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Math Help - Geometry(?) Problem

  1. #1
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    Geometry(?) Problem

    I'm pretty sure I've done this in grade 8, but sadly I threw out last year's Math notes... Reminder anybody? Please and thank you's.

    In the diagram, triangle XYZ is right-angled at X, with YX = 60 and XZ = 80. W is the point on YZ so that WX is perpendicular to YZ. Determine the length of WZ.

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  2. #2
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    Hello, anh1tran!

    In the diagram, triangle XYZ is right-angled at X,
    with YX = 60 and XZ = 80.
    XW is perpendicular to YZ.
    Determine the length of WZ.
    Code:
                X
                *
               *|   *
           60 * |       *   80
             *  |           *
            *   |               *
           *    |                   *
          *  *  *  *  *  *  *  *  *  *  *
          Y     W                       Z
          : - - - - - 100 - - - - - - - :

    Using Pythagorus: . YZ^2 \:=\:60^2+80^2 \:=\:100,000\quad\Rightarrow\quad YZ = 100

    Since \Delta WXZ \sim \Delta XYZ, we have: . \frac{WZ}{XZ} \:=\:\frac{XZ}{YZ}

    Hence: . \frac{WZ}{80}\:=\:\frac{80}{100}\quad\Rightarrow\q  uad WZ \:=\:\frac{6400}{100} \:=\:\boxed{64}

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  3. #3
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    Oh my.. Thank you so much!
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