Hello, ihmth!

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

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: $\displaystyle AC = \tfrac{9}{8}$

. . And we find that: .$\displaystyle AB \:=\:\frac{9\sqrt{3}}{8}$

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

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

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

The top and bottom of the box are equilateral triangles of side $\displaystyle s.$

The area of an equilateral triangle of side $\displaystyle s$ is: .$\displaystyle \frac{\sqrt{3}}{4}s^2$

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

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