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Complicated Math Question.....

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Gotta remember this one theorem... something about inscribed angles and such..
Is that radius 6?

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actually, I don't know for sure.
Hmm...we cant answer the question unless we know what the radius is =/. Lets just operate assuming its 6.
Then the total circumference is 12pi. This tells us that the measure of angle EDN (i think thats a D in the middle) is:\[\frac{2\pi}{12\pi}\cdot 360=60\]degrees. This means that the area of that slice is:\[\frac{60}{360}\cdot \pi (6)^2=6\pi\]The same can be said of that second slice opposite EDN. So now we have to find the are of the last piece.
So we have EDN and BDA so far. 6pi each for a total of 12 pi.
To get the area of the last piece, we need the area of the slice ADN, minus the area of the triangle ADN.
Since the angle EDN is 60 degrees, it turns out that angle EAN is half of that, 30 degree (because it hits the same arc EN).
So now looking at that triangle (ADN), we see that:|dw:1364788041228:dw|the area is going to be\[\frac{1}{2}\cdot 6\cdot 6\sin(120)=\frac {1}{2}\cdot 36\cdot \frac{\sqrt{3}}{2}=9\sqrt{3}\]
The area of arc ADN is:\[\frac{120}{360}\cdot \pi (6)^2=12\pi\]So the area of the region is 12pi-9root(3). Putting everything together you get:\[6\pi+6\pi+12\pi-9\sqrt{3}=24\pi-9\sqrt{3}\]A fair warning, arithmetic may be off, check the math to make sure its correct.
24 pi - 9 root 3 is the full answer?

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