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
= (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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