Question

The sum of the squares of the sides of a rhombus is equal to the sum of the squares of its

• A. diagonals
• B. angles
• C. opposite sides
• D. none of these

Answer 1

The correct option is A diagonals

The diagonals of a rhombus bisect each other at ${90}^{\circ }$ so triangle AOB,AOD,BOC and COD are right angled triangle.

In $△AOB$

$A{O}^2+O{B}^2=A{B}^2$(using Pythagoras theorem)

$\Rightarrow {\left(\frac{AC}{2}\right)}^2+{\left(\frac{BD}{2}\right)}^2=A{B}^2$

$\Rightarrow \frac{A{C}^2}{4}+\frac{B{D}^2}{4}=A{B}^2$

$\Rightarrow A{C}^2+B{D}^2=4A{B}^2$

Similarly $A{C}^2+B{D}^2=4B{C}^2$

$A{C}^2+B{D}^2=4C{D}^2$

$B{D}^2+A{C}^2=4A{D}^2$

Adding all these we get

$\Rightarrow 4(A{B}^2+B{C}^2+C{D}^2+A{D}^2)=4(A{C}^2+B{D}^2)$

Hence the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals .