Question

Find the area of a rhombus, each side of which measures $20cm$ and one of whose diagonals is $24cm$.

Answer 1

Step : Find the area of rhombus

Given : Side of the rhombus = $20cm$
Length of one of the diagonals $=24cm$

Let $ABCD$ be the rhombus

We know that,
Diagonals of a rhombus bisect at right angles.

$\Rightarrow AO=\frac{1}{2}\left(AC\right)=\frac{1}{2}\left(24\right)=12cm$

$\therefore In\Delta AOB,$
Applying Pythagoras theorem, we get
$(AB{)}^2=(OB{)}^2+(AO{)}^2$

$\Rightarrow (20{)}^2=(OB{)}^2+(12{)}^2$
$\Rightarrow (OB{)}^2=400-144=256$
$\Rightarrow OB=16cm$

As, $BD=2\left(OB\right)$
$\Rightarrow BD=2\left(16\right)=32cm$

Now,
$AC=24cm$
$BD=32cm$

We know that
Area of rhombus $=\frac{1}{2}\times$ (Product of diagonals)

$=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times 24\times 32$

$=384c{m}^2$

Hence, the area of rhombus $ABCD$ is $384c{m}^2.$