Question : 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?
Option 1: $441\sqrt{3}-693$
Option 2: $882\sqrt{3}-693$
Option 3: $882\sqrt{3}-462$
Option 4: $441\sqrt{3}-462$
Correct Answer: $441\sqrt{3}-693$
Solution : The side length of the equilateral triangle $=2 \times 21 = 42\;\mathrm{cm}$ The area of the equilateral triangle $=\frac{\sqrt{3}}{4} \times (42)^2=441\sqrt3\;\mathrm{cm^2}$ Each angle of an equilateral triangle $=60^{\circ}$ The area of each sector $=\frac{60^{\circ}}{360^{\circ}} \times \pi \times (21)^2=231\;\mathrm{cm^2}$ The area of the portion enclosed by the three circles $=(441\sqrt3 - 3 \times231)\;\mathrm{cm^2}$ The area of the portion enclosed by the three circles $=(441\sqrt3 - 693)\;\mathrm{cm^2}$ Hence, the correct answer is $441\sqrt3 - 693$.
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