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geometry

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Square ABCD and circle T have equal areas and share the same center O.The circle intersects side AB at points E and F.Given that EF=√(1600-400π),what is the radius of T ?

  • geometry - ,

    Hmmm. Let's set up a diagram.

    Drop a perpendicular from O to AB to intersect at P.
    Let OA intersect circle T at Q.
    Now, let
    b = AE
    h = OP
    a = AQ
    OA is the radius r of T.
    EF = 20√(4-π)
    let d be the diagonal of the square, s√2

    Now, since ABCD and T have the same area,

    s^2 = πr^2
    s = r√π

    c = 1/2 EF = 10√(4-π)
    b+c = s/2 = r/2 √π, so
    b = r/2 √π - 10√(4-π)
    a+r = d/2 = s/√2, so
    a = r - r√(π/2)

    since we have two secants from A,
    b(s-b) = a(d-a)
    plug it all in, and you get r=20.

  • geometry - ,

    I sketched the situation so that, for the square, point A is in quadrant I and B in in quad II

    let the radius of the circle be r
    area of circle = πr^2
    then the area of square is πr^2
    and each side of the square is r√π
    we can call A(r√π/2 , r√π/2)
    EA = r√π/2 - (1/2)EF
    = r√π/2 - (1/2)√(1600 - 400π)
    = r√π/2 - (1/2)(20)√(4 - π)
    = r√π/2 - 10√(4-π)

    so now:
    r^2 = (r√π/2)^2 + (r√π/2 - 10√(4-π) )^2
    arghhhh!!!!!
    I then multiplied by 2, simplified and got

    r^2(π-2) - 20π√(4-√) + 800-200π = 0

    which is a quadratic in r, with
    a=π-2
    b=-20π√(4-π)
    c = 800-200π

    I used my calculator and the quadratic formula to get

    r = 47.85 or r = 3.14285

    checking:
    if r = 3.14285..
    area of circle = 31.0311...
    side of square = r√π = 3.14285√π = 5.5705..
    area of square = 31.0311... YEAHHHHH!

    if r = 47.85..
    area of circle = appr 7193
    side of square = 47.85√π = 84.81..
    area of square = appr. 7193

    both answers are valid,
    r = appr 47.85 or r = 3.14285

  • geometry - ,

    Hmmm. I musta blown it somewhere. Of course, if r=3.14, there's no way that EF could be √(1600-400π) = 18.53

  • Go with Steve's - geometry - ,

    "overthinking" again.

  • geometry - ,

    Steve, my hunch is that you are correct,
    your answer came out "too nice" and clearly mine does not work.

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