Ignore the marked answer, accidentally clicked it

From the Pythagoras's Theorem, we know that:

(Hypotenuse)² = (base)² + (perpendicular)²

We are Given:

We are given that the 2 legs (Base and Perpendicular) of the triangle are (m²-n² and 2mn)

Solving for the Hypotenuse:

replacing the variables in the Pythagoras's theorem, we get:

Hypotenuse² = (m²-n²)² + (2mn)²

Hypotenuse² = (m⁴ -2m²n² + n⁴) + (4m²n²)                  [(a-b)² = a² - 2ab - b²]

Hypotenuse² = m⁴ -2m²n² + n⁴ + 4m²n²

Hypotenuse² = m⁴ + 2m²n² + n⁴

Hypotenuse² = (m²)² + 2m²n² + (n²)²

Hypotenuse² = (m² + n²)²                                              [a² + 2ab + b² = (a+b)²]

taking the square root of both the sides

Hypotenuse = m² + n²