Squares And Square Roots : Exercise 6.2 (Mathematics NCERT Class 8th)

Q.1 Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93  (v) 71 (vi) 46
Sol. (i) 32
322 = (30 + 2)2
30(30 + 2) + 2(30 + 2)
302 + 30 x 2 + 2 x 30 + 22
900 + 60 +60 + 4
1024

(ii) 35
352 = (30 + 5)2
30(30 + 5) + 5(30 + 5)
302 + 30 x 5 + 5 x 30 + 52
900 + 150 +150 + 25
1225

(iii) 86
862 = (80 + 6)2
80(80 + 6) + 6(80 + 6)
802 + 80 x 6 + 6 x 80 + 62
6400 + 480 +480 + 36
7396

(iv) 93
932 = (90 + 3)2
90(90 + 3) + 3(90 + 3)
902 + 90 x 3 + 3 x 90 + 32
8100 + 270 +270 + 9
8649

(v) 71
712 = (70 + 1)2
70(70 + 1) + 1(70 + 1)
702 + 70 x 1 + 1 x 70 + 12
4900 + 70 +70 + 1
5041

(vi) 46
462 = (40 + 6)2
40(40 + 6) + 6(40 + 6)
402 + 40 x 6 + 6 x 40 + 62
1600 + 240 +240 + 36
2116

Q.2 Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv)18
Sol. We know that, for any natural number m > 1, we have (2m) 2 + (m2 – 1)2 = (m2 + 1)2 where, 2m, m2 – 1 and m2 + 1 forms a Pythagorean triplet.

(i) 6
For 6 to be member of Pythagorean triplet we select 2m = 6.
Therefore, m = 3.
Second number = (m2 – 1) = (32 – 1) = 9 – 1 = 8
Third number = (m2 + 1) = (32 + 1)2 = 9 + 1 = 10
Thus, the Pythagorean triplet is (6, 8, 10).

(ii) 14
For 14 to be member of Pythagorean triplet we select 2m = 14.
Therefore, m = 7.
Second number = (m2 – 1) = (72 – 1) = 49 – 1 = 48
Third number = (m2 + 1) = (72 + 1) = 49 + 1 = 50
Thus, the Pythagorean triplet is (14, 48, 50).

(iii) 16
For 16 to be member of Pythagorean triplet we select 2m = 16.
Therefore, m = 8.
Second number = (m2 – 1) = (82 – 1) = 64 – 1 = 63
Third number = (m2 + 1) = (82 + 1) = 64 + 1 = 65
Thus, the Pythagorean triplet is (16, 63, 65).

(iv)18
For 18 to be member of Pythagorean triplet we select 2m = 18.
Therefore, m = 9.
Second number = (m2 – 1) = (92 – 1) = 81 – 1 = 80
Third number = (m2 + 1) = (92 + 1) = 81 + 1 = 82
Thus, the Pythagorean triplet is (18, 80, 82).