(iv) Prove that the sum of the squares of the diagonals of a rhombus is equal to the sum of the squares
of the sides.
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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 ] .
Hence, it is proved.
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