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grayp

  • 3 years ago

Complicated Math Question.....

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  1. grayp
    • 3 years ago
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  2. wio
    • 3 years ago
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    Gotta remember this one theorem... something about inscribed angles and such..

  3. joemath314159
    • 3 years ago
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    Is that radius 6?

  4. grayp
    • 3 years ago
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    actually, I don't know for sure.

  5. joemath314159
    • 3 years ago
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    Hmm...we cant answer the question unless we know what the radius is =/. Lets just operate assuming its 6.

  6. joemath314159
    • 3 years ago
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    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
    • 3 years ago
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    |dw:1364787772939:dw|

  8. joemath314159
    • 3 years ago
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    So we have EDN and BDA so far. 6pi each for a total of 12 pi.

  9. joemath314159
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    24 pi - 9 root 3 is the full answer?

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