# Problem of the week #34 - November, 19th 2012

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#### Jameson

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Congratulations to the following members for their correct solutions:

1) MarkFL
2) SuperSonic4
3) soroban
4) Sudharaka
6) caffeinemachine

Solution (from soroban):
Let $O$ be the center of the circle.

The area of sector $AOB \text{ }$ is: .$\frac{1}{6}\pi r^2 \:=\:\frac{\pi}{6}(6^2) \:=\:6\pi$

The area of equilateral triangle $AOB \text{ }$ is: .$\frac{\sqrt{3}}{4}(6^2) \:=\:9\sqrt{3}$

Therefore, area of the segment is: .$6\pi - 9\sqrt{3} \;\approx\;3.261$

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