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Math Help - Super Urgent Circles Help!

  1. #1
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    Super Urgent Circles Help!



    In the figure above, CD is the diameter which meets the chord AB in E such that AE=BE=4cm. If CE = 3cm, find the radius of the circle.

    Please help... It's super urgent super super urgently needed... PLEASE
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  2. #2
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    Quote Originally Posted by Ruler of Hell View Post


    In the figure above, CD is the diameter which meets the chord AB in E such that AE=BE=4cm. If CE = 3cm, find the radius of the circle.

    Please help... It's super urgent super super urgently needed... PLEASE
    Hello,

    I've modified your drawing (see attachment)
    You are dealing actually with a right triangle. The height divide the hypotenuse into 2 parts, called p and q. Then the product of these parts equals the square of the height (height theorem of a right triangle(?))

    h^2=p \cdot q. Plug in the values you know: h = 4, q = 3

    4^2=p \cdot 3 \Longleftrightarrow p = \frac{16}{3}. Therefore the hypotenuse c = p + q:

    c = \frac{25}{3}

    EB
    Attached Thumbnails Attached Thumbnails Super Urgent Circles Help!-hoehensatz1.gif  
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  3. #3
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    Hello, Ruler of Hell!

    Another approach . . .
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          *         O         *
          *         *         *
          *       / | \       *
                /   |   \R
           *  /    E|  4  \  *
            * - - - + - - - *A
              *     |3    *
                  * * *
                    D

    We have: . OA = OD = R . and: . OE = R - 3

    In right triangle OEA\!:\;\;(R-3)^2 + 4^2 \:=\:R^2\quad\Rightarrow\quad R^2 - 6R + 9 + 16 \:=\:R^2

    Therefore: . 6R \,=\,25\quad\Rightarrow\quad\boxed{R\,=\,\frac{25}  {6}}

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  4. #4
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    @Soroban: After trying a lot, I also got the same answer.. Thanks a lot!
    @earboth: Thanks a lot nonetheless... I never knew about this method... It'll be helpful to me..

    Thanks both of you people!
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