Question : ABC is an equilateral triangle. If the area of the triangle is $36 \sqrt{3}$, then what is the radius of the circle circumscribing the $\triangle ABC$?
Option 1: $2 \sqrt{3}$
Option 2: $3 \sqrt{3}$
Option 3: $4 \sqrt{3}$
Option 4: $6 \sqrt{3}$
Correct Answer: $4 \sqrt{3}$
Solution :
Given: ABC is an equilateral triangle.
The area of the triangle is $36 \sqrt{3}$.
The area of the equilateral triangle $=\frac{\sqrt3}{4}\times (s)^2$, where $s$ is the sides of the triangle.
The radius of the circle circumscribing the equilateral triangle $=\frac{s}{\sqrt3}$.
⇒ $\frac{\sqrt3}{4}\times (s)^2= 36 \sqrt{3}$
⇒ $(s)^2=4\times 36$
⇒ $s = 2\times 6=12$ units
The radius of the circle circumscribing the $\triangle ABC$ $=\frac{12}{\sqrt3}=4\sqrt3$.
Hence, the correct answer is $4 \sqrt{3}$.
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