Question

prove that sum of the square of diagonal of rhombus is equal to sum of square of the sides.​

Answer 1

Step-by-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 have

AB² = OA² + OB² [ by Pythagoras' theorem ]

. 

$\bf { \implies {AB}^{2} = (\frac{1}{2} AC}^{2}) + ( { \frac{1}{2}BD) }^{2} .⟹AB2=(21AC2)+(21BD)2. </p><p>$

\bf { \implies {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }⟹AB2=41(AC2+BD2) 

$\bf { \implies {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }⟹AB2=41(AC2+BD2) </p><p>$

==> 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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