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Math Help - geometry extra credit help

  1. #1
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    geometry extra credit help

    help?
    Last edited by natbat77; November 16th 2008 at 08:12 PM.
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  2. #2
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    no congruent triangles ... you have some similar triangles, though.

    remember that the sides of similar triangles are proportional.
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  3. #3
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    so theyre similar and right (they're both perpendicular to the ground)
    but where do i go from there?
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  4. #4
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    Hello, natbat77!

    On a level plot of ground there stand two vertical posts, one 6 ft tall, the other 10 ft tall.
    From the top of each pole a rope is stretched to the foot of the other pole.
    How far above ground is the point where the ropes cross?
    Code:
                                  * C
                                * |
                              *   |
                            *     |
                          *       |
                        *         |
        A *           *           | 10
          |   *   P *             |
          |       *               |
        6 |     * :   *           |
          |   *   :h      *       |
          | *     :           *   |
        B * - - - * - - - - - - - * D
              a   Q       b

    The poles are: . AB = 6,\;\;CD = 10
    The ropes AD and BC cross at P.
    PQ = h is perpendicular to BD.
    Let a = BQ,\;\;b = QD

    In similar right triangles PQD\text{ and }ABD\!:\;\;\frac{h}{b} \:=\: \frac{6}{a+b} \quad\Rightarrow\quad b \:=\:\frac{ah}{6-h} .[1]

    In similar right triangles PQB \text{ and }CDB\!:\;\;\frac{h}{a} \:=\:\frac{10}{a+b} \quad\Rightarrow\quad b \:=\:\frac{10a - ah}{h} .[2]


    Equate [1] and [2]: . \frac{ah}{6-h} \:=\:\frac{10a-ah}{a} \quad\Rightarrow\quad ah^2 \:=\:60a - 6ah - 10ah + ah^2

    . . 16ah \:=\:60a \quad\Rightarrow\quad h \:=\:\frac{60a}{16a} \quad\Rightarrow\quad\boxed{ h \:=\:\frac{15}{4}\text{ ft}}

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  5. #5
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    thank you so much!
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