Question

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


solve full solution.

Answer 1

Given : ABCD is a rombos

To proof : AB^2 + BC^2 + CD^2 +

AD^2 = AC^2 + BD^2

proof: : we know diagonal of rombos bisect at right angle

AB ^2 = OA^2 + OB^2

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

= 1/4 AC^2 + 1/4 BD^2

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

or, 4AB^2 = AC^2 + BD ^2

so, AB^2 + BC^2 + CD^2 + AD^2 =

AC^2 +BD ^2 (Proved)

hope it helps