## MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.2

**MP Board Class 8 Maths Chapter 6 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) (30 + 2)

= 30 (30 + 2) + 2 (30 + 2)

= 900 + 60 + 60 + 4 = 1024.

(ii) 35^{2} = (30 + 5)^{2} = (30 + 5) (30 + 5)

= 30(30 + 5) + 5 (30 + 5)

= 900 + 150 + 150 + 25 = 1225.

(iii) 86^{2} = (80 + 6)^{2} = (80 + 6) (80 + 6)

= 80(80 + 6) + 6(80 + 6)

= 6400 + 480 + 480 + 36 = 7396.

(iv) 93^{2} = (90 + 3)^{2} = (90 + 3) (90 + 3)

= 90(90 + 3) + 3(90 + 3)

= 8100 + 270 + 270 + 9 = 8649.

(v) 71^{2} = (70+ 1)^{2} = (70 + 1) (70 + 1)

= 70(70 + 1) + 1 (70 + 1)

= 4900 + 70 + 70 + 1 = 5041.

(vi) 46^{2} = (40 + 6)^{2} = (40 + 6) (40 + 6)

= 40(40 + 6) + 6(40 + 6)

= 1600 + 240 + 240 + 36 = 2116.

**8th Class Maths Chapter 6 Exercise 6.2 Question 2.**

Write a Pythagorean triplet whose one member is

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solution:

We can get Pythagorean triplet by using general form 2m, m^{2} – 1, m^{2} + 1.

(i) Let us take 2m = 6 ⇒ m = 3

Thus, m^{2} – 1 = 3^{2} – 1 = 9 – 1 = 8 and m^{2} + 1 = 3^{2} + 1 = 9 + 1 = 10.

∴ The required triplet is 6, 8, 10

(ii) Let us take 2m = 14 ⇒ m = 7

Thus, m^{2} – 1 = 7^{2} – 1 = 49 – 1 = 48

and m^{2} + 1 = 7^{2} + 1 = 49 + 1 = 50

∴ The required triplet is 14, 48, 50.

(iii) Let us take 2m = 16 ⇒ m = 8

Thus, m^{2} – 1 = 8^{2} – 1 = 64 – 1 = 63

and m^{2} + 1 = 8^{2} + 1 = 64 + 1 = 65.

∴ The required triplet is 16, 63, 65.

(iv) Let us take 2m = 18 ⇒ m = 9

Thus, m^{2} – 1 = 9^{2} – 1 = 81 – 1 = 80

and m^{2} + 1 = 9^{2} + 1 = 81 + 1 = 82

∴ The required triplet is 18, 80, 82.