Question

prove that the sum of the square of the side of a rhombus is equal to the sum of the square of its diagonal

Answer 1

In rhombus ABCD, AB = BC = CD = DA

We know that diagonals of a rhombus bisect each other perpendicularly.

That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB

AB2 = OA2 + OB2   [By Pythagoras theorem]

⇒  4AB2 = AC2+ BD2

⇒  AB2 + AB2 + AB2 + AB2 = AC2+ BD2

∴ 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.

Answer 2

given

rhombus ABCD,

WITH DIAGONALS AC&BD INTERSECTING AT O

TO PROVE,

SUM OF SQUARES OF ALL SIDES=SUM OF SQUARES  OF ITS DIAGONALS

⇒AB²+BC²+CD²+AD²=AC²+BD²

SINCE SIDES OF RHOMBUS ARE EQUAL

AB=BC=CD=AD