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Lethal

Geometry question w/ a Circle. find area of a shaded region. (Harder problem)

  • one year ago
  • one year ago

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  1. Lethal
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    |dw:1345779977023:dw|

    • one year ago
  2. Lethal
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    I think what you have to do is to bisect the triangle into 60 degree up top and then the bottom left and right are still 30 degrees. then you do the 30 60 90 triangle thing.

    • one year ago
  3. Lethal
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    i am stuck from there

    • one year ago
  4. alexwee123
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    i did it a different way...

    • one year ago
  5. Lethal
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    well explain.

    • one year ago
  6. alexwee123
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    with integrals...

    • one year ago
  7. Lethal
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    well. it shouldn't be that complicated im only doing geometry.

    • one year ago
  8. Mimi_x3
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    Area of minor segment \[A = \frac{1}{2} r^2 \left(\theta-\sin\theta\right) \] But the only problem is \(r\) is not given..

    • one year ago
  9. Hero
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    I think I meant to post the area of a sector of a circle

    • one year ago
  10. Lethal
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    you guys are thinking too hard its only like subtracting a sector

    • one year ago
  11. Mimi_x3
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    Area of sector - area of triangle

    • one year ago
  12. Hero
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    Yeah that

    • one year ago
  13. Mimi_x3
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    But \(r\) is not given!!! Im thinking of using the cosine rule and sine rule to find \(r\) would it work?

    • one year ago
  14. Lethal
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    here ill take it a bit further. |dw:1345780578744:dw| does that help?

    • one year ago
  15. Hero
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    It will not be hard to find \(r\)

    • one year ago
  16. Hero
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    Is the vertex of angle 120 is the center of the circle?

    • one year ago
  17. Lethal
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    yep

    • one year ago
  18. Mimi_x3
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    |dw:1345780720301:dw| looks like you can use the sine rule..

    • one year ago
  19. Hero
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    Okay, so yeah just do Area of Sector - Area of Triangle like what @Mimi_x3 suggested

    • one year ago
  20. Lethal
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    soo all we need is the area of the circle and then we can use the 120/360 * pi r2 or something and subtract the area from that.

    • one year ago
  21. Mimi_x3
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    lets find \(r\) first

    • one year ago
  22. Lethal
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    |dw:1345780842471:dw|

    • one year ago
  23. Mimi_x3
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    I got 8 as well using the sine rule

    • one year ago
  24. Mimi_x3
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    Now its easy..you can do it yourself

    • one year ago
  25. Lethal
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    ok so 64 pi is the area of the circle

    • one year ago
  26. Lethal
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    and the area of the triangle left side is 1/2 * 4sqroot3 * 4

    • one year ago
  27. Lethal
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    *2 to make up for the other side.

    • one year ago
  28. Mimi_x3
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    I will just use this formula\[A = \frac{1}{2} r^2 \left(\theta-\sin\theta\right) \] No need to derive it

    • one year ago
  29. Lethal
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    ok

    • one year ago
  30. Mimi_x3
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    and it must be in radians..so convert it..

    • one year ago
  31. Lethal
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    oh heck nah jose

    • one year ago
  32. alexwee123
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    you guys are thinking too hard o.0

    • one year ago
  33. Lethal
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    oh wait i got the answer. just do 120/360 which is 1/3 and then i did 1/3 * 64pi which is the area of the circle and it is 64pi/3 and now subtract it by the area of the triangle which is

    • one year ago
  34. Lethal
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    1/2 4sqrt3 * 4 or 8sqrt3 * 2 for the other side

    • one year ago
  35. alexwee123
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    |dw:1345781211630:dw| |dw:1345781273195:dw|

    • one year ago
  36. alexwee123
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    |dw:1345781325118:dw|

    • one year ago
  37. Lethal
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    so the final answer is \[64\pi/3 - 16\sqrt{3}\]

    • one year ago
  38. alexwee123
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    yes good job @Lethal

    • one year ago
  39. Lethal
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    yep, i would best response both of u but i can only do one. thx alex.

    • one year ago
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