Results 1 to 2 of 2

Math Help - Billiards

  1. #1
    Junior Member ihmth's Avatar
    Joined
    Jan 2008
    Posts
    44

    Billiards

    A set of billiard balls contains 15 numbered balls each having a diameter of 2.25 inches. If the container for the 15 billiard balls will be in the form of a right triangular prism, find how much cardboard is needed in order to make the container.

    Need help to this problem. But I already try to solve it but I do not know if its correct.

    -tnx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, ihmth!

    You could have shown us your answer, you know . . .


    A set of billiard balls has 15 balls each having a diameter of 2 inches.
    If the container for the 15 balls will be in the form of a right triangular prism,
    find how much cardboard is needed in order to make the container.

    I must assume that the balls are in a triangular array, like in the rack,
    . . and the box is a triangular "pizza box".

    Along the edge of the box (an equilateral triangle), there are five balls.

    At each end, there is this arrangement:
    Code:
                        \
                         \
                  * * *   \
              *           *\
            *               *
           *                 *
                              \ 
          *         C         *\
          *         *         * \
          *         :         *  \
                    :             \
           *        :        *     \
            *       :       *       \
              *     :     *      60 \
        - - - - - * * * - - - - - - - *
                    A                 B

    The radius is: AC = \tfrac{9}{8}
    . . And we find that: . AB \:=\:\frac{9\sqrt{3}}{8}


    Make a sketch of the five balls in a row and the edges of the box.

    The length of the edge is: . s \;=\;\tfrac{9\sqrt{3}}{8} + \tfrac{9}{8} + \tfrac{9}{4} + \tfrac{9}{4} + \tfrac{9}{4} + \tfrac{9}{8} + \tfrac{9\sqrt{3}}{8}

    . . Hence: . s\;=\;\tfrac{9}{4}(4 + \sqrt{3})



    The top and bottom of the box are equilateral triangles of side s.
    The area of an equilateral triangle of side s is: . \frac{\sqrt{3}}{4}s^2

    The sides of the box are three rectangles of dimensions: . 2\tfrac{1}{4} \times s


    I'll let you crank out the arithmetic . . .

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Geometry applied in Billiards
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 2nd 2009, 03:59 AM

Search Tags


/mathhelpforum @mathhelpforum