Find area of the shaded region of the circle

The area of a circular sector is given by: \(\displaystyle \L\\\frac{1}{2}r^{2}({\theta}-sin{\theta})\)

You have \(\displaystyle \L\\\frac{1}{2}(12)^{2}(\frac{\pi}{3}-sin(\frac{\pi}{3}))=24{\pi}-36\sqrt{3}\approx{13.04}\)
 
or ...

(1/6 area of circle) - (area of equilateral triangle) =

\(\displaystyle \L \frac{\pi \cdot 12^2}{6} - \frac{\sqrt{3}}{4} \cdot 12^2 =\)

\(\displaystyle \L 144\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) \approx 13.04\)
 
thanks, i asked my teacher about it, and she said i was right when i got 13. she just graded my test wrong. :oops:
 
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