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Thread: Annulus Area Question

  1. #1
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    Annulus Area Question

    The given chord length 2X is tangent to the inner circle What is the area of the annulus?"

    How can I go about understanding this? There's an image of a circle but not sure how to appropriately solve it. I'd appreciate any feedback. Thanks!
    Last edited by mr fantastic; May 19th 2010 at 07:27 PM. Reason: Restored deleted question.
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  2. #2
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    Hello, LydiaK67!

    The given chord length $\displaystyle 2x$ is tangent to the inner circle.
    What is the area of the annulus?
    Code:
                        * * * 
                    *           *
                *   x     C     x   *
            A o - - - - * o * - - - - o B
                    *     |     *   *
           *      *       |       *      *
                 *       r|     *R *
          *               |   *           *
                *         | *       *
          * - - * - - - - o - - - - * - - *
                          O

    $\displaystyle AB$ is a chord of circle $\displaystyle O$ with radius $\displaystyle R.$
    It is tangent to the inner circle of radius $\displaystyle r$ at $\displaystyle C.$
    . . $\displaystyle AC \,=\,CB\,=\,x$


    The area of the annulus is: .$\displaystyle A \;=\;\pi R^2 - \pi r^2 \;=\;\pi(R^2-r^2)$ .[1]


    In right triangle $\displaystyle OCB\!:\;\;r^2 + x^2 \:=\:R^2 \quad\Rightarrow\quad R^2 - r^2 \:=\:x^2$ .[2]


    Substitute [2] into [1]: .$\displaystyle \boxed{A \;=\;\pi x^2}$

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