Question

Prove that the sum of squares of the sides of a rhombus is equal.

Answer 1

Answer:

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 AC AND OB=1 BD

       2                       2

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

⇒AB²= 1 (AC²+BD²)

            4

==> 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

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