# If there are 4 volleyballs within a cube, that each have a radius of 4 inches, what is the volume...

## Question:

If there are 4 volleyballs within a cube, that each have a radius of 4 inches, what is the volume of the cube?

## Three Dimensional Measure:

It is confusing dealing with a three-dimensional measure compared to a one-dimensional measure. Length is in one dimension measured from some starting point across to an ending point in a single direction. Volume is a three-dimensional measure measured from a starting point in three separate directions that are perpendicular to one another. As such, to measure of one foot in length one measures across an object's length from a starting point to an ending point one foot later in one dimension. A one cubic foot measure reflects measuring across an object's length, width and thickness dimensions where their product equals one cubic foot. Because this measure is the product of three other measures, each individual can be any value as long as when multiplied together they equal one.

## Answer and Explanation:

This question does not give a number of volleyballs to be discussed. The smallest number of volleyballs in the cube would be one. Since cubes are equal on all sides (length, width and thickness) and the volleyball's radius is 4 in. (which would have a diameter of 8 in.), then the cube would have to be 8 In. long, 8 in. wide and 8 in. thick. This means the cube would have to be 512 cubic inches (8in x 8in x 8in).

But since the plural for volleyballs was used in the problem, it was intended for multiple volleyballs to be discussed. The least number of volleyballs in a cube (again of equal sides) would be two across, two wide and two high. Again, the volleyballs have a radius of 4 in. or diameter of 8 in., which two across would be 16 in. So the cube would have to be 16 in. by 16 in. by 16 in. This would have a volume of 4096 inches cubed.

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