Question

ABCD is the rombus prove that AC²+BD²=4AB²​

Answer 1

Answer:

AC²+BD²=4AB²

Step-by-step explanation:

the diagonal of a rhombus intersect at O

AC is perpendicular to BD

then OA=OC and OB=OD (diagonal of the rhombus bisect each other at right angled )

let consider the traingle AOB and applying pythagoras theorem

AB²=OA²+OB²

AB²=(AC/2)²+(BD/2)²

AB²=AC²/4+BD²/4

4AB²=AC²+BD²

AC²+BD²=4AC²

Answer 2

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O.

To Prove :- 4AB² = ( 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 ).

It is Proved