Question

There is a Pythagorean triplet whose one member is $6$ and other member is $10$

• A. True
• B. False

Answer 1

The correct option is A True

$\Rightarrow$ $2m,m^2-1,m^2+1$ forms a Pythagorean triplet for any number.

Let $2m=6$

$\Rightarrow$ $m=\frac{6}{2}=3$

$\Rightarrow$ $m^2-1=(3{)}^2-1$

$=9-1$

$=8$

$\Rightarrow$ $m^2+1=(3{)}^2+1$

$=9+1$

$=10$

$\therefore$ The triplets are $6,8$ and $10.$

So it is true that there is a Pythagorean triplet whose one member is $6$ and other member is $10.$