An equilateral triangle ABC is inscribed in a circle of radius $$20\sqrt{3}cm$$. The centroid of the
triangle ABC is at a distance d from the vertex A. Then d is equal to?

An equilateral triangle ABC is inscribed in a circle of radius $$20\sqrt{3}cm$$, thus the circumradius will be $$20\sqrt{3}cm$$.

Area of Triangle = $$\dfrac{abc}{4R}$$, where a, b, c are the sides of the triangle and R is the circumradius. 

$$\dfrac{\sqrt{3}}{4}\times a^2=\dfrac{a\times a\times a}{4\times20\sqrt{3}}$$

$$a=60$$ cm. 

Now, the altitude of the equilateral triangle (h) = $$\dfrac{\sqrt{3}a}{2}$$

The centroid of the triangle ABC is at a distance $$\dfrac{2h}{3}$$ from the vertex A.

$$d=\dfrac{2}{3}\times\dfrac{\sqrt{3}}{2}\times60$$

$$d=20\sqrt{3}$$ cm.