Question

Prove that sum of the squares on the diagonals of a rhombus is equal to four times the square on the side of the rhombus?

Answer 1

Please see below.

Explanation:

In the absence of any figure I have drawn it as given below.

In a rhombus diagonals intersect each other at right angles. In the given rhombus `ABCD` diagonal bisect each other at `O` hence

`BC^2+AD^2=(2BO)^2+(2AO)^2=4BO^2+4AO^2`

= `4(BO^2+AO^2)=4AB^2`