Question

Find the area of a rhombus having each side equal to $15cm$ and one of whose diagonals is $24cm$.

Answer 1

Let $ABCD$ be the rhombus where diagonals intersect at $O$.

Then, $AB=15cm$ and $AC=24cm$.
The diagonals of a rhombus bisect each other at right angles.
Therefore, $\Delta AOB$ is a right-angled triangle, right angled at $O$ such that
$\Rightarrow OA=\frac{1}{2}AC=12cm$ and $AB=15cm.$
By Pythagoras theorem, we have,
$(AB{)}^2=(OA{)}^2+(OB{)}^2$
$(15{)}^2=(12{)}^2+(OB{)}^2$
$(OB{)}^2=(15{)}^2-(12{)}^2$
$(OB{)}^2=225-144=81$
$(OB{)}^2=(9{)}^2$
$OB=9cm$
$\therefore BD=2\times OB=2\times 9cm=18cm$
Hence,
Area of the rhombus $ABCD=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times 24\times 18=216c{m}^2$