If sum of the areas of the circumcircle and the incircle of an equilateral triangle is $$770 cm^2$$, then what is the area $$(in cm^2)$$ of the triangle?

Let the side of equilateral triangle = $$a$$ cm

=> Circumradius of equilateral triangle = $$R=\frac{a}{\sqrt3}$$ cm and inradius = $$r=\frac{a}{2\sqrt3}$$ cm

Thus, sum of areas of circumcircle and incircle = $$(\pi R^2)+(\pi r^2)=770$$

=> $$\frac{22}{7} [(\frac{a}{\sqrt3})^2+(\frac{a}{2\sqrt3})^2]=770$$

=> $$\frac{a^2}{3}+\frac{a^2}{12}=770\times\frac{7}{22}$$  

=> $$\frac{5a^2}{12}=245$$

=> $$a^2=245\times\frac{12}{5}=588$$

$$\therefore$$ Area of equilateral triangle = $$\frac{\sqrt3}{4}a^2$$

= $$\frac{\sqrt3}{4}\times588=147\sqrt3$$ $$cm^2$$

=> Ans - (B)