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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