Question

Write a pythagorean triplet whose one member is $18$.

• A. $(18,80,82)$
• B. $(18,82,84)$
• C. $(18,80,84)$
• D. $(18,82,86)$

Answer 1

Answer: (A)

$(18,80,82)$

We have pythagorean triplet $(2n,n^2-1,n^2+1)$ where $n>1$ (natural number)
Let $n^2-1=18$
$n^2=19$
$n=\sqrt{19}$ (not an integer)
$n^2+1=18$
$n^2=18-1$
$n^2=17$
$n=\sqrt{17}$ (is not an integer)
So, let us take $2n=18$
$n=9$
$\therefore$ The required triplet is (an integer)
$(18,9^2-1,9^2+1)=(18,81-1,81+1)$
i.e., $(18,80,82)$