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Math Help - SoYouThinkYouAreAGenius?

  1. #1
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    SoYouThinkYouAreAGenius?

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  2. #2
    MHF Contributor
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    How about stubborn?

    Quote Originally Posted by hotgal24 View Post
    Okay, let us call AB =x; AD = y; CE = u and CF = v

    In triangle ABE:
    (1/2)(x)(y-u) = 4
    y-u = 8/x
    u = y -(8/x) ---------**

    In triangle ADF:
    (1/2)(y)(x-v) = 5
    x-v = 10/y
    v = x -(10/y) --------**

    In triangle CEF:
    (1/2)(u)(v) = 3
    u*v = 6
    Substitutions,
    [y -(8/x)]*[x -(10/y)] = 6
    xy -10 -8 +80/(xy) = 6
    xy +80/(xy) = 24
    Multiply both sides by xy,
    (xy)^2 +80 = 24(xy)
    (xy)^2 -24(xy) +80 = 0
    Factoring that,
    (xy -20)(xy -4) = 0
    So,
    xy = 20 or 4

    xy cannot be 4, so xy = 20.

    Therefore, in rectangle ABCD,
    Let K = area of triangle AEF,
    x*y = 3 +4 +5 +K
    K = 20 -3-4-5 = 8 -----------answer.
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