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Lethal Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Lethal Group Title
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    |dw:1345779977023:dw|

    • 2 years ago
  2. Lethal Group Title
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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.

    • 2 years ago
  3. Lethal Group Title
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    i am stuck from there

    • 2 years ago
  4. alexwee123 Group Title
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    i did it a different way...

    • 2 years ago
  5. Lethal Group Title
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    well explain.

    • 2 years ago
  6. alexwee123 Group Title
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    with integrals...

    • 2 years ago
  7. Lethal Group Title
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    well. it shouldn't be that complicated im only doing geometry.

    • 2 years ago
  8. Mimi_x3 Group Title
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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..

    • 2 years ago
  9. Hero Group Title
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    I think I meant to post the area of a sector of a circle

    • 2 years ago
  10. Lethal Group Title
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    you guys are thinking too hard its only like subtracting a sector

    • 2 years ago
  11. Mimi_x3 Group Title
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    Area of sector - area of triangle

    • 2 years ago
  12. Hero Group Title
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    Yeah that

    • 2 years ago
  13. Mimi_x3 Group Title
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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?

    • 2 years ago
  14. Lethal Group Title
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    here ill take it a bit further. |dw:1345780578744:dw| does that help?

    • 2 years ago
  15. Hero Group Title
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    It will not be hard to find \(r\)

    • 2 years ago
  16. Hero Group Title
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    Is the vertex of angle 120 is the center of the circle?

    • 2 years ago
  17. Lethal Group Title
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    yep

    • 2 years ago
  18. Mimi_x3 Group Title
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    |dw:1345780720301:dw| looks like you can use the sine rule..

    • 2 years ago
  19. Hero Group Title
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    Okay, so yeah just do Area of Sector - Area of Triangle like what @Mimi_x3 suggested

    • 2 years ago
  20. Lethal Group Title
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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.

    • 2 years ago
  21. Mimi_x3 Group Title
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    lets find \(r\) first

    • 2 years ago
  22. Lethal Group Title
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    |dw:1345780842471:dw|

    • 2 years ago
  23. Mimi_x3 Group Title
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    I got 8 as well using the sine rule

    • 2 years ago
  24. Mimi_x3 Group Title
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    Now its easy..you can do it yourself

    • 2 years ago
  25. Lethal Group Title
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    ok so 64 pi is the area of the circle

    • 2 years ago
  26. Lethal Group Title
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    and the area of the triangle left side is 1/2 * 4sqroot3 * 4

    • 2 years ago
  27. Lethal Group Title
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    *2 to make up for the other side.

    • 2 years ago
  28. Mimi_x3 Group Title
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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

    • 2 years ago
  29. Lethal Group Title
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    ok

    • 2 years ago
  30. Mimi_x3 Group Title
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    and it must be in radians..so convert it..

    • 2 years ago
  31. Lethal Group Title
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    oh heck nah jose

    • 2 years ago
  32. alexwee123 Group Title
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    you guys are thinking too hard o.0

    • 2 years ago
  33. Lethal Group Title
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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

    • 2 years ago
  34. Lethal Group Title
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    1/2 4sqrt3 * 4 or 8sqrt3 * 2 for the other side

    • 2 years ago
  35. alexwee123 Group Title
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    |dw:1345781211630:dw| |dw:1345781273195:dw|

    • 2 years ago
  36. alexwee123 Group Title
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    |dw:1345781325118:dw|

    • 2 years ago
  37. Lethal Group Title
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    so the final answer is \[64\pi/3 - 16\sqrt{3}\]

    • 2 years ago
  38. alexwee123 Group Title
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    yes good job @Lethal

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

    • 2 years ago
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