## grayp Group Title Complicated Math Question..... one year ago one year ago

1. grayp Group Title

2. wio Group Title

Gotta remember this one theorem... something about inscribed angles and such..

3. joemath314159 Group Title

4. grayp Group Title

actually, I don't know for sure.

5. joemath314159 Group Title

Hmm...we cant answer the question unless we know what the radius is =/. Lets just operate assuming its 6.

6. joemath314159 Group Title

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.

7. joemath314159 Group Title

|dw:1364787772939:dw|

8. joemath314159 Group Title

So we have EDN and BDA so far. 6pi each for a total of 12 pi.

9. joemath314159 Group Title

To get the area of the last piece, we need the area of the slice ADN, minus the area of the triangle ADN.

10. joemath314159 Group Title

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).

11. joemath314159 Group Title

So now looking at that triangle (ADN), we see that:|dw:1364788041228:dw|the area is going to be$\frac{1}{2}\cdot 6\cdot 6\sin(120)=\frac {1}{2}\cdot 36\cdot \frac{\sqrt{3}}{2}=9\sqrt{3}$

12. joemath314159 Group Title

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.

13. grayp Group Title

24 pi - 9 root 3 is the full answer?