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Math Help - Sector area and probability questions

  1. #1
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    Sector area and probability questions

    Hello. For the first question refer to this:

    http://img693.imageshack.us/img693/8362/46035151.jpg - the outer numbers are degrees

    What I want to do is find the area in the middle there (where the two and arrows are). Not quite sure how to go about it though. I was thinking divide the sections into triangle but I'd need a radius. Anyway the answer is 4pi + 6\sqrt{3} over 3.


    Second question: "A dart is thrown at a board 12m long and 5m wide. Attached to the board are 30 balloons, each with radius 10cm. Assuming each balloon lies entirely on the board, find the probability that a dart that hits the board will also hit a balloon."

    I'm not even sure where to start here.

    The answer is pi/200 is about 0.016.

    Basically I just want to see how each was worked out. Thanks.
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  2. #2
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    Lexington, MA (USA)
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    Hello, Phil!

    Find the area of the middle region.

    Answer: . \dfrac{4\pi + 6\sqrt{3}}{3}
    Code:
                  * * *
              *           *
          A ♥ - - - + - - - ♥ B
           *  *     |     *  *
                *   |1  *
          *       * | *       *
          *         ♥O        *
          *       * | *       *
                *   |1  *
           *  *     |     *  *
            * - - - + - - - *
              *           *
                  * * *

    Note all the 60^o angles.

    We find that the radius is 2.
    The area of the circle is: . \pi r^2 \:=\:\pi(2^2) \:=\:4\pi


    The area of sector OAB is: . \frac{1}{3}(4\pi)

    The area of \Delta OAB\:=\:\frac{1}{2}(2)(2)\sin120^o \;=\;\sqrt{3}

    The area of the segment is: . \dfrac{4\pi}{3} - \sqrt{3}


    The area of the middle region is: . 4\pi - 2\left(\dfrac{4\pi}{3} - \sqrt{3}\right)

    . . =\;\;4\pi - \dfrac{8\pi}{3} + 2\sqrt{3} \;\;=\;\;\dfrac{4\pi}{3} + 2\sqrt{3} \;\;=\;\;\dfrac{4\pi + 6\sqrt{3}}{3}
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  3. #3
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    Thank you - was very helpful.
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