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Math Help - Computing the distance

  1. #1
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    Computing the distance

    Triangle ABC has side lengths AB = 231;BC = 160; and AC = 281: Point D is constructed on
    the opposite side of line AC as point B such that AD = 178 and CD = 153: How to compute the distance
    from B to the midpoint of segment AD
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  2. #2
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    Re: Computing the distance

    Hello, Mhmh96!

    I assume you made a sketch . . .


    Triangle ABC has side lengths AB = 231, BC = 160, and AC = 281.
    Point D is constructed on the opposite side of line AC as point B such that AD = 178 and CD = 153.
    Compute the distance from B to the midpoint of segment AD.

    The diagram seems to be something like this.
    Code:
                    A
                    o     178
                   *  *   *      D
                  *     *       o
                 *        *      *
            231 *           *     *
               *          281 *    *
              *                 *   * 153
             *                    *  *
            *                       * *
           *                          **
        B o  *  *  *  *  *  *  *  *  *  o C
                        160
    Draw segment AD.

    We find that: . 160^2 + 231^2 \:=\:281^2
    . . Hence: . \angle B \,=\,90^o.

    Now the diagram looks like this:
    Code:
        A o
          * **
          * 1 * * 178
          *     *  *
          *       *   *
          *         *    *   D
      231 *           *     o
          *         281 *    *
          *               *   *
          *                 * 2* 153
          *                   * *
          *                     **
        B o  *  *  *  *  *  *  *  o C
                    160
    Let \angle 1 = \angle BAC,\;\angle2 = \angle ACD


    Law of Cosines:

    \cos(\angle 1) \:=\:\frac{231^2 + 281^2 - 160^2}{2(231)(281)} \:=\:\frac{106,\!722}{129,\!822} \:=\:\frac{231}{281}

    \cos(\angle 2) \:=\:\frac{281^2 + 153^2 - 178^2}{2(281)(153)} \:=\:\frac{70,\!686}{85,\!986} \:=\:\frac{231}{281}

    Hence: . \angle 1 = \angle 2 \quad\Rightarrow\quad CD \parallel AB \quad\Rightarrow\quad \angle BCD = 90^o


    Finally, the diagram looks like this:

    Code:
        A o
          * * 89
          *   *  M
       78 * - - o
          *     : * 89
          *     :   *  D
          o - - : - - o
          *     :     *
          *     :     *
      153 *     :     * 153
          *     :     *
          *     :     *
          *     :     *
        B o - - o - - o C
            80  N  80
    Draw segment BM.

    We see that MN\:=\:39 + 153 \:=\:192

    BM is the hypotenuse of right triangle M\!N\!B.

    . . Hence: . B\!M^2 \:=\:80^2 + 192^2 \:=\:43,\!264

    Therefore: . B\!M \:=\:208
    Thanks from Mhmh96
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