Class 8 | NCERT Solution Maths Chapter 6 | Squares and Square Roots | Exercise 6.2

 Exercise 6.2 Page: 98

1. Find the square of the following numbers.

i. 32

ii. 35

iii. 86

iv. 93

v. 71

we. 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

ii. (35)2

= (30+5 )2

= (30)2 + (5)2 + 2×30×5 [Since, (a+b)2 = a2+b2 +2ab]

= 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 [Since, (a+b)2 = a2+b2 +2ab]

= 8100 + 9 + 540

= 8649

v. (71) 2

= (70+1 )2

= (70)2 + (1)2 +2×70×1 [Since, (a+b)2 = a2+b2 +2ab]

= 4900 + 1 + 140

= 5041

we. (46) 2

= (50 -4 )2

= (50)2 + (4)2 – 2×50×4 [Since, (a+b)2 = a2+b2 +2ab]

= 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 2m, m2–1, m2+1 is a Pythagorean triplet.

i. 2m = 6

⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

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

∴ (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

∴ (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

∴ (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

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

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