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Math Help - Finding the length of line

  1. #1
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    Finding the length of line

    If you know that AC=3,AB=4,XY=1,then what is XW=?

    WXYZ and ABCD are rectangulars
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  2. #2
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    Re: Finding the length of line

    Hello, Mhmh96!

    \text{Given: }\:AC=3,\;AB=4,\;XY=1.
    . . . . . . W\!XYZ\text{ and }ABDC\text{ are rectangles.}

    \text{Find the length of }XW.


    Draw diagonal BC.
    Triangle ABC is a 3-4-5 right triangle.

    We find that XW and Y\!Z are parallel to BC.
    And all the right triangles are similar.

    \text{In right triangle }X\!BY\!:\:XY = 1,\;B\!X = \tfrac{3}{5},\;BY = \tfrac{4}{5}

    Can you finish the problem?

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    Re: Finding the length of line

    Actually no Soroban,can you explain more .
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    Re: Finding the length of line

    Quote Originally Posted by Soroban View Post
    Hello, Mhmh96!


    Draw diagonal BC.
    Triangle ABC is a 3-4-5 right triangle.

    We find that XW and Y\!Z are parallel to BC.
    And all the right triangles are similar.

    \text{In right triangle }X\!BY\!:\:XY = 1,\;B\!X = \tfrac{3}{5},\;BY = \tfrac{4}{5}

    Can you finish the problem?

    XW and YZ are not necessarily parallel to BC.
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    Re: Finding the length of line

    Quote Originally Posted by Mhmh96 View Post
    If you know that AC=3,AB=4,XY=1,then what is XW=?

    WXYZ and ABCD are rectangulars
    I'm going to call length \displaystyle \begin{align*} CZ = XB = x \end{align*}, length \displaystyle \begin{align*} WC = BY = y \end{align*} and \displaystyle \begin{align*} WX = YZ = z  \end{align*}.

    From the smallest triangles, we can see that \displaystyle \begin{align*} x^2 + y^2 = 1 \end{align*} by Pythagoras.

    We can also see from the larger triangles that

    \displaystyle \begin{align*} (4 - x)^2 + (3 - y)^2 &= z^2 \\ 16 - 8x + x^2 + 9 - 6y + y^2 &= z^2 \\ 16 - 8x + 9 - 6y + 1 &= z^2 \textrm{ since }x^2 + y^2 = 1 \\ 26 - 8x - 6y &= z^2 \\ z &= \sqrt{26 - 8x - 6y} \end{align*}


    Now note that the area of the largest rectangle is \displaystyle \begin{align*} 12 \end{align*}. If we evaluate the area of each smaller piece, we find

    \displaystyle \begin{align*} (3 - y)(4 -x ) + \frac{1}{2}xy + (3 - y)(4 - x) + \frac{1}{2}xy + \sqrt{26 - 8x - 6y} &= 12 \\ 2\left(12 - 3x - 4y + xy\right) + xy + \sqrt{26 - 8x - 6y} &= 12 \\ 24 - 6x - 8y + 2xy + xy + \sqrt{26 - 8x - 6y} &= 12 \\ 24 - 6x - 8y + 3xy + \sqrt{26 - 8x - 6y} &= 12 \end{align*}

    Substituting \displaystyle \begin{align*} x = \sqrt{1 - y^2} \end{align*}, (since we know \displaystyle \begin{align*} x^2 + y^2 = 1 \end{align*}) will enable solving for \displaystyle \begin{align*} y \end{align*}, which will enable solving for \displaystyle \begin{align*} x \end{align*}, and finally solving for \displaystyle \begin{align*} z \end{align*}.

    See how you go.
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  6. #6
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    Re: Finding the length of line

    Following Soroban's hint I calculated

    WX= 4.25= ( 17/5) / cos 36.87
    ZY = 3.66 = (11/5 /sin 36.87

    WXYZ is not a rectangle but a trapezoid
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    Re: Finding the length of line

    The rectangle has to be in a particular position, the angle AXW being approximately 31.7501 degrees.
    WX turns out to be approximately 4.08512.
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