Question

The diagonals of a rhombus measure $16$ $cm$ and $30$ $cm$. Find its perimeter.

Answer 1

Let $ABCD$ be a rhombus (all sides are of equal length) and its diagonals, $AC$ and $BD$, are intersecting each other at point $O$.
Diagonals in a rhombus bisect each other at $90 \degree$.
It can be observed that

$AO= \frac{AC}{2}=\frac{16}{2}=8\;cm$

$BO = \frac{BD}{2}=\frac{30}{2}=15\;cm$

By applying Pythagoras theorem in $Δ$ $AOB$,
$OA^2 + OB^2= AB^2$
$8^ 2 + 15^2 = AB^2$
$64 + 225 = AB^2$
$289 = AB^2$
$AB = 17$
Therefore, the length of the side of rhombus is $17$ $cm$.
Perimeter of rhombus = $4 × Side \;of\; the\; rhombus$
=$4 × 17 = 68$ $cm$