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

Solution (By Examveda Team)

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