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Class 8 NCERT Solutions – Chapter 6 Squares and Square Roots – Exercise 6.2

  • Last Updated : 09 Nov, 2020

Question 1. Find the square of the following numbers.

(i) 32

(32)2

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= (30 + 2)2

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

= 900 + 4 + 120

= 1024

(ii) 35

(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

(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 



(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

(71)2

= (70 + 1)2

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

= 4900 + 1 + 140

= 5041

(vi) 46

(46)2

= (50 – 4)2

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

= 2500 + 16 – 400

= 2116

Question 2. Write a Pythagorean triplet whose one member is.

(i) 6 

For any natural number m, we know that 2m, m2 – 1, m2 + 1 is a Pythagorean triplet.

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) 14

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) 16 

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) 18

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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