Question

Prove that $a^2+b^2=c^2$ where a , b and c are sides of a right angled triangle.

Answer 1

In a right angle triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the "Pythagorean equation"

In a right angled triangle

Four such triangles can be arranged as shown below

Area of Whole Square

It is a big square, with each side having a length of a+b, so the total area is:

A = (a+b)(a+b)

Area of The Pieces
Now let's add up the areas of all the smaller pieces:
Adding up the tilted square and the four triangles $=c^2+2ab$
The area of the large square is equal to the area of the tilted square and the 4 triangles. This can be written as:
$(a+b)(a+b)=c^2+2ab$
Now, let us rearrange
$a^2+b^2=c^2$