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What is the area of the segment formed by a chord in a circle of radius 10 cm when the angle subtended at the centre is $120^\circ$?
- $\frac{100\pi}{3}-25\sqrt{3}$
- $\frac{100\pi}{3}-50\sqrt{3}$
- $50\pi-25\sqrt{3}$
- $50\pi-50\sqrt{3}$
Correct answer: $\frac{100\pi}{3}-25\sqrt{3}$
Solution
The area of the sector is $\frac{120}{360}\pi(10)^2=\frac{100\pi}{3}$. The area of the triangle formed by the two radii is $\frac12\cdot 10\cdot 10\cdot \sin 120^\circ=25\sqrt{3}$. Subtracting gives the segment area $\frac{100\pi}{3}-25\sqrt{3}$.
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