From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are √3, cm 2√3 cm and 5√3 cm. The perimeter (in cm) of the triangle is

Answer: Option C

Let P be the point inside the equilateral ΔABC
Let, PD = √3, PE = 2√3, PF = 5√3 and AB = BC = AC = $$x$$
$$\begin{array}{rl}& \text{ar}\text{.}\Delta \text{ABC}=\text{ar}\text{.}\Delta \text{ABP}+\text{ar}\text{.}\Delta \text{ACP}+\text{ar}\text{.}\Delta \text{BCP}\\ & \frac{\sqrt{3}}{4}{x}^2=\frac{1}{2}\times x\times \sqrt{3}+\frac{1}{2}\times x\times 2\sqrt{3}+\frac{1}{2}\times x\times 5\sqrt{3}\\ & \sqrt{3}x=2\sqrt{3}+4\sqrt{3}+10\sqrt{3}\\ & x=16\end{array}$$
∴ Perimeter of triangle = 3$$x$$ = 3 × 16 = 48 cm