# How do you show that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides ?

Jan 28, 2017

See explanation...

#### Explanation:

Consider this diagram:

Each of the triangles is a right angled triangle with sides $a , b , c$ and area $\frac{a b}{2}$ (since it is half of an $a \times b$ rectangle)

The area of the large square is:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

The area of the smaller square plus the area of the triangles is:

${c}^{2} + 4 \left(\frac{a b}{2}\right) = {c}^{2} + 2 a b$

These two expressions must be equal. So we have:

${a}^{2} + 2 a b + {b}^{2} = {c}^{2} + 2 a b$

Subtracting $2 a b$ from both sides, we find:

${a}^{2} + {b}^{2} = {c}^{2}$

This is Pythagoras' Theorem:

In a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.