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Math Help - Puzzle

  1. #1
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    Puzzle

    What is wrong with the following 'solution' to the following problem?

    Let ABCD be a square with side lengh of six. Circular arcs of radius six are drawn with centers at B and D. Find the area of the intersection of the areas in the square bounded by the arcs.

    The area of the region of the square that is above the oval shaped intersection is 36-1/4 the area of one of the circle. Now there are 2 of these. Hence, the area of the oval shaped intersection is 36-18\pi
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  2. #2
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    Quote Originally Posted by Chris11 View Post
    What is wrong with the following 'solution' to the following problem?

    Let ABCD be a square with side lengh of six. Circular arcs of radius six are drawn with centers at B and D. Find the area of the intersection of the areas in the square bounded by the arcs.

    The area of the region of the square that is above the oval shaped intersection is 36-1/4 the area of one of the circle. Now there are 2 of these. Hence, the area of the oval shaped intersection is 36-18\pi
    1. Your considerations are correct.

    The area which doesn't belong to the red "2-gon" is calculated by:

    2\left(36- \frac14 \cdot \pi \cdot 6^2\right) = 72\left(1-\frac14 \pi\right)

    2. Now the area of the red "2-gon" is calculated by:

    36- 72\left(1-\frac14 \pi\right) = -36+18\pi
    Attached Thumbnails Attached Thumbnails Puzzle-zweieckinquadrat.png  
    Last edited by earboth; July 8th 2010 at 03:51 AM. Reason: typo detected
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  3. #3
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    Hello, Chris11!

    Another approach . . .
    Code:
        D * - - - - - * - * C
          |     *:::::::* |
          |   *:::::::*   |
          |  *::::::*     |
          | ::::::*       | 6
          |*::::*         |
          |:::*           |
          *:*             |
        A * - - - - - - - * B
                  6

    The shaded region is:

    . . \begin{array}{cccccc}<br />
\text{a quarter-circle:} & \frac{1}{4}\pi(6^2) &=& 9\pi \\<br />
\text{minus} \\<br />
\text{half the square:} & \frac{1}{2}(6^2) &=& 18 \end{array}

    Hence: . 9\pi - 18 \;=\;9(\pi-2)


    Therefore, the lens-shaped region has area:

    . . . . 2\cdot9(\pi - 2) \:=\:18(\pi - 2)
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