Question

prove that the sum of the square of the sides of a rhombus is equal to the sum of the square of its diagonals (can u give diagram also?)




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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°,

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

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

AB²=(1/2AC)² +(1/2 BD)²

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.

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