Question

Question 2 (iv)
Write a Pythagorean Triplet whose one member is:

18

Answer 1

There are three numbers $2m,m^2-1and{m}^2+1,$ in a Pythagorean Triplet$(for,m>1)$.

$Here,2m=18\Rightarrow m=\frac{18}{2}=9\therefore \text{Second number}(m^2-1)=(9{)}^2-1=81-1=80\text{And third number}(m^2+1)=(9{)}^2+1=81+1=82$

Hence, Pythagorean triplet is (18, 80, 82).