Question

Prove That the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals...

Answer 1

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 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

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.