The area of a rhombus is $384\ m^2$ and one of its diagonals is $24\ m$. What is the perimeter of the rhombus?

Given:

The area of a rhombus is $384\ m^2$ and one of its diagonals is $24\ m$.

To do:

We have to find the perimeter of the rhombus.

Solution:

Let $d_1$ and $d_2$ be the two diagonals of a rhombus.

Then, $d_1=24\ m$

We know that,

Area of Rhombus $=\frac{d_1\times d_2}{2}$

$384​=\frac{(24)(d_2)}{2}$

​$d_2=32$

Side of the rhombus $s=\sqrt{(\frac{d_1}{2})^2+(\frac{d_2}{2})^2}$

$=\sqrt{(\frac{24}{2})^2+(\frac{32}{2})^2}$

$=\sqrt{(12)^2+(16)^2}$

$=\sqrt{144+256}$

$=\sqrt{400}$

$=20\ m$

Perimeter $=4\times s$

$=4\times20\ m$

$=80\ m$

The perimeter of the rhombus is 84 m.

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Updated on: 10-Oct-2022

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