Question

if one numver is pythogrean triplet is 6 then find the triplet​

Answer 1

Answer:

The least no. (m) = 6

Pythagorean Triplet = (2m,m²+1,m²-1)

= (2×6, 6² + 1, 6²-1)

= (12, 36+1,36-1)

= (12,36,35)

Answer 2

Explanation:

The simple method to find a Pythagorean Triplet is to have three numbers $2n, n^2+1,n^2-1$ where n can be any integer.

Since one of the numbers is given as 6, we can say that

$2n=6$

∴ $n=3$

∴ The triplet is

$2n=6$

$n^2-1=3^2-1=9-1=8$

$n^2+1=3^2+1=9+1=10$

Verification:

In this given triplet, n^2+1 is the greatest, and so, shall be the hypotenuse.

$LHS=10^2=100$

$RHS=6^2+8^2=36+64=100$

∴$LHS=RHS$

Answer: 6, 8 and 10

I hope this helps. :D