Triangle Base Pyramid Volume: Formula & Calculator

This calculator works out the volume of a pyramid with a triangular base (a tetrahedron) from the base edge lengths and the pyramid’s height. You enter the three side lengths of the triangular base and the perpendicular height from the base to the apex; you get back the volume in cubic units. The volume equals one-third of the base area times the height, so the calculator first finds the base area using Heron’s formula.

Formula

The volume \(V\) of any pyramid is given by:

\(V = \frac{1}{3} \times B \times h\)

where:

• \(B\) = area of the base (for a triangle, \(B = \sqrt{s(s-a)(s-b)(s-c)}\) with \(s = \frac{a+b+c}{2}\), Heron’s formula)
• \(h\) = perpendicular height from the base plane to the apex (not the slant height)
• \(a, b, c\) = lengths of the three sides of the triangular base

All lengths must be in the same unit (e.g., meters, feet). The volume comes out in that unit cubed (m³, ft³).

Worked Example

Using the default values: a = 3 m, b = 4 m, c = 5 m, h = 6 m.

1. Compute the semiperimeter: \(s = (3+4+5)/2 = 6\) m.
2. Apply Heron’s formula: \(B = \sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6\) m².
3. Multiply by height and divide by 3: \(V = \frac{1}{3} \times 6 \times 6 = 12\) m³.

The calculator shows 12.00 m³ for the default inputs.

When to Use This Calculation

You need the volume of a triangular-base pyramid in geometry problems, architecture, or construction when estimating material for a sloped roof, a tent, or a decorative solid. It also appears in packing problems and in 3D modeling. The formula works for any triangle—right, equilateral, or scalene—as long as you know the three side lengths and the true height.

How to Read the Result

The volume is always a positive number in cubic units. If you measure in meters, the volume is in cubic meters; in feet, cubic feet. For a real-world object, the result tells you how much space it occupies. For example, a pyramid with a base area of 6 m² and height 6 m occupies 12 m³, which is about the volume of 12,000 liters of water.

Common Mistakes and Pitfalls

• Using slant height instead of perpendicular height: The height must be measured straight down from the apex to the base plane, not along a slanted edge.
• Mixing units: All three sides and the height must be in the same unit. Convert first (e.g., 1 ft = 0.3048 m) before calculating.
• Forgetting the 1/3 factor: The volume is one-third of base area times height, not base area times height.
• Invalid triangle sides: The sum of any two sides must be greater than the third. If not, Heron’s formula gives a negative under the square root.

Limits of the Formula

The formula assumes a perfectly flat triangular base and a straight height. For real-world objects with curved or irregular bases, it gives an approximation. Also, if you only know the edge lengths from the apex to the base vertices (not the height), you cannot directly use this formula—you would need to compute the height from those lengths, which requires additional geometry.

Frequently Asked Questions

What is a triangular base pyramid called?

It is called a tetrahedron. When all four faces are equilateral triangles, it is a regular tetrahedron.

Can I use this formula for a pyramid with a non-right triangle base?

Yes. Heron’s formula works for any triangle, and the 1/3 × base area × height formula applies to all pyramids regardless of base shape.

What if I only know the base area and height?

Then you don’t need the side lengths. Just multiply the base area by the height and divide by 3. The calculator requires side lengths because it computes the area for you.

How do I convert the volume to liters?

1 cubic meter = 1000 liters. So multiply the volume in m³ by 1000. For cubic feet, 1 ft³ ≈ 28.317 liters.

What is the difference between height and slant height?

Height (or altitude) is the perpendicular distance from the base plane to the apex. Slant height is the distance along a triangular face from the apex to the midpoint of a base edge. They are equal only in special cases, like a regular pyramid with a square base.

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in a | Side a | m | 3 | 0
in b | Side b | m | 4 | 0
in c | Side c | m | 5 | 0
in h | Height | m | 6 | 0
out V | Volume | m³ | (1/3) * (sqrt((a+b+c)/2 * ((a+b+c)/2 - a) * ((a+b+c)/2 - b) * ((a+b+c)/2 - c))) * h | 2
out base_area | Base area | m² | sqrt((a+b+c)/2 * ((a+b+c)/2 - a) * ((a+b+c)/2 - b) * ((a+b+c)/2 - c)) | 2