Mathematics – Class 8 – Chapter 6 – Squares and Square root – Exercise 6.2 – NCERT Exercise Solution

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) 2 + 2×30×2 [Since, (a+b) 2 = a2+b2 +2ab]

= 900 + 4 + 120

= 1024

Hence, (32)2 = 1024

ii. (35) 2

= (30+5) 2

= (30) 2 + (5) 2 + 2×30×5

= 900 + 25 + 300

= 1225

iii. (86) 2

= (90 – 4) 2

= (90) 2+ (4) 2 – 2×90×4 [Since, (a-b) 2 = a2+b2 – 2ab]

= 8100 + 16 – 720

= 8116 – 720

= 7396

iv. (93) 2

= (90+3 ) 2

= (90) 2 + (3) 2 + 2×90×3

= 8100 + 9 + 540

= 8649

v. (71) 2

= (70+1 ) 2

= (70) 2 + (1) 2 +2×70×1

= 4900 + 1 + 140

= 5041

vi. (46) 2

= (50 -4 ) 2

= (50) 2 + (4) 2 – 2×50×4

= 2500 + 16 – 400

= 2116

2. Write a Pythagorean triplet whose one member is.

i. 6

ii. 14

iii. 16

iv. 18

Solution:

For any natural number m, we know that triplet are in the form of 2m, m2–1, and m2+1

Now,

i. 2m = 6

m = 6/2 = 3

m2–1 = 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

Hence, (6, 8, 10) is a Pythagorean triplet.

ii. 2m = 14

m = 14/2 = 7

m2 – 1= 72 – 1 = 49 – 1 = 48

m2 + 1 = 72 + 1 = 49 + 1 = 50

Hence, (14, 48, 50) is not a Pythagorean triplet.

iii. 2m = 16

m = 16/2 = 8

m2 – 1 = 82 – 1 = 64 – 1 = 63

m2 + 1 = 82 + 1 = 64+1 = 65

Hence, (16, 63, 65) is a Pythagorean triplet.

iv. 2m = 18

m = 18/2 = 9

m2 – 1 = 92 – 1 = 81–1 = 80

m2 + 1 = 92 + 1 = 81+1 = 82

Hence, (18, 80, 82) is a Pythagorean triplet.

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