The Pythagorean triplet of number 18 is 18,80,82
State true or false.

True

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False

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Solution

The correct option is A True
$2 k$ , $(k^{2} + 1)$ and $(k^{2} - 1)$ form a Pythagorean triplet for any number, k > 1
Let $2 k = 18$
Hence, $k = 9$
and, $k^{2} + 1 = 9^{2} + 1 = 81 + 1 = 82$
and, $k^{2} - 1 = 9^{2} - 1 = 81 - 1 = 80$
Check the answers: $18^{2} + 80^{2} = 324 + 6400 = 6724 = 82^{2}$
Hence, the triplet is $18, 80, 82$

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