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Math Help - Right Triangles

  1. #1
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    Right Triangles

    Find the value of x and y:

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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by icebreaker09 View Post
    Find the value of x and y:

    I'm sorry, but I find this to be a bit ambiguous. Which segments are 5, 12, and 13? (I've got what I think is a good guess, but I need to be sure.)

    -Dan
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    Sorry, but that's all it said: I can post the first part though (already solved it): A frame in the shape of the simple scissors truss shown a the right below can be used to support a peaked roof. The weight of the roof compresses some parts of the frame, while other parts are in tension. A frame made with s segments joined at j points is stable is s is greater to or less than 2j-3. In this truss, 9 segments connect 6 points. Verify that the truss is stable. Then find the values of x and y.
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  4. #4
    Forum Admin topsquark's Avatar
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    For convenience I'm going to label the points top down and left to right with A, B, C, etc.

    There are two right triangles: triangle BDE and CDG. The right angles are angles DBE and DCG, though the diagram doesn't make it look like that.

    The problem I am having here is that the apex angle EAG is not specified. We may have a number of possible apex angles for this construction, all of which give different x and y values.

    -Dan
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  5. #5
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    Hello, icebreaker!

    Find the value of x and y.
    Code:
                    A
                    *
                   /*\
                  / * \
               y /  *  \ y
                /   *   \
               /    *    \
              /     *     \
           F *  5   *   5  * E
            /   *   *   *   \
        12 /        O        \ 12
          /  13 *   *   * 13  \
         /  *       *       *  \
      B * - - - - - * - - - - - * C
              x     D     x

    Triangles OFB and OEC have sides: 5, 12, 13.
    . . Since 5^2 + 12^2 \,=\,13^2, they are right triangles.
    Then: \angle OFB = \angle OEC = 90^o

    In right triangle BEC\!:\;\;BC^2 \:=\:BE^2 + EC^2
    . . That is: . (2x)^2\:=\:18^2+12^2\quad\Rightarrow\quad4x^2\,=\,  468\quad\Rightarrow\quad x^2\,=\,117
    Therefore: . x \:=\:\sqrt{117}\quad\Rightarrow\quad\boxed{x \,=\,3\sqrt{13}}

    In right triangle BEA\!:\;\;FA^2 \:=\:BE^2 + AE^2
    . . That is: . (y+12)^2\:=\:18^2 + y^2\quad\Rightarrow\quad y^2 + 24y + 144\:=\:324 + y^2
    Therefore: . 24y \,=\,180\quad\Rightarrow\quad\boxed{y \,=\,\frac{15}{2}}

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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, icebreaker!


    Triangles OFB and OEC have sides: 5, 12, 13.
    . . Since 5^2 + 12^2 \,=\,13^2, they are right triangles.
    Then: \angle OFB = \angle OEC = 90^o

    In right triangle BEC\!:\;\;BC^2 \:=\:BE^2 + EC^2
    . . That is: . (2x)^2\:=\:18^2+12^2\quad\Rightarrow\quad4x^2\,=\,  468\quad\Rightarrow\quad x^2\,=\,117
    Therefore: . x \:=\:\sqrt{117}\quad\Rightarrow\quad\boxed{x \,=\,3\sqrt{13}}

    In right triangle BEA\!:\;\;FA^2 \:=\:BE^2 + AE^2
    . . That is: . (y+12)^2\:=\:18^2 + y^2\quad\Rightarrow\quad y^2 + 24y + 144\:=\:324 + y^2
    Therefore: . 24y \,=\,180\quad\Rightarrow\quad\boxed{y \,=\,\frac{15}{2}}

    Am I being too suspiscious here? How do we know that BO and OE form a straight line when connected?

    -Dan
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