Question

22. Prove that the sum of the squares of the sides of rhombus is equal to the sum of
squares of its diagonals ?​

Answer 1

Answer:

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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 ) .

hence, In a rhombus , all sides are equal