Question

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

Answer 1

Let $ABCD$ be the rhombus, such that, $AC=24$cm and $AD=20$ cm

We know that diagonals of a rhombus bisect each other at ${90}^{\circ }$.

Therefore, $AO=CO=12$ cm

By Pythagoras Theorem in $\Delta AOD$

$\Rightarrow A{D}^2=A{O}^2+D{O}^2$

$\Rightarrow {20}^2={12}^2+D{O}^2$

$\Rightarrow 400-144=D{O}^2$

$\Rightarrow {16}^2=D{O}^2$

$\therefore DO=16cm$

Hence, $BD=2\times 16=32$ cm

Thus, Area of $ABCD=\frac{1}{2}\times AC\times BD$

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

$=384c{m}^2$