Three circles of radius 63 cm are placed in such a way that each circle touches the other two. What is the area of the portion enclosed by the three circles?

Radius of each circle = $$r=63$$ cm

=> AC = $$r+r=126$$ cm

Similarly, AB = 126 cm and BC = 126 cm

=> $$\triangle$$ ABC is an equilateral triangle having $$\angle A=\angle B=\angle C=60^\circ$$

Thus, area of shaded portion = (Area of $$\triangle$$ ABC) - ($$3\times$$ Area of each sector)

= $$(\frac{\sqrt3}{4}\times s^2)-(3\times\frac{\theta}{360^\circ}\times\pi r^2)$$

= $$(\frac{\sqrt3}{4}\times 126\times126)-(3\times\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times63\times63)$$

= $$(3969\sqrt3)-(11\times9\times63)$$

= $$(3969\sqrt3-6237)$$ $$cm^2$$

=> Ans - (D)