Question

1. 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

Answer:

Step-by-step explanation:

▶ Answer :-

▶ 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°,

OA= 1/2AC & OB=1/2 BD

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

==> AB²=(1/2AC²)+(1/2BD² )

==>  AB²=1/4( AC² + BD² )

==> 4AB² = ( AC² + BD² )

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

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

[ In a rhombus , all sides are equal ]

Hence, it is proved.