Question

Using the pythagorean theorem prove that in rhombus abcd , 4ab square = ac2 + bd2

Answer 1

Hey !!

Here is your answer...

Given :- ◻ABCD is a rhombus.

To Prove :- 4AB^2 = AC^2 + BD^2

Proof :- As we know, diagonals of the rhombus bisect each other at right angle.

OA = 1/2 AC
OB = 1/2 BD

and /_ AOB = 90°

Now, In right ∆AOB using prthagorean triplet.

AB^2 = OA^2 + OB^2
AB^2 = (1/2 AC^2 ) + ( 1/2 BD^2 )
AB^2 = 1/4( AC^2 + BD^2 )
4AB^2 = AC^2 + BD^2 .... ( proved )

Thanks.. ^-^

Answer 2

Answer:

Given :- ◻ABCD is a rhombus.

To Prove :- 4AB^2 = AC^2 + BD^2

Proof :- As we know, diagonals of the rhombus bisect each other at right angle.

OA = 1/2 AC

OB = 1/2 BD

and /_ AOB = 90°

Now, In right ∆AOB using prthagorean triplet.

AB^2 = OA^2 + OB^2

AB^2 = (1/2 AC^2 ) + ( 1/2 BD^2 )

AB^2 = 1/4( AC^2 + BD^2 )

4AB^2 = AC^2 + BD^2 .... ( proved )