Question

Prove that, in a right-angled triangle, the square of hypotenuse is equal to the sum of the square of remaining two sides.

Answer 1

Consider a right angled triangle with length of hypotenuse as $c$ and the remaining two sides as $a$ and $b$ respectively.

We arrange 4 such triangles in such a fashion that they form a square of side $a+b$ (as shown in the figure).

Area of the figure $A=$ $(a+b)(a+b)$ $.......\left(1\right)$

Also area of the figure can be written as, $A=4\times ar\left(Bluerighttriangle\right)+ar\left(centralyellowsquare\right)$

$A=4\times \frac{1}{2}ab+c^2$ $.......\left(2\right)$

Equating (1) and (2) we get,

$c^2+2ab=a^2+b^2+2ab$

Consequently, $c^2=a^2+b^2$. Hence, Pythagoras theorem is proved.