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Math Help - Area of semicircle + isoceles triangle

  1. #1
    Member realintegerz's Avatar
    Joined
    Aug 2008
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    Area of semicircle + isoceles triangle




    I'm trying to find the formula to get the area as a function of that length....

    So far I got

    Area =

    1/2[sq. root(2L^2)][sq. root(L^2 - [(sq. root((2L^2)/2)]^2)] + [pi([(sq. root((2L^2)/2)^2]]/2

    The question asks...find the length of l for area 30 cm^2
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  2. #2
    MHF Contributor
    Joined
    Apr 2008
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    The radius r of the semicircle is r = \frac{\sqrt{2}}{2}L.

    Hence, the area of the semicircle is \frac{1}{2}\pi{r^2} = \frac{1}{2}\pi \cdot \left(\frac{1}{2}L^2\right) = \frac{1}{4}\pi{L^2}.

    The area of the triangle is \frac{1}{2}L^2.

    So the total area A is A = \left(\frac{1}{2} + \frac{\pi}{4}\right)L^2.

    Solve A = 30 for L.
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