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Truncated pyramid volume calculator

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What is a truncated pyramid?

A truncated pyramid (frustum), is a three-dimensional geometric shape formed by cutting off the top of a pyramid with a plane parallel to its base. This results in two parallel polygonal bases (the original base and the truncated top) connected by trapezoidal faces. Truncated pyramids are commonly encountered in architecture, engineering, and everyday objects like buckets or lampshades.

Formula for volume of a truncated pyramid

The volume V V of a truncated pyramid can be calculated using the areas of the two bases and the height (the perpendicular distance between the bases). The formula is:

V=13h(A1+A2+A1A2)V = \frac{1}{3} \cdot h \cdot \left( A_1 + A_2 + \sqrt{A_1 \cdot A_2} \right)

Where:

  • A1 A_1 = Area of the lower base
  • A2 A_2 = Area of the upper base
  • h h = Height of the truncated pyramid

This formula applies only if the truncation is parallel to the base and both bases are similar in shape (e.g., both squares or both rectangles).

Step-by-step calculation examples

Example 1: Square bases

Problem:
A truncated pyramid has a lower base area of 100cm2 100 \, \text{cm}^2, an upper base area of 25cm2 25 \, \text{cm}^2, and a height of 12cm 12 \, \text{cm}. Calculate its volume.

Solution:

  1. Substitute values into the formula: V=1312(100+25+10025)V = \frac{1}{3} \cdot 12 \cdot \left( 100 + 25 + \sqrt{100 \cdot 25} \right)
  2. Simplify the square root term: 10025=2500=50\sqrt{100 \cdot 25} = \sqrt{2500} = 50
  3. Combine terms: V=1312(100+25+50)=4175=700cm3V = \frac{1}{3} \cdot 12 \cdot (100 + 25 + 50) = 4 \cdot 175 = 700 \, \text{cm}^3

Example 2: Rectangular bases

Problem:
A frustum has a lower base of 8m×6m 8 \, \text{m} \times 6 \, \text{m} and an upper base of 4m×3m 4 \, \text{m} \times 3 \, \text{m}. The height is 5m 5 \, \text{m}. Find its volume.

Solution:

  1. Calculate the areas: A1=86=48m2,A2=43=12m2A_1 = 8 \cdot 6 = 48 \, \text{m}^2, \quad A_2 = 4 \cdot 3 = 12 \, \text{m}^2
  2. Substitute into the formula: V=135(48+12+4812)V = \frac{1}{3} \cdot 5 \cdot \left( 48 + 12 + \sqrt{48 \cdot 12} \right)
  3. Simplify the square root term: 576=24\sqrt{576} = 24
  4. Combine terms: V=13584=140m3V = \frac{1}{3} \cdot 5 \cdot 84 = 140 \, \text{m}^3

Historical context and applications

The concept of truncated pyramids dates back to ancient civilizations. For example:

  • Egyptian pyramids were often constructed with truncated tops for religious or structural reasons.
  • Mesopotamian ziggurats resembled stepped truncated pyramids.

Modern applications include:

  • Architecture: Designing skylights or atriums.
  • Engineering: Calculating material volumes for components like chimneys or pipelines.
  • 3D Modeling: Creating tapered shapes in computer graphics.

Common mistakes to avoid

  1. Confusing height with slant height: The height h h is the perpendicular distance between the bases, not the length of the lateral face.
  2. Non-parallel bases: The formula assumes the bases are parallel. If they are not, the shape is not a frustum, and the formula does not apply.
  3. Inconsistent units: Ensure all measurements (areas and height) use the same unit system.

Area of the bases

For the calculation of the area of the bases of a truncated pyramid, you can use the following calculators:

Frequently Asked Questions

How to convert units before calculation?

Convert all measurements to the same unit. For example, if A1=2m2 A_1 = 2 \, \text{m}^2, A2=1500cm2 A_2 = 1500 \, \text{cm}^2, convert A2 A_2 to 0.15m2 0.15 \, \text{m}^2 before applying the formula. For the conversion of units of area, use our converter area units converter.

Why is there a square root in the formula?

The term A1A2 \sqrt{A_1 \cdot A_2} geometrically represents the “average” of the two base areas, accounting for the linear scaling between them due to the height.

What is the volume of a truncated pyramid with bases 10x10 cm and 5x5 cm and height 7 cm?

V=137(100+25+10025)=137175=408.33cm3V = \frac{1}{3} \cdot 7 \cdot \left( 100 + 25 + \sqrt{100 \cdot 25} \right) = \frac{1}{3} \cdot 7 \cdot 175 = 408.33 \, \text{cm}^3

The volume of the truncated pyramid is 408.33 cm³.