Question

The area of a rhombus is $28c{m}^2$ and one of its diagonals is $4cm$. Its perimeter is

• A. $4\sqrt{53}$ $cm$
• B. $36$ $cm$
• C. $2\sqrt{53}$ $cm$
• D. None of these

Answer 1

The correct option is A $4\sqrt{53}$ $cm$

Area of a rhombus $=\frac{1}{2}\times (\text{diagonal 1}\times \text{diagonal 2})$

Given, Area $=28{cm}^2$

$\therefore \frac{1}{2}\times 4\times \text{diagonal 2}=28{cm}^2$

Hence, $\text{diagonal 2}=14cm$

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

Referring the diagram, if $PR=4cm,SQ=14cm$, then $OP=2cm,OQ=7cm$

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

${PQ}^2={OP}^2+{OQ}^2\Rightarrow 2^2+7^2=53$
$PQ=\sqrt{53}cm$
Hence, the length of the sides of the rhombus $=\sqrt{53}cm$.
Since the length of all sides of a rhombus are equal, Perimeter $=4\times$ length of one side $=4\times \sqrt{53}cm$