Exercise 6.2 Page: 98
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
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
vi. (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.