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Question

Write a Pythagorean triplet whose one member is 6 ,14 ,16 and 18

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Solution

i)

Let 2m = 6
m = 3


m²+ 1 = 3² + 1= 9 + 1 = 10


m²- 1 = 3 2 - 1 = 9 - 1 = 8



check:

6 ² + 8² = 36 + 64 = 100 = 10²


Hence, the triplet is 6, 8 & 10



ii)

Let 2 m = 14

m = 7

m² + 1 = 7²+ 1 = 49 + 1 = 50


m²- 1 = 7² - 1 = 49 - 1 = 48



check:

14² + 48² = 196 + 1304 = 2500
= 50²


Hence, the triplet is 14, 48, and 50



iii)

Let 2 m = 16,

m = 8


m² + 1 = 8² + 1 = 64 + 1 = 65


m²- 1 = 8 2 - 1 = 64 - 1 = 63



check:

16² + 63² = 256 + 3969 = 4225 =
65²


Hence, the triplet is 16, 63 & 65



iv)

Let 2 m = 18

m = 9

m² + 1 = 9² + 1 = 81 + 1 = 82

m² - 1 = 9² - 1 = 81 - 1 = 80

check:
18²+ 80² = 324 + 6400 = 6724 = 82²

Hence, the triplet is 18, 80 & 82

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Hope this will help you....


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