# Volume of a Truncated Pyramid -- Pyramidal Frustum

A truncated square pyramid, also called a square pyramidal frustum, is a solid shape with six faces that can best be described as a pyramid with its apex sliced off parallel to the base. The base and top are squares of unequal sizes, and the four sides between the bottom and the top are trapezoids. The volume of a truncated pyramid is calculated from three measurements: the height, base width, and top width. If you know these three dimensions, you can compute the capacity of any square pyramidal frustum.

Below we give the volume formula, along with several example problems to help you work out real-life geometry problems.

## Formula for Volume

Call the height of the pyramidal frustum H, the length of the edge along base B, and the length of the edge along the top T. Then the volume of the truncated square pyramid is

Volume = H*(B^2 + TB + T^2)/3

The factor (B^2 + TB + T^2)/3 is the Heronian mean of the areas of the top and bottom squares. For comparison, the Heronian mean is less than the arithmetic mean and greater than the geometric mean. This average is then multiplied by the height of the figure to give the volume.

## Example 1

A truncated pyramid has a square base with a side length of 24 cm, and a square top with a side length of 13 cm. The height of the shape is 15 cm. What is the volume?

Here we have B = 24, T = 13, and H = 15. Therefore, the volume is

15*(24^2 + 13*24 + 13^2)/3

= 15*(576 + 312 + 169)/3

= 15*1057/3

= 5285 cm^3

So the volume is 5285 cubic centimeters, or equivalently 5.285 liters.

## Example 2

A square bowl is shaped as a pyramidal frustum. The open top of the bowl has an edge length of 6 inches and the square base has an edge length of 3 inches. The depth of the bowl is 4.5 inches. What is its capacity in cubic inches?

Here, B = 3, T = 6, and H = 4.5. The volume is then

4.5*(3^2 + 6*3 + 6^2)/3

= 4.5*(9 + 18 + 36)/3

= 4.5*63/3

= 94.5 in^3

The capacity is therefore 94.5 cubic inches, or equivalently 0.4091 gallons.

## Example 3

A truncated square pyramid has a volume of 1014 cm^3. The depth is 18 cm and the area of the base is 64 cm^2. What is the area of the top?

For this problem, we have volume = 1014 and H = 18. If the area of the base is 64, then the side length of the square is 8, so B = 8. Now we just need to solve for T. Putting all the information together gives us the equation

1014 = 18*(8^2 + 8T + T^2)/3

169 = 64 + 8T + T^2

0 = T^2 + 8T - 105

0 = (T - 7)(T + 15)

T = 7 and T = -15

The only solution that makes physical sense in the context of the problem is T = 7 cm. Therefore, the area of the top is 49 cm^2.

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