Question

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

• A. $72\sqrt{3}$
• B. $48\sqrt{3}$
• C. $72$
• D. $36$

Answer 1

Answer: (C)

$72$

Area of circle $=48\pi$

$\Rightarrow \pi r^2=48\pi$

$\Rightarrow r=\sqrt{48}=4\sqrt{3}units$

Now, $ON\perp BC$. So, $ON$ is the height of $△OBC$ correspoding to $BC$.

Area of $△ABC$ $=$ Area $△OBC$ + Area $△OCA$ + Area $△OAB$

$\frac{\sqrt{3}}{4}{a}^2=3\times area△OBC=3\times \frac{1}{2}\times BC\times ON$

$\frac{\sqrt{3}}{4}{a}^2=3\times \frac{1}{2}\times a\times r=3\times \frac{1}{2}\times 0\times 4\sqrt{3}$

$\frac{\sqrt{3}}{4}{a}^2=\frac{3}{2}\times 4\sqrt{3}\times a$

$\Rightarrow a=24$

$\therefore$ Perimeter of equilateral $△ABC=3a=3\times 24=72$ units