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Lethal

  • 2 years ago

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

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  1. Lethal
    • 2 years ago
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    |dw:1345779977023:dw|

  2. Lethal
    • 2 years ago
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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.

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

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

  5. Lethal
    • 2 years ago
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    well explain.

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

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

  8. Mimi_x3
    • 2 years ago
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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..

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

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

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

  12. Hero
    • 2 years ago
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    Yeah that

  13. Mimi_x3
    • 2 years ago
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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?

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

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

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

  17. Lethal
    • 2 years ago
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    yep

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

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

  20. Lethal
    • 2 years ago
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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.

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

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

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

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

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

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

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

  28. Mimi_x3
    • 2 years ago
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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

  29. Lethal
    • 2 years ago
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    ok

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

  31. Lethal
    • 2 years ago
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    oh heck nah jose

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

  33. Lethal
    • 2 years ago
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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

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

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

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

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

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

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

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