Question

Find the remaining numbers of the Pythagorean triplet whose largest number is 122.

• A. 22, 121
• B. 21, 121
• C. 22, 123
• D. 22, 120

Answer 1

The correct option is D 22, 120

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

$Here,m^2+1=122$
$\Rightarrow m=\sqrt{122-1}\Rightarrow m=\sqrt{121}\Rightarrow m=11\therefore \text{Second number}(m^2-1)=(11{)}^2-1=121-1=120\text{And third number}2m=2(11)=22$

Hence, the Pythagorean triplet is (22, 120, 122) and the correct option is D.