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
