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Math Help - Quadrilateral ABCD

  1. #1
    MHF Contributor alexmahone's Avatar
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    Quadrilateral ABCD

    Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
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  2. #2
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    Quote Originally Posted by alexmahone View Post
    Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
    Unfortunately I can't provide you with an algebraic solution.

    In the attachment you see the only possible dimensions of the quadrilateral with approximative values.

    Maybe you can use the drawing to control your exact result.

    (The radius is of course r = \dfrac h2 \approx 0.87
    Attached Thumbnails Attached Thumbnails Quadrilateral ABCD-trapez_mitinkreis.png  
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  3. #3
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    Hello, alexmahone!

    I have a start on this problem, but haven't finished it.


    Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD.
    AB = 3CD, and the area of the quadrilateral is 4.
    If a circle can be inscribed in the quadrilateral, find its radius.
    Code:
          D - - - x - - - C
       -  o-------*-*-*---o
       :  |   *     :     *\
       :  | *       :       *
       :  |*        :r       *
       :  |         :         \
       :  *         :         *\
      2r  *         *         * \
       :  *         :         *  \
       :  |         :             \
       :  |*        :r       *     \
       :  | *       :       *       \
       :  |   *     :     *          \ 
       -  o-------*-*-*---------------o
          A - - - - - - 3x  - - - - - B
    ABCD is a trapezoid with two right angles.

    Its height is 2r, its bases are AB = 3x,\:CD = x.

    The area of a trapezoid is: . A \;=\;\frac{h}{2}(b_1 + b_2)

    So we have: . A \;=\;\frac{2r}{2}(x + 3x) \:=\:4 \quad\Rightarrow\quad rx \:=\:1


    Now where can I find another equation?

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  4. #4
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    Quote Originally Posted by alexmahone View Post
    Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
    I hope that I now found the solution:

    For convenience I set CD = t.

    Your quadrilateral is a trapezium with t and 3 t as the lengths of the parallel sides. The area is calculated by:

    A = \dfrac{t+3t}2 \cdot h = 4 which yield h = \dfrac2t and therefore r = \dfrac1t

    The length of x = t-  \dfrac1t and

    y = 3t- \dfrac1t

    Since FB = 2t you can calculate the sides of the right triangle CFB:

    \left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2

    Expand the brackets and collect like terms:

    \dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{  t=\sqrt{\dfrac43}}

    Re-substitute this value and you'll get r = \dfrac12 \cdot \sqrt{3}


    EDIT: Too late
    Attached Thumbnails Attached Thumbnails Quadrilateral ABCD-trapez_mitinkreis2.png  
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by earboth View Post
    I hope that I now found the solution:

    For convenience I set CD = t.

    Your quadrilateral is a trapezium with t and 3 t as the lengths of the parallel sides. The area is calculated by:

    A = \dfrac{t+3t}2 \cdot h = 4 which yield h = \dfrac2t and therefore r = \dfrac1t

    The length of x = t- \dfrac1t and

    y = 3t- \dfrac1t

    Since FB = 2t you can calculate the sides of the right triangle CFB:

    \left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2

    Expand the brackets and collect like terms:

    \dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{  t=\sqrt{\dfrac43}}

    Re-substitute this value and you'll get r = \dfrac12 \cdot \sqrt{3}


    EDIT: Too late
    Thanks so much!
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