Question

prove that the sum of squares of the sides of a rhombus is equal to the sum of the square of its diagonals

Answer 1

this is the answer......

Answer 2

▶ 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°,


$\bf { OA = \frac{1}{2} AC \: and \: OB = \frac{1}{2} BD . }$

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

$\bf { \implies {AB}^{2} = (\frac{1}{2} AC}^{2}) + ( { \frac{1}{2}BD) }^{2} .$

$\bf { \implies {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }$

==> 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.