If a is the length of equilateral triangle, height is $$\frac{\sqrt{3}a}{2}$$
Inradius =Â $$\frac{\sqrt{\ 3}a}{2}\left(\frac{1}{3}\right)=\frac{a}{2\sqrt{3}}$$
It is given, R =Â $$\frac{a}{2\sqrt{3}}$$
Area of triangle = Inradius * semiperimeter =Â $$R\left(\frac{3a}{2}\right)$$ =Â $$\frac{3R}{2}\left(2\sqrt{3}R\right)=3\sqrt{3}R^2$$
The answer is option D.
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