Pythagorean triplet of given number $6$ is

6, 8, 10

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6, 9, 10

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6, 8, 11

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6, 7, 10

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Solution

The correct option is D $6, 8, 10$
$2 k$ , $(k^{2} + 1)$ and $(k^{2} - 1)$ form a Pythagorean triplet for any number, k > 1
Let $2 k = 6$
Hence, $k = 3$
and, $k^{2} + 1 = 3^{2} + 1 = 9 + 1 = 10$
and, $k^{2} - 1 = 3^{2} - 1 = 9 - 1 = 8$
Check the answers: $6^{2} + 8^{2} = 36 + 64 = 100 = 10^{2}$

Hence, the triplet is $6, 8, 10$

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