Question

Prove that sum of square of sides of rohumbus is equal to th esum of square of its diogonal

Answer 1

In rhombus ABCD, AB = BC = CD = DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC is perpendicular to BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB

=> AB²= OA² + OB²   [By Pythagoras theorem]

=> AB² = (AC/2)²+(BD/2)²

=> AB² = AC²/4 + BD²/4

=> AB² = (AC²+BD²)/4


=>  4AB² = AC²+ BD²

=> AB² + AB² + AB²+ AB² = AC²+ BD²
therefore , AB² + BC² + CD² + DA² = AC²+ BD²


Thus , the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.


hope this helps