Question

The diagonals of a rhombus are 10 centimeters

and 24 centimeters. Find the

perimeter of the rhombus.

Answer 1

In any rhombus, the diagonals  bisect each other at right angles (90°). 

Assume $a$ is the side of the rhombus, $d_1$ and $d_2$ are its diagonals

$d_1=10\ cm, d_2=24\ cm.$

By the Pythagorean Theorem

$$a^2=(\dfrac{d_1}{2})^2+(\dfrac{d_2}{2})^2$$

$$a=\sqrt{(\dfrac{10}{2})^2+(\dfrac{24}{2})^2}=13(cm)$$

Find the perimeter of the perimeter

$$P=4a=4(13 \ cm)=52\ cm$$

The perimeter of the rhombus is 52 centimeters.