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) 32^{2}

= (30 + 2)^{2}

= 30^{2} + 2 (30) (2) + 2^{2} [∵ (a + b)^{2} = a^{2} + 2ab + b^{2} ]

= 900 + 120 + 4

= 1024

(ii) 35^{2}

= (30 + 5)^{2}

= 30^{2} + 2 (30) (5) + 5^{2}

= 900 + 300 + 25

= 1225

(iii) 86^{2}

= (80 + 6)^{2}

= 80^{2} + 2 (80) (6) + 6^{2}

= 6400 + 960 + 36

= 7396

(iv) 93^{2}

= (90 + 3)^{2}

= 90^{2} + 2 (90) (3) + 3^{2}

= 8100 + 540 + 9

= 8649

(v) 712

= (70 + 1)^{2}

= 70^{2} + 2(70) (1) + 1^{2}

= 4900 + 140 + 1

= 5041

(vi) 46^{2}

= (40 + 6)^{2}

= 40^{2} + 2 (40) (6) + 6^{2}

= 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

n^{2} – 1

= 3^{2} – 1

= 8

n^{2} + 1

= 3^{2} + 1

= 10

∴ The required Pythagorean triplet is 6, 8 and 10.

(ii) Let 2n = 14

n = \(\frac{14}{2}\) = 7

n^{2} – 1

= 7^{2} – 1

= 49 – 1

= 48

n^{2} + 1

= 7^{2} + 1

= 49 + 1

= 50

∴ Thus, the required Pythagorean triplet is 14, 48 and 50.

(iii) Let 2n = 16

n = \(\frac{16}{2}\) = 8

n^{2} – 1

= 8^{2} – 1

= 64 – 1

= 63

n^{2} + 1

= 8^{2} + 1

= 64 + 1

= 65

∴ The required Pythagorean triplet is 16, 63 and 65.

(iv) Let 2n = 18

n = \(\frac{18}{2}\) = 9

n^{2} – 1

= 9^{2} – 1

= 81 – 1

= 80

n^{2} + 1

= 9^{2} + 1

= 81 + 1

= 82

∴ The required Pythagorean triplet is 18, 80 and 82.