Question

prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagnols ?​

Answer 1

$\implies$In rhombus ABCD, AB = BC = CD = DA

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]

$\implies$4AB2 = AC2+ BD2

$\implies$ AB2 + AB2 + AB2 + AB2 = AC2+ BD2

∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2

$\implies$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

Step-by-step explanation:

In rhombus ABCD, AB = BC = CD = DA

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]

\implies⟹ 4AB2 = AC2+ BD2

\implies⟹ AB2 + AB2 + AB2 + AB2 = AC2+ BD2

∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2

\implies⟹ Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.