Question : Three circles of radius $63\;\mathrm{cm}$ each 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? (in $\mathrm{cm^2}$)
Option 1: $7938\sqrt{3}-4158$
Option 2: $3969\sqrt{3}-4158$
Option 3: $7938\sqrt{3}-6237$
Option 4: $3969\sqrt{3}-6237$
Correct Answer: $3969\sqrt{3}-6237$
Solution :
The side length of the equilateral triangle $=2 \times 63 = 126\;\mathrm{cm}$.
The area of an equilateral triangle with side length ($a$) $=\frac{\sqrt{3}}{4}a^2$
The area of the equilateral triangle $=\frac{\sqrt{3}}{4} \times (126)^2=3969\sqrt3\;\mathrm{cm^2}$
Each sector cut out by the triangle is one-sixth of a circle (since the angle at the centre is $60^\circ$ and a full circle is $360^\circ$).
The area of a circle with radius ($r$) $=\pi r^2$
So, the area of each sector $=\frac{1}{6} \times \pi \times (63)^2\;\mathrm{cm^2}$
Since there are three sectors, the total area of the sectors $=3 \times \frac{1}{6} \times \pi \times (63)^2=6236.9\approx6237\;\mathrm{cm^2}$
The area of the portion enclosed by the three circles $=(3969\sqrt3-6237)\;\mathrm{cm^2}$
Hence, the correct answer is $3969\sqrt3-6237$.
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