Question

Pythagorean triplet, whose one member is $16,$ is$16,63,65.$
State true or false.

• A. True
• B. False

Answer 1

The correct option is A True

$2k,$ $(k^2+1)$ and $(k^2-1)$ form a Pythagorean triplet for any number, $k>1$
Let $2k=16$
Hence, $k=8$
and, $k^2+1=8^2+1=64+1=65$
and, $k^2-1=8^2-1=64-1=63$
Check the answers: ${63}^2+{16}^2=3969+256=4225={65}^2$
Hence, the triplet is $16,63,65$