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Math Help - Area of a circle

  1. #1
    Branden
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    Exclamation Area of a circle

    I'm having trouble solving the following problem:

    A local toy factory is making a new type of doll and are taking great care in creating the eyes. The crosshairs are centered at the center of the eye. The black part of the eye is tangent to the edge of the eye, and the horizontal crosshair crosses 9 millimeters of the white part of the eye, and the vertical crosshair crosses 5 mm of the white part of the eye on both the top and bottom. What is the area, in square millimeters, of the white part of the eye?

    Here is the diagram for the problem:
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  2. #2
    Super Member malaygoel's Avatar
    Joined
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    Quote Originally Posted by Branden
    I'm having trouble solving the following problem:

    A local toy factory is making a new type of doll and are taking great care in creating the eyes. The crosshairs are centered at the center of the eye. The black part of the eye is tangent to the edge of the eye, and the horizontal crosshair crosses 9 millimeters of the white part of the eye, and the vertical crosshair crosses 5 mm of the white part of the eye on both the top and bottom. What is the area, in square millimeters, of the white part of the eye?

    Here is the diagram for the problem:
    Let us first label the diagram
    the horizontal crosshair has total FOUR intersections, name them as A,B, C and D from the left side. the vertical crosshair has five intesections, we will need only top three of them.One of them(third from top) has been named as C.LEt the top two intersections be E and F staring from top.
    Let r and R be the radii of the smaller and bigger circle respectively
    Let the centre of smaller circle be G(centre of bigger circle is C)
    AD = AB +BD
    2R = 9 + 2r ......(1)
    CG = CD -GD
    CG = R - r = 9/2
    BC=BG-CG=r - 9/2
    Join G and F
    GF=r
    FC^2 = GF^2 - CG^2
    FC = (r^2 - 81/4)^(1/2)
    CE = CA
    CF +FE = CB +BA
    (r^2 - 81/4)^(1/2) + 5 = r - 9/2 + 9
    (r^2 - 81/4)^(1/2) = r - 1/2
    (r^2 - 81/4) = r^2 - r + 1/4
    r = 41/2
    R= 25
    Required Area = pi*(R-r)(R+r)
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