Question

Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonal.

Answer 1

In rhombus ABCD, AB = BC = CD = DAWe know that diagonals of a rhombus bisect each other perpendicularly.That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° andConsider right angled triangle AOBAB2= OA2+ OB2 [By Pythagoras theorem]⇒ 4AB2= AC2+ BD2⇒ AB2+ AB2+ AB2+ AB2= AC2+ BD2∴ AB2+ BC2+ CD2+ DA2= AC2+ BD2Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer 2

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