Question

prove that in any Rhombus the sum of square of side is equal to the sum of square of the diagonals​

Answer 1

Step-by-step explanation:

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O.

To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ).

Proof :- 

➡ We know that the diagonals of a rhombus bisect each other at right angles.

==>∠AOB = ∠BOC = ∠COD = ∠DOA = 90°,

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

==> 4AB² = ( AC² + BD² ) .

==> AB² + AB² + AB² + AB² = ( AC² + BD² ) .

•°• AB² + BC² + CD² + DA² = ( AC² + BD² ) .

[ In a rhombus , all sides are equal ] .

Answer 2

Answer:

All sides are equal in Rhombus and the diagnols intersect each other at 90.....