Question

prove that the sum of the square of the sides of a rhombus is equal to the sum of the square of its diagonals.

Answer 1

Step-by-step explanation:

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

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