## MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 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 = (30 + 2) (30 + 2)
= 30 (30 + 2) + 2 (30 + 2)
= 900 + 60 + 60 + 4 = 1024.
(ii) 352 = (30 + 5)2 = (30 + 5) (30 + 5)
= 30(30 + 5) + 5 (30 + 5)
= 900 + 150 + 150 + 25 = 1225.
(iii) 862 = (80 + 6)2 = (80 + 6) (80 + 6)
= 80(80 + 6) + 6(80 + 6)
= 6400 + 480 + 480 + 36 = 7396.
(iv) 932 = (90 + 3)2 = (90 + 3) (90 + 3)
= 90(90 + 3) + 3(90 + 3)
= 8100 + 270 + 270 + 9 = 8649.
(v) 712 = (70+ 1)2 = (70 + 1) (70 + 1)
= 70(70 + 1) + 1 (70 + 1)
= 4900 + 70 + 70 + 1 = 5041.
(vi) 462 = (40 + 6)2 = (40 + 6) (40 + 6)
= 40(40 + 6) + 6(40 + 6)
= 1600 + 240 + 240 + 36 = 2116. 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, m2 – 1, m2 + 1.
(i) Let us take 2m = 6 ⇒ m = 3
Thus, m2 – 1 = 32 – 1 = 9 – 1 = 8 and m2 + 1 = 32 + 1 = 9 + 1 = 10.
∴ The required triplet is 6, 8, 10

(ii) Let us take 2m = 14 ⇒ m = 7
Thus, m2 – 1 = 72 – 1 = 49 – 1 = 48
and m2 + 1 = 72 + 1 = 49 + 1 = 50
∴ The required triplet is 14, 48, 50.

(iii) Let us take 2m = 16 ⇒ m = 8
Thus, m2 – 1 = 82 – 1 = 64 – 1 = 63
and m2 + 1 = 82 + 1 = 64 + 1 = 65.
∴ The required triplet is 16, 63, 65.

(iv) Let us take 2m = 18 ⇒ m = 9
Thus, m2 – 1 = 92 – 1 = 81 – 1 = 80
and m2 + 1 = 92 + 1 = 81 + 1 = 82
∴ The required triplet is 18, 80, 82.