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Prove that sum of the square of diagonal of rhombus is equal to sum of square of the sides. |
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Answer» -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°, From right ∆AOB , we haveAB² = OA² + OB² [ by Pythagoras' theorem ]. \bf { \implies {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }⟹AB2=41(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.plzz mark me as a brainlist answer |
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