Question

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

plz solve this its urgent

Answer 1

hey mate
refer to attachment

Answer 2

Refer the attachment for figure.


Name the rhombus as ABCD mark the intersecting point of diagonals as O

Let first diagonal be x and second be y for convenience. Let the sides be a


So We know that diagonals bisect at right angles

Hence, in ∆ AOB,

AB will be hypotenuse and AO and BO will be perpendicular or base.

=> AB² = AO² + BO²

Now,

AO = half of first diagonal

=> AO = (x/2)

Similarly, BO = (y/2)

AB is the side

=> AB = a

Hence,

a² = (x/2)² + (y/2)²

=> a² = x²/4 + y²/4

Similarly, in all other three triangles,
a² will be equal to x²/4 + y²/4

Now, to prove that sum of squares of side = sum of squares of diagonals

Sum of squares of side = a² + a² + a² + a²

=> 4a²

But we found that,

a² = x²/4 + y²/4

=> 4a² = 4(x²/4 + y²/4)

After opening brackets,

=> 4a² = x² + y²

=> a² + a² + a² + a² = x² + y²

Hence Proved,