Pyramidal Frustum — Definition, Formula & Examples

A pyramidal frustum is the solid that remains when you slice a pyramid with a plane parallel to its base and remove the smaller pyramid on top. It has two parallel polygonal faces (the original base and the cut cross-section) connected by trapezoidal lateral faces.

Given a pyramid with polygonal base $B_1$ and apex $A$ , a pyramidal frustum is the portion of the pyramid between $B_1$ and a plane parallel to $B_1$ that intersects the lateral edges, producing a smaller similar polygon $B_2$ . The perpendicular distance between the two parallel planes is the height $h$ of the frustum.

Key Formula

V = \frac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 \cdot A_2}\right)

Where:

$V$ = Volume of the frustum
$h$ = Perpendicular height between the two parallel bases
$A_1$ = Area of the larger base
$A_2$ = Area of the smaller base

Worked Example

Problem: A square pyramid is cut parallel to its base, forming a frustum with height 9 cm. The larger base is a square with side 10 cm and the smaller base is a square with side 4 cm. Find the volume.

Find base areas: Compute the area of each square base.

A_1 = 10^2 = 100 \text{ cm}^2, \quad A_2 = 4^2 = 16 \text{ cm}^2

Compute the geometric mean term: The formula requires the square root of the product of the two base areas.

\sqrt{A_1 \cdot A_2} = \sqrt{100 \cdot 16} = \sqrt{1600} = 40 \text{ cm}^2

Apply the volume formula: Substitute all values into the frustum volume formula.

V = \frac{9}{3}(100 + 16 + 40) = 3 \times 156 = 468 \text{ cm}^3

Answer: The volume of the frustum is 468 cm³.

Why It Matters

Frustum calculations appear in architecture and engineering whenever structures taper — think of the base of an obelisk or a truncated roof. Mastering this formula also prepares you for integral-based volume methods in calculus, where frustums approximate solids of revolution.

Common Mistakes

Mistake: Using the average of the two base areas instead of the full three-term expression (A₁ + A₂ + √(A₁·A₂)).

Correction: The volume formula specifically includes the geometric mean term √(A₁·A₂). Averaging the bases alone underestimates the volume. Always use all three terms inside the parentheses.