × 104 = (100 + 3)(100 + 4) = (100)
2
+ (3 + 4) (100) + 3 × 4 = 10000 + 700 + 12 = 10712
(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5)
2
+ (0.1 + 0.2) (5) + 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52
(iii) 103 × 98 = (100 + 3) (100 – 2) = (100)
2
+ (3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)
2
– (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06
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