Question

Prove That the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer 1

In rhombus ABCD, AB = BC = CD = DA

We know that diagonals of a rhombus bisect each other perpendicularly.

That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB

AB2 = OA2 + OB2   [By Pythagoras theorem]

⇒  4AB2 = AC2+ BD2

⇒  AB2 + AB2 + AB2 + AB2 = AC2+ BD2

∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2

Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer 2

$<b> <I>Hey there !!$

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.

$=> \bf { \angle AOB = \angle BOC = \angle COD = \angle DOA = 90°, }$

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

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

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

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

=> 4AB² = ( AC² + BD² ) ........(1).

Similarly, we have:

4BC² = ( AC² + BD² ) .........(2).

4CD² = ( AC² + BD² ) .........(3).

4DA² = ( AC² + BD² ) ..........(4).

On adding equation (1) , (2), (3), and (4), we get

( AB² + BC² + CD² + DA² ) = ( AC² + BD² ).

✔✔ Hence, it is proved ✅✅.

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