Notes
Worksheets
Ncert Solutions
In this page we have Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 for
Exercise 9.5. Hope you like them and do not forget to like , social share
and comment at the end of the page.
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-1/2)(3a-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) [ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
(x) (7a – 9b) (7a – 9b)
Answer:
We will be using below identities in these question
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a – b)(a + b) = a2 – b2
1) This question can be solved using the first identify
(x + 3) (x + 3)
=x
2 + 6x + 9
2) This question can be solved using the first identify
(2y + 5) (2y + 5)
=4y
2 + 20y + 25
3) This question can be solved using the second identify
(2a – 7) (2a – 7)
= 4a
2 – 28a + 49
4) This question can be solved using the second identify
(3a-1/2)(3a-1/2)= 9a
2 -3a+(1/4)
5) This question can be solved using the third identity
(1.1m – 0.4) (1.1m + 0.4)
= 1.21m
2 – 0.16
5) This question can be solved using the third identity
(a
2+ b
2) (– a
2+ b
2)
= (b
2 + a
2 ) (b
2 – a
2)
= a
4 - b
4
6) This question can be solved using the third identity
(6x – 7) (6x + 7)
=36x
2 - 49
7) This question can be solved using the third identity
= c
2 - a
2
8) This question can be solved using the first identity
[ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
=(x
2/4) + (9y
2/16) +(3xy/4)
9) This question can be solved using second identity
= 49a
2 – 126ab + 81b
2
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)
Answer:
1) x
2 + (3+7)x + 21
= x
2 + 10x + 21
2) (4x + 5) (4x + 1)
= 16x
2 + (5 + 1)4x + 5
= 16x
2 + 24x + 5
3) (4x – 5) (4x – 1)
= 16x
2 + (-5-1)4x + 5
= 16x
2 - 20x + 5
4) (4x + 5) (4x – 1)
= 16x
2 + (5-1)4x - 5
= 16x
2 +16x - 5
5) (2x + 5y) (2x + 3y)
= 4x
2 + (5y + 3y)4x + 15y
2
= 4x
2 + 32xy + 15y
2
6) (2a
2+ 9) (2a
2+ 5)
= 4a
4 + (9+5)2a
2 + 45
= 4a
4 + 28a
2 + 45
7) (xyz – 4) (xyz – 2)
= x
2y
2z
2 + (-4 -2)xyz - 8
= x
2y
2z
2 - 6xyz – 8
Question 3
Find the following squares by using the identities.
(i) (b – 7)
2
(ii) (xy + 3z)
2
(iii) (6x
2– 5y)
2
iv) [(2m/3}) + (3n/2)]
2
(v) (0.4p – 0.5q)
2
(vi) (2xy + 5y)
Answer:
- b2 - 14b + 49
- x2y2 + 6xyz + 9z2
- 36x4 - 60x2y + 25y2
- (4m2/9) +(9n2/4) +2mn
- 0.16p2 - 0.4pq + 0.25q2
- 4x2y2 + 20xy2 + 25y2
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– 2ab²c
(vii) (m
2 – n
2m)
2 + 2m
3n
2
Answer:
1) a
4 - b
4
2) (2x + 5)
2 – (2x – 5)
2
= 4x
2 + 20x +25 - (4x
2 - 20x + 25)
= 4x
2 + 20x + 25 - 4x
2 + 20x - 25
= 40
3) (7m – 8n)
2+ (7m + 8n)
2
= 49m
2 - 112mn + 64n
2 + 49m
2 + 112mn + 49n
2
= 98m
2 + 128n
2
4) (4m + 5n)
2 + (5m + 4n)
2
= 16m
2 + 40mn + 25n
2 + 25m
2 + 40mn + 16n
2
= 41m
2 + 80mn + 41n
2
5) (2.5p – 1.5q)
2 – (1.5p – 2.5q)
2
= 6.25p
2 - 7.5pq + 2.25q
2 - 2.25p
2 + 7.5pq - 6.25q
2
= 4p
2 - 4q
2
6) (ab + bc)
2– 2ab²c
= a
2b
2 + 2ab
2c + b
2c
2 - 2ab
2c
= a
2b
2 + b
2c
2
7) (m
2 – n
2m)
2 + 2m
3n
2
= m
4 - 2m
3n
2 + m
2n
4 + 2m
3n
2
= m
4 + m
2n
4
Question 5
Show that.
(i) (3x + 7)
2 – 84x = (3x – 7)
2
(ii) (9p – 5q)
2+ 180pq = (9p + 5q)
2
iv) (4pq + 3q)
2– (4pq – 3q)
2 = 48pq
2
v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Answer:
1) LHS = 9x
2 + 42x + 49 - 84x
= 9x
2 - 42x + 49
RHS = 9x
2 - 42x + 49
LHS = RHS
2) LHS = 91p
2 - 90pq + 25q
2 + 180pq
= 91p
2 + 90pq + 25q
2
RHS = 91p
2 + 90pq + 25q
2
4) LHS= 16p
2q
2 + 24pq
2 + 9q
2 - 16p
2q
2 + 24pq
2 - 9q
2
= 48pq
2
5) LHS= a
2 - b
2 + b
2 - c
2 + c
2 - a
2
= 0
Question 6
Using identities, evaluate.
(i) 71²
(ii)
99²
(iii) 102
2
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.9
2
(ix) 10.5 x 9.5
Answer:
We will be using below identities in these question
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a – b)(a + b) = a2 – b2
1) 71
2 = (70+1)
2
= 70
2 + 140 + 1
2
= 4900 + 140 +1= 5041
2) 99²
= (100 -1)2
= 1002 - 200 + 12
= 10000 - 200 + 1
= 9801
3) 102
2= (100 + 2)
2
= 100
2 + 400 + 2
2
= 10000 + 400 + 4
= 10404
4) 998
2= (1000 - 2)
2
= 1000
2 - 4000 + 2
2
= 1000000 - 4000 + 4
= 996004
5) 5.2
2= (5 + 0.2)
2
= 5
2 + 2 + 0.2
2
= 25 + 2 + 0.4
= 27.4
6) = (300 - 3 )(300 + 3)
= 300
2 - 3
2
= 90000 - 9
= 89991
7) = (80 - 2)(80 + 2)
= 80
2 - 2
2
= 6400 - 4
= 6396
8) 8.9
2= (9 - 0.1)
2
= 9
2 - 1.8 + 0.1
2
= 81 - 1.8 + 0.01
= 79.21
9) = (10 + 0.5)(10 - 0.5)
= 10
2 - 0.5
2
= 100 - 0.25
= 99.75
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
Answer:
1) = (51 + 49)(51 - 49)
= 100 x 2
= 200
2) = (1.02 + 0.98)(1.02 - 0.98)
= 2 x 0.04
= 0.08
3) = (153 + 147)(153 - 147)
= 300 x 6
= 1800
4) = (12.1 + 7.9)(12.1 - 7.9)
= 20 x 4.2
= 84
Question 8
Using (x + a) (x + b) = x
2+ (a + b) x + ab, find
(i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8
Answer:
1) = (100 + 3)(100 + 4)
= 100
2 + (3 + 4)100 + 12
= 10000 + 1200 + 12
= 11212
2) = (5 + 0.1)(5 + 0.2)
= 5
2 + (0.1+0.2)5 + 0.02
= 25 + 1.5 + 0.02
= 26.52
3) = (100 + 3)(100 - 2)
= 100
2 + (3-2)100 - 6
= 10000 + 100 - 6
= 10094
4) = (9 + 0.7 )(9 + 0.8)
= 9
2 + (0.7 + 0.8)9 + 0.63
= 81 + 13.5 + 0.63
= 95.13
Download Exercise 9.5 as pdf
Class 8 Maths
Class 8 Science
