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Math Help - area

  1. #1
    Newbie parmis's Avatar
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    area

    Area of black Square?
    Attached Thumbnails Attached Thumbnails area-gggg.jpg  
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  2. #2
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    Re: area

    We need more information before we can help you. Surely the problem gives you more detail than this.
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  3. #3
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    Re: area

    Hello, parmis!

    I will assume that the side of the square is trisected.


    Area of black square?

    Code:
          A   x   E   x   F   x   B
          * - - - * - - - * - - - *
          | *       *   *   *   * |
          |   *      .*.      *   |
          |     *  .*:::*.  *  J  |
       3x |       *:::::::*       | 3x
          |     *   *:::*   *     |
          |   *       *       *   |
          | *θ      *θ  *       * |
          * - - - * - - - * - - - *
          D   x   G   x   H   x   C

    Let the side of square ABCD be 3x.
    Let \angle FDC = \angle BGC = \theta

    Note the right triangle FJB at the upper-right.

    Code:
                x
        F * * * * * * * B
            *      θ *
              *     *
                *  *
                  *
                  J
    FB = x,\;\angle B = \theta
    Hence: . FJ = x\sin\theta
    This is the side of the shaded square.

    Area of shaded square: . A_1 \:=\:(x\sin\theta)^2 \:=\:x^2\sin^2\!\theta\;\;[1]

    In right triangle BCG\!:\;\tan\theta \,=\,\tfrac{3x}{2x} \,=\,\tfrac{3}{2}
    Hence: . \sin\theta \,=\,\tfrac{3}{\sqrt{13}}

    Substitute into [1]: . A_1 \:=\:x^2\left(\frac{3}{\sqrt{13}}\right)^2 \:=\:\frac{9x^2}{13}

    The area of square ABCD\!:\;A \,=\,(3x)^2 \,=\,9x^2


    Therefore: . \frac{A_1}{A} \:=\:\dfrac{\frac{9x^2}{13}}{9x^2} \:=\:\frac{1}{13}

    The area of the shaded square is one-thirteenth that of the original square.
    Thanks from parmis, johng and LimpSpider
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  4. #4
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    Re: area

    I agree with second poster.
    Using Sorobans diagram where the slopes of diagonals are3/2,Iget a rhombus in the middle area = 1/12 of square
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  5. #5
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    Re: area

    Hi all,
    I agree with the first response. More information should have been given. Also bjhopper is correct in asserting that the inner quadrilateral is a non-square rhombus when the outer quadrilateral is square. The attachment shows what I think is a reasonable interpretation of the problem and a solution for the area of the inner square.

    area-mhfgeometry43.png
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