Question

If the area of a rhombus be $24{cm}^2$ and one of its diagonals be $4cm$, find the perimeter of the rhombus.

Answer 1

If $d_1$ and $d_2$ are the two diagonals of the rhombus then,

Area of rhombus $=\frac{1}{2}\times d_1\times d_2$

It is given that, area of rhombus $=24{cm}^2$

One of the diagonal $\left(d_1\right)=4cm$

$\Rightarrow \frac{1}{2}\times 4cm\times d_2=24{cm}^2$

$\Rightarrow 2cm\times d_2=24{cm}^2$

$\Rightarrow d_2=\frac{24{cm}^2}{2cm}$

$\Rightarrow d_2=12cm$

Since, the diagonals of the rhombus are perpendicularly bisects each other.

lets draw the rough diagram of the given rhombus.

In rhombus $ABCD$, lets take $\Delta COD$

In $\Delta COD,\angle COD={90}^{\circ }$

$\Rightarrow C{D}^2=C{O}^2+D{O}^2$ [Since, by Pythagoras theorem]

$\Rightarrow C{D}^2=(2cm{)}^2+(6cm{)}^2$

$\Rightarrow C{D}^2=4c{m}^2+36c{m}^2$

$\Rightarrow C{D}^2=40c{m}^2$

$\Rightarrow CD=\sqrt{40c{m}^2}$

$\Rightarrow CD=\sqrt{4\times 10c{m}^2}$

$\Rightarrow CD=2\sqrt{10}cm$

i.e., Side of the rhombus $=2\sqrt{10}cm$

$\Rightarrow$ Perimeter of the rhombus $=4\times$ Side

$=4\times 2\sqrt{10}cm$

$=8\sqrt{10}cm$

$=8\times 3.16cm$ (Approximately) [Since, $\sqrt{10}=3.16$ (Approximately)

$=25.28cm$ (Approximately)