Question

The perimeter of a rhombus is $52$ cm. If one diagonal is $24$ cm, then find its area.

Answer 1

Let $PQRS$ be the rhombus as shown in the figure.

Since the length of all sides of a rhombus are equal, perimeter $=4\times$ length of one side $⟹4\times PQ=52$ cm $\Rightarrow PQ=13$ cm

Hence, length of each side of the rhombus $=13$ cm

In a rhombus, the diagonals bisect each other at ${90}^0$

Referring the diagram, if $PR$ $=24$ cm, then $OP$ $=12$ cm

$OQ$ $=\frac{SQ}{2}$

Since, triangle $POQ$ is a right angled triangle,

${PQ}^2={OP}^2+{OQ}^2$

$\Rightarrow {13}^2={12}^2+{\left(\frac{SQ}{2}\right)}^2$

$\Rightarrow {\left(\frac{SQ}{2}\right)}^2={13}^2-{12}^2$

$\Rightarrow \left(\frac{SQ}{2}\right)=\sqrt{25}=5$ cm

$\Rightarrow SQ=10$ cm

Hence, length of the other diagonal $=10$ cm

Area of a rhombus $=\frac{1}{2}\times \text{(diagonal 1}\times \text{diagonal 2)}=\frac{1}{2}\times 24\times 10=120{cm}^2$