Question

The area of a circle inscribed in an equilateral triangle is $48\mathrm{\pi }$ square units. Then the perimeter of the triangle is (in units)

• A. $72\surd 3$
• B. $48\surd 3$
• C. $72$
• D. $36$

Answer 1

The correct option is C

$72$

Step 1: Drawing the diagram and finding the radius

Given the area of a circle $=48\mathrm{\pi }$

$$\begin{gathered} \Rightarrow {\mathrm{\pi r}}^2=48\mathrm{\pi } \\ \Rightarrow {\mathrm{r}}^2=48 \\ \Rightarrow \mathrm{r}=4\sqrt{3}\mathrm{units} \end{gathered}$$

Step 2: Finding the length of sides

Let $a$ be the side of the equilateral triangle $ABC$

From the figure $ON\perp BC$

In the triangle $OBC$, $ON$ is the height corresponding to $BC$

$ON=r,BC=AB=AC=a$

Area of $∆ABC$ = area of $∆OBC$ $+$ area of $∆OCA+$ area of $∆OAB$

$\Rightarrow \left(\frac{\sqrt{3}}{4}\right)a^2=3\times$area of $△OBC$ [Triangle is equilateral triangle]

$$\begin{gathered} \Rightarrow \left(\frac{\sqrt{3}}{4}\right)a^2=3\times \frac{1}{2}\times a\times r\because \left(areaof△OBC=\frac{1}{2}\times a\times r\right) \\ \Rightarrow \left(\frac{\sqrt{3}}{4}\right)a^2=3\times \frac{1}{2}\times a\times 4\sqrt{3}\because \left(r=4\sqrt{3}units\right) \\ \Rightarrow \left(\frac{\sqrt{3}}{4}\right)a=3\times \frac{1}{2}\times 4\sqrt{3} \\ \Rightarrow a=\frac{3\times 4\sqrt{3}}{2}\times \frac{4}{\sqrt{3}} \\ \Rightarrow a=24units \end{gathered}$$

Step 3: Finding the perimeter

The perimeter of the triangle $=3\times a$

$$\begin{gathered} =3\times 24 \\ =72units \end{gathered}$$

Hence, option C is the answer.