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Math Help - area in a circle?

  1. #1
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    area in a circle?



    how do i find the area of the shaded regionnnn?
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  2. #2
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    Quote Originally Posted by hellothere View Post


    how do i find the area of the shaded regionnnn?
    Is anything else known about the triangle?
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  3. #3
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    I'm assuming the triangle is equilateral. You should be able to find the radius of the circle HA easily, since it's part of a 30-60-90 triangle. Then you can find the area of the the circle, subtract the area of the triangle, and divide by 3.
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  4. #4
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    Assuming the triangle {\Delta}ABC is equilateral we can find the radius by finding the length of HA.
    We can create two 30/60/90 triangles within {\Delta}ABC:{\Delta}AHD and {\Delta}HCD, which forms {\Delta}AHC.
    We know {\angle}DAH is equal to 30. Thus, we can let sin(30)=\frac{5}{h} and solve for h. Doing so, we get h=\frac{5}{sin(30)}=10.
    Now, we use Pythagoreans theorem to find the radius (length HA),
    10^2=5^2+d^2 \longrightarrow d^2=10^2-5^2 \longrightarrow d=5\sqrt{3} \longrightarrow 2d=10\sqrt{3}
    So, the area of {\Delta}AHC=\frac{1}{2}(10\sqrt{3})(5)=25\sqrt{3}
    Since we know {\Delta}AHD and {\Delta}HCD are both 30/60/90 triangles, we know {\Delta}AHC gives us {\angle}AHC=120.
    Therefore, the area of the sector HADF=\frac{120}{360}\pi(10^2)=\frac{100\pi}{3}.
    Finally, we just subtract the area of the sector from the area of {\Delta}AHC to get \frac{100\pi}{3}-25\sqrt{3}=61.41

    Sorry, I shoulda drawn a picture...may be confusing.
    Last edited by Danneedshelp; May 28th 2009 at 08:01 PM.
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  5. #5
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    Correct.

    Edit. Oops. I thought you were the original poster.
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