# Billiards

• Dec 1st 2008, 07:07 AM
ihmth
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:)
• Dec 1st 2008, 08:24 AM
Soroban
Hello, ihmth!

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

Quote:

A set of billiard balls has 15 balls each having a diameter of 2¼ 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 . . .