Write a Pythagorean triplet whose one member is
a) $6$
b) $14$

6, 8, 10; 14, 48, 50

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6, 8, 12; 14, 48, 52

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6, 8, 10; 15, 45, 50

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6, 9, 13; 14, 48, 50

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Solution

The correct option is C $6, 8, 10; 14, 48, 50$
Put $m = 3$ in $2 m, m^{2} - 1, m^{2} + 1$
$∴ 2 m = 6, m^{2} - 1 = 3^{2} - 1 = 9 - 1 = 8$
and $m^{2} + 1 = 9 + 1 = 10$
Thus, $6, 8 a n d 10$ are Pythagorean triplets.
Put $m = 7$ in $2 m, m^{2} - 1, m^{2} + 1$
$∴ 2 m = 14, m^{2} - 1 = 7^{2} - 1 = 49 - 1 = 48$
and $m^{2} + 1 = 7^{2} + 1 = 49 + 1 = 50$
Thus, $14, 48 a n d 50$ are Pythagorean triplets.

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