How to Find the Volume of a Tetrahedron

A tetrahedron is a three-dimensional solid with four triangular faces. Its volume is found much like the volume of any pyramid: use a triangular face as the base and the perpendicular distance from that face to the opposite vertex as the height.

Use a triangular base and perpendicular height

Let $B$ be the area of one triangular face and let $h$ be the perpendicular distance from the opposite vertex to the plane containing that face. Then

$$V = \frac{1}{3}Bh.$$

You can choose any of the four faces as the base. The result is the same, provided you use the matching perpendicular height. The height is not generally the length of an edge joining the base to the opposite vertex: it is the shortest distance from that vertex to the base plane, meeting the plane at a right angle.

To use this method, find the area of the chosen triangle, determine its perpendicular height, multiply those values, and divide by 3. If the base triangle has base length $b$ and altitude $a$, its area is $B=\frac{1}{2}ba$. Here, $a$ is the altitude within the triangular face, while $h$ is the tetrahedron’s height; these are different measurements.

Worked example with a triangular base

Suppose a tetrahedron has a right-triangular base with perpendicular sides of lengths 4 units and 3 units. The opposite vertex is 5 units above the plane of that base.

First find the base area:

$$B = \frac{1}{2}(4)(3)=6\text{ square units}.$$

Now apply the tetrahedron volume formula:

$$V = \frac{1}{3}(6)(5)=10\text{ cubic units}.$$

The volume is 10 cubic units. The 5-unit measurement must be perpendicular to the base plane; a slanted edge would not work unless it happened to be perpendicular.

Find the volume from vertex coordinates

If the four vertices are given by coordinates, use a determinant. Call the vertices $A$, $B$, $C$, and $D$. Form three vectors from the same vertex, such as

$$\mathbf{u}=B-A,\qquad \mathbf{v}=C-A,\qquad \mathbf{w}=D-A.$$

The absolute value of the determinant of these vectors gives the volume of the parallelepiped they span. A tetrahedron occupies one-sixth of that parallelepiped, so

$$V=\frac{1}{6}\left|\det\begin{pmatrix} u_x & v_x & w_x \\ u_y & v_y & w_y \\ u_z & v_z & w_z \end{pmatrix}\right|.$$

The absolute value matters because the determinant can be positive or negative depending on the order of the vertices, but volume cannot be negative.

For example, take

$$A=(0,0,0),\quad B=(4,0,0),\quad C=(0,3,0),\quad D=(1,1,5).$$

Using vectors from $A$ gives $\mathbf{u}=(4,0,0)$, $\mathbf{v}=(0,3,0)$, and $\mathbf{w}=(1,1,5)$. Therefore,

$$V=\frac{1}{6}\left|\det\begin{pmatrix} 4 & 0 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & 5 \end{pmatrix}\right| =\frac{1}{6}|60|=10\text{ cubic units}.$$

This agrees with the base-and-height calculation: the first three points make a right-triangular base of area 6 in the plane $z=0$, and the fourth point is 5 units above that plane.

Special case: a regular tetrahedron

A regular tetrahedron has six equal edges. If each edge has length $a$, its volume is

$$V=\frac{\sqrt{2}}{12}a^3.$$

To see where this comes from, use an equilateral triangular face as the base. Its area is $\frac{\sqrt{3}}{4}a^2$. The opposite vertex lies above the center of the face; using a right triangle gives the tetrahedron height $h=a\sqrt{\frac{2}{3}}$. Substituting into $V=\frac{1}{3}Bh$ gives the stated formula. For an edge length of 6 units, for instance, the volume is $\frac{\sqrt{2}}{12}(6^3)=18\sqrt{2}$ cubic units.

Common mistakes

• Using a slanted edge as the height: The height must be perpendicular to the base plane. If the opposite vertex is not directly above the base, use its perpendicular distance to the plane.
• Forgetting the factor of one-third: A tetrahedron is a pyramid, not a triangular prism. Its volume is one-third of the base area times the height.
• Confusing the two triangle heights: The altitude used to calculate the base triangle’s area lies in that triangle. The tetrahedron height is perpendicular to the whole base plane.
• Leaving out the determinant’s absolute value: The sign depends on vertex order; take the absolute value before dividing by 6.
• Using the wrong units: If lengths are in centimeters, the volume is in cubic centimeters. Area units multiplied by length units produce cubic units.