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in Pythagoras Theorem by (47.6k points)
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Assign different values to a and b and obtain 5 Pythagorean triplets.

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i. Let a = 2, b = 1 

a2 + b2 = 22 + 12 = 4 + 1 = 5 

a2 – b2 = 22 – 12 = 4 – 1 = 3 

2ab = 2 × 2 × 1 = 4 

∴ (5, 3, 4) is a Pythagorean triplet.

ii. Let a = 4,b = 3 

a2 + b2 = 42 + 32 = 16 + 9 = 25 

a2 – b2 = 42 – 32 = 16 – 9 = 7 

2ab = 2 × 4 × 3 = 24 

∴ (25, 7, 24) is a Pythagorean triplet

iii. Let a = 5, b = 2 

a2 + b2 = 52 + 22 = 25 + 4 = 29 

a2 – b2 = 52 – 22 = 25 – 4 = 21 

2ab = 2 × 5 × 2 = 20 

∴ (29, 21, 20) is a Pythagorean triplet.

iv. Let a = 4,b = 1 

a2 + b2 = 42 + 12 = 16 + 1 = 17 

a2 – b2 = 42 – 12 = 16 – 1 = 15

2ab = 2 × 4 × 1 = 8 

∴ (17, 15, 8) is a Pythagorean triplet. 

v. Let a = 9, b = 7 

a2 + b2 = 92 + 72 = 81 + 49 = 130 

a2 – b2 = 92 – 72 = 81 – 49 = 32 

2ab = 2 × 9 × 7 = 126 

∴ (130,32,126) is a Pythagorean triplet.

Note: Numbers in Pythagorean triplet can be written in any order.

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