Class 8 Maths NCERT Solutions for Chapter – 6 Squares and Square Roots Ex – 6.2

Squares and Square Roots

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 (30 + 2) + 2 (30 + 2)
= 302 +30 x 2 + 2 x 30 + 22
= 900 + 60 + 60 + 4 = 1024

(ii) = (30 + 5)2 = 30 (30 + 5) + 5 (30 + 5)
= 30 2 + 30 x 5 + 5 x 30 + 52
= 900 +150 +150 + 25 = 1225

(iii) = (80 + 6)2 = 80 (80 + 6) + 6 (80 + 6)
= 802 +80 x 6 + 6 x 80 + 62
= 6400 + 480 + 480 + 36 = 7396

(iv) = (90 + 3)2 = 90 (90 + 3) + 3 (90 + 3)
= 902 + 90 x 3 + 3 x 90 + 92
= 8100 + 270 + 270 + 81 = 8649

(v) = (70 +1)2 = 70 (70 +1) +1 (70 +1)
= 702 +70 x 1 + 1 x 70 + 12
= 4900 + 70 + 70 +1 = 5041

(vi)
 = (40 + 6)2 = 40 (40 + 6) + 6 (40 + 6)
= 402 + 40 x 6 + 6 x 40 +6 x 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:
(i) Put m = 3 in 2m, m2 -1, m2 +1
∴ 2m=6, m2 – l =32 – 1= 9 – 1 = 8
and m2 + 1=9 + 1=10
Thus, 6, 8 and 10 are Pythagorean triplets.

(ii) Put m = 7 in 2m, m2 -1, m2 +1
∴ 2m = 14, m2 – 1 = 72 – 1 = 49 – 1 = 48
and m2 + 1 = 72 +1 = 49+1 = 50
Thus, 14, 48 and 50 are Pythagorean triplets.

(iii) Put m = 8 in 2m, m2 – 1, m2 +1
∴ 2m =16, m2 – 1 = 82 – 1 = 64 – 1 =63
and m2 + 1 =82 + 1 =64 + 1 =65
Thus, 16, 63 and 65 are Pythagorean triplets.

(iv) Put m = 9 in 2m, m2 – 1, m2 +1
∴ 2m =18, m2 – 1 = 92 – 1=81 – 1 = 80,
m2+ 1 =92 + 1 =81 + 1 =82
Thus, 18, 80 and 82 are Pythagorean triplets.

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