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Thread: stuck on these 3 problems

  1. #1
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    stuck on these 3 problems

    solved.
    Last edited by oohhoohh; Mar 2nd 2010 at 02:46 PM.
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  2. #2
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    Hello, oohhoohh!

    2. Find the radius of a circle in which a 96 cm chord
    is 16 cm closer to the center than an 80 cm chord.
    Code:
                 40       E
        C * - - - - - - - o - - - - - - - . D
            *             |             .
              *           |16         .
              R *        D|       48.
        A . - - - * - - - o - - - . - - - * B
              .     *     |     .     *
                  .   *   |x  .   * R
                      . * | . *
                          *
                          O

    The center of the circle is $\displaystyle O.$
    The radius is: .$\displaystyle R \:=\:OA = OB = OC = OD.$
    The chords are: .$\displaystyle AB = 96,\;CD = 80$
    Their distance is $\displaystyle DE = 16.$
    Let $\displaystyle x = OD.$

    In right triangle $\displaystyle ODB\!:\;\;x^2 + 48^2 \:=\:R^2$ .[1]

    In right triangle $\displaystyle OEC\!:\;\;(x+16)^2 + 40^2 \:=\:R^2$ .[2]


    Equate [1] and [2]: .$\displaystyle x^2 + 48^2 \:=\x+16)^2 + 40^2 \quad\Rightarrow\quad 32x \:=\:448 \quad\Rightarrow\quad x \:=\:14$

    Substitute into [1]: .$\displaystyle 14^2 + 48^2 \:=\:R^2 \quad\Rightarrow\quad R^2 \:=\:2500$

    . . Therefore: . $\displaystyle R \:=\:50\text{ cm}$

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