These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.2
Question 1.
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution:
(i) 322
= (30 + 2)2
= 302 + 2 (30) (2) + 22 [∵ (a + b)2 = a2 + 2ab + b2 ]
= 900 + 120 + 4
= 1024
(ii) 352
= (30 + 5)2
= 302 + 2 (30) (5) + 52
= 900 + 300 + 25
= 1225
(iii) 862
= (80 + 6)2
= 802 + 2 (80) (6) + 62
= 6400 + 960 + 36
= 7396
(iv) 932
= (90 + 3)2
= 902 + 2 (90) (3) + 32
= 8100 + 540 + 9
= 8649
(v) 712
= (70 + 1)2
= 702 + 2(70) (1) + 12
= 4900 + 140 + 1
= 5041
(vi) 462
= (40 + 6)2
= 402 + 2 (40) (6) + 62
= 1600 + 480 + 36
= 2116
Question 2.
Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
(i) Let 2n = 6
n = \(\frac{6}{2}\) = 3
n2 – 1
= 32 – 1
= 8
n2 + 1
= 32 + 1
= 10
∴ The required Pythagorean triplet is 6, 8 and 10.
(ii) Let 2n = 14
n = \(\frac{14}{2}\) = 7
n2 – 1
= 72 – 1
= 49 – 1
= 48
n2 + 1
= 72 + 1
= 49 + 1
= 50
∴ Thus, the required Pythagorean triplet is 14, 48 and 50.
(iii) Let 2n = 16
n = \(\frac{16}{2}\) = 8
n2 – 1
= 82 – 1
= 64 – 1
= 63
n2 + 1
= 82 + 1
= 64 + 1
= 65
∴ The required Pythagorean triplet is 16, 63 and 65.
(iv) Let 2n = 18
n = \(\frac{18}{2}\) = 9
n2 – 1
= 92 – 1
= 81 – 1
= 80
n2 + 1
= 92 + 1
= 81 + 1
= 82
∴ The required Pythagorean triplet is 18, 80 and 82.