Question
Write a Pythagorean triplet whose one number is $18$
✓
Answer
Here, $2m = 18$
$\therefore m = \frac{{18}}{2} = 9$
$ \therefore m^2-1=9^2-1=81-1=80 $
and, $ m^2+1=9^2+1=81+1=82 $
So, a Pythagorean triplet, whose one member is $18$, is $18, 80, 82.$
$\therefore m = \frac{{18}}{2} = 9$
$ \therefore m^2-1=9^2-1=81-1=80 $
and, $ m^2+1=9^2+1=81+1=82 $
So, a Pythagorean triplet, whose one member is $18$, is $18, 80, 82.$
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