Using identities, evaluate
i) 712 ii) 992 iii) 1022 iv) 9982 v) 5.22
vi) 297 × 303 vii) 78 × 82 viii) 8.92 ix) 10.5 × 9.5
Solution:
The three basic algebraic identities, which we will be using to evaluate the expressions are as follows.
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b)(a - b) = a2 - b2
(i) 712
= (70 + 1)2
= (70)2 + 2(70)(1) + (1)2 [Since, (a + b)2 = a2 + 2ab + b2]
= 4900 + 140 + 1
= 5041
(ii) 992
= (100 - 1)2
= (100)2 - 2(100)(1) + (1)2 [Since, (a - b)2 = a2 - 2ab + b2]
= 10000 - 200 + 1
= 9801
(iii) 1022
= (100 + 2)2
= (100)2 + 2(100)(2) + (2)2 [Since, (a + b)2 = a2 + 2ab + b2]
= 10000 + 400 + 4
= 10404
(iv) 9982
= (1000 - 2)2
= (1000)2 - 2(1000)(2) + (2)2 [Since, (a - b)2 = a2 - 2ab + b2]
= 1000000 - 4000 + 4
= 996004
(v) (5.2)2
= (5.0 + 0.2)2
= (5.0)2 + 2(5.0)(0.2) + (0.2)2 [Since, (a + b)2 = a2 + 2ab + b2]
= 25 + 2 + 0.04
= 27.04
(vi) 297 × 303
= (300 - 3) × (300 + 3) [Since, (a + b)(a - b) = a2 - b2]
= (300)2 - (3)2
= 90000 - 9
= 89991
(vii) 78 × 82
= (80 - 2) × (80 + 2) [Since, (a + b)(a - b) = a2 - b2]
= (80)2 - (2)2
= 6400 - 4
= 6396
(viii) 8.92
= (9.0 - 0.1)2
= (9.0)2 - 2(9.0)(0.1) + (0.1)2 [Since, (a - b)2 = a2 - 2ab + b2]
= 81 - 1.8 + 0.01
= 79.21
(ix) 10.5 × 9.5
= (10 + 0.5) × (10 - 0.5)
= 102 - 0.52 [Since, (a + b)(a - b) = a2 - b2]
= 100 - 0.25
= 99.75
☛ Check: NCERT Solutions for Class 8 Maths Chapter 9
Video Solution:
Using identities, evaluate. i) 71² ii) 99² iii) 102² iv) 998² v) (5.2)² vi) 297 × 303 vii) 78 × 82 viii) 8.9² ix) 10.5 × 9.5
NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.5 Question 6
Summary:
Using identities, the following expressions i) 712 ii) 992 iii) 1022 iv) 9982 v) (5.2)2 vi) 297 × 303 vii) 78 × 82 viii) 8.92 ix) 10.5 × 9.5 are evaluated as follows i) 5041 ii) 9801 iii) 10404 iv) 996004 v) 27.04 vi) 89991 vii) 6396 viii) 79.21 ix) 99.75
☛ Related Questions:
- 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) (1.1 m - 0.4)(1.1 m + 0.4) (vi)(a2 + b2)(-a2 + b2) (vii) (6x - 7)(6x + 7) (viii) (-a + c)(-a + c) (ix) (x/2 + 3y/4)(x/2 + 3y/4) (x) (7a - 9b)(7a - 9b)
- 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 + 3 y) (vi) (2a2 + 9)(2a2 + 5) (vii) (xyz - 4)(xyz - 2)
- Find the following squares by using the identities. (i)(b - 7)2 (ii) (xy + 3z)2 (iii) (6x2 - 5 y)2 (iv) (2m/3 + 3n/2)2 (v) (0.4 p - 0.5q)2 (vi) (2xy + 5 y)2
- Simplify (i)(a2 - b2)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) (m2 - n2m)2 + 2m3n2
visual curriculum