Evaluate using suitable identities:
(i) (48)²
(ii) 181² - 19²
(iii) 497 × 505
(iv) 2.07 × 1.93
Solution:
(i) (48)²
We can write (48)² as (50 - 2)²
We have (a - b)² = a² - 2ab + b²
Here a = 50 and b = 2
∴ (50 - 2)² = (50)² - (2 × 50 × 2) + (2)²
= 2500 - 200 + 4
= 2504 - 200
= 2304
(ii) 181² - 19²
We have the identity: a² - b² = (a - b) (a + b)
Here a = 181 and b = 19
∴ 181² - 19² = (181 - 19) (181 + 19)
= 162 × 200
= 32400
(iii) 497 × 505
We can write 497 × 505 as (500 - 3) (500 + 5)
Using the identity (x + a) (x + b) = x² + (a + b) x + ab
= 500² + [(-3 + 5) × 500] + [(-3) (5)]
= 250000 + 1000 - 15
= 250985
(iv) 2.07 × 1.93
We can write 2.07 × 1.93 as (2 + 0.07) (2 - 0.07)
Using the identity (a + b) (a - b) = a² - b²
Here a = 2 and b = 0.07
∴ (2 + 0.07) (2 - 0.07)
= 2² - (0.07)²
= 3.9951
✦ Try This: Evaluate using suitable identities: (i) 271² - 29², (ii) 294 × 306
(i) 271² - 29²
We have the identity: a² - b² = (a - b) (a + b)
Here a = 271 and b = 29
∴ 271² - 29² = (271 - 29) (271 + 29)
= 242 × 300
= 72600
(ii) 291 × 306
We can write 291 × 306 as (300 - 9) (300 + 6)
Using the identity (x + a) (x + b) = x² + (a + b) x + ab
= 300² + [(-9 + 6) × 300] + [(-9) (6)]
= 90000 - 900 - 54
= 89046
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9
NCERT Exemplar Class 8 Maths Chapter 7 Sample Problem 13
Evaluate using suitable identities: (i) (48)² (ii) 181² - 19² (iii) 497 × 505 (iv) 2.07 × 1.93
Summary:
Evaluating (i) (48)², (ii) 181² - 19², (iii) 497 × 505, (iv) 2.07 × 1.93, using identities we get 2304, 32400, 250985 and 3.9951 respectively
☛ Related Questions:
visual curriculum