Question

Prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides.Using the above result show that sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.​

Answer 1

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YOUR ANSWER IS ATTACHED.

Answer 2

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=$\frac{1}{2}$AC AND OB=$\frac{1}{2}$BD

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

AB²=$\frac{1}{2}$AC²+$\frac{1}{2}$BD²

AB²=$\frac{1}{4 }$ (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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