Volume of a regular tetrahedron

The formula

• V = a³√2 / 12

a is the edge, the same for all six edges of a regular tetrahedron: the pyramid whose four faces are equal equilateral triangles. √2 / 12 ≈ 0.1179, so the volume is a little under an eighth of the cube on the same edge.

A worked example

An edge of 3 cm: V = 27 · 1.4142 / 12 = 3.18 cm³. A tetrahedral die with 2 cm edges holds 8 · 1.4142 / 12 ≈ 0.94 cm³, and a cube die of the same edge eight times as much, which is why the four-sided die feels so small.

Where the √2 comes from

A tetrahedron is a pyramid, so V = Bh / 3. The base is an equilateral triangle of area a²√3 / 4. The apex sits above the base’s centroid, which is a / √3 from each corner, and Pythagoras gives the height: h = √(a² − a² / 3) = a√(2/3). Multiplying, V = (a²√3 / 4)(a√2 / √3) / 3 = a³√2 / 12. The √3 of the base cancels the √3 of the height and the √2 is what remains.

The edge from a volume

Turned round, a = ∛(12V / √2) = ∛(6√2 · V). A tetrahedron holding 100 cm³ has an edge of ∛848 ≈ 9.5 cm.

Common mistakes

• Using the cube’s volume, a³. The tetrahedron is about 12% of that.
• Taking a face’s area for the base and forgetting the third. Bh / 3, as for every pyramid.
• Using the slant height for h. The height drops to the centroid, not to an edge.

Frequently asked questions