Three circles of radius 21 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 = $$r=21$$ cm

=> AC = $$r+r=42$$ cm

Similarly, AB = 42 cm and BC = 42 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)

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

= $$(\frac{\sqrt3}{4}\times 42\times42)-(3\times\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times21\times21)$$

= $$(441\sqrt3)-(11\times3\times21)$$

= $$(441\sqrt3-693)$$ $$cm^2$$

=> Ans - (A)