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Math Help - Rectangle in a circle

  1. #1
    srh
    srh is offline
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    Rectangle in a circle

    Hi guys, I'm not sure if this is the right place for this question but here goes..

    A washer is made from a circular disc of radius 5.5cm by removing a rectangular area of 18cm2. If a line is drawn from the centre of the washer through a corner of the rectangular area to the outer radius of the washer, then the distance from the edge to the outer radius of the washer is 0.65cm.

    Determine the dimensions of the rectangular area removed.

    If someone could please help me with this?

    I need to know the answer but more importantly I want to know how/why the method used works.

    Thanks
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  2. #2
    Super Member

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    Re: Rectangle in a circle

    Hello, srh!

    Did you make a sketch?


    A washer is made from a circular disc of radius 5.5 cm by removing a rectangular area of 18 cm2.
    If a line is drawn from the centre of the washer through a corner of the rectangular area to the outer radius
    of the washer, then the distance from the edge to the outer radius of the washer is 0.65cm.

    Determine the dimensions of the rectangular area removed.

    Code:
                  * * *
              *           *
            *               *
           *                 *
                *-------*
          *     |     x |     *
          *     |  O*---+A    *
          *     |     * |y    *
                *-------*
           *            B *  *
            *               *
              *           *  P
                  * * *
    The center of the circle is O.
    The radius is OP = 5.5
    Distance BP = 0.65
    Hence:. OB = 4.85
    Let x = OA,\;y = AB.

    In right triangle OAB, we have: . x^2 + y^2 \:=\:(4.85)^2 .[1]

    The area is 18: . (2x)(2y) \:=\:18 \quad\Rightarrow\quad y \:=\:\tfrac{9}{2x}

    Substitute into [1]: . x^2 + \left(\tfrac{9}{2x}\right)^2 \:=\:(4.85)^2 \quad\Rightarrow\quad 4x^4 - 94.09x^2 + 81 \:=\:0

    Quadratic Formula: . x^2 \;=\;\frac{94.09 \pm \sqrt{7556.9281}}{8} \;=\;\begin{Bmatrix}22.6275742 \\ 0.894925803\end{array}

    Then: . x \;=\;\begin{Bmatrix}4.756844984 & \text{too big} \\ 0.946005181 \end{array}

    The dimensions of the rectangle are: . \begin{Bmatrix}2x &\approx&1.892 \\ 2y &\approx& 9.514 \end{Bmatrix}
    Thanks from Goku and srh
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