Given the edges of a tetrahedron, its volume can be calculated, as any of its heights can be determined. (The thought process leading to this can start with rotating the other three faces of the tetrahedron into the plane of one of its faces). Let's perform the calculation in the following example: the edge lengths of the tetrahedron $A B C D$ are as follows: $A B=6, A C=7, A D=8, C D=9$, $B D=10, B C=11$ units. (We can choose the base as, for example, the triangle $A B C$.)
Let's create a paper model of the 3 faces meeting at vertex $A$ of the tetrahedron. From the sides of these faces, we will construct the triangle $ABC$ on the paper, then the triangle $ABD_1$ congruent to $ABD$, and the triangle $ACD_2$ congruent to $ACD$, such that the latter two are connected to the first one along the common edges $AB$ and $AC$, and all three triangles are the mirror images of the corresponding faces of the given tetrahedron (with opposite orientation, as we would see the faces from the inside, see Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_4493c756a7f64913d899g-1.jpg?height=628&width=610&top_left_y=342&top_left_x=736)

Let $D'$ be the intersection of the perpendiculars dropped from $D_1$ to $AB$ and from $D_2$ to $AC$.

After cutting out the combined 3 faces, fold the $ABD_1$ face around $AB$ as the axis above the base plane $ABC$, so $D_1$ passes above the line $D_1D'$, as it traces an arc in the plane perpendicular to $AB$, and the projection of this entire plane onto the base plane is the drawn line $D_1D'$. Similarly, folding $ACD_2$ around $AC$, $D_2$ passes above the line $D_2D'$, and $D_1$ and $D_2$ meet at the point $D$ above $D'$, since $AD_1 = AD_2 = AD$.

Thus, in the original tetrahedron, $D'$ is the projection of vertex $D$ onto the base plane $ABC$, and $DD' = m$ is the height of the body perpendicular to this base.

Accordingly, if we determine the distance of $D'$ from one of the vertices $A, B, C$, then $m$ can be calculated from the triangles $DD'A$, $DD'B$, or $DD'C$. For example, from the first one,

$$
m^2 = DA^2 - D'A^2
$$

we choose this one, as in calculating $D'A$, we can advantageously use the observation that $\angle DAB = 90^\circ$, since $DB^2 = DA^2 + AB^2$.

Thus, $D_1D'$ passes through $A$, and if $D_2'$ is the intersection of $D_2D'$ and $AC$, and $C'$ is the projection of $C$ onto $AB$, then the triangles $D'A D_2'$ and $AC'C$ are similar right triangles, from which

$$
D'A = AC \frac{D_2'A}{C'C}
$$

For this, from the right triangles $D_2D_2'A$ and $D_2D_2'C$,

$$
\begin{gathered}
D_2'D_2^2 = D_2A^2 - D_2'A^2 = D_2C^2 - D_2'C^2 = D_2C^2 - (AC - D_2'A)^2 \\
D_2'A = \frac{AC^2 + AD^2 - CD^2}{2AC} = \frac{16}{7}
\end{gathered}
$$

With the same reasoning, from the right triangles $CC'B$ and $CC'A$,

$$
\begin{gathered}
C'C^2 = BC^2 - C'B^2 = AC^2 - (BC' - AB)^2 \\
BC' = \frac{BC^2 + AB^2 - AC^2}{2AB} = 9, \quad \text{so} \quad CC' = \sqrt{40}
\end{gathered}
$$

Therefore, from (2) and (1),

$$
D'A = \frac{8}{\sqrt{10}}, \quad m^2 = 64 - \frac{64}{10} = \frac{9 \cdot 64}{10}, \quad m = \frac{24}{\sqrt{10}}
$$

With $CC'$ known, the area of the base triangle $ABC$ can also be calculated, and thus the volume of the tetrahedron:

$$
V = \frac{m}{3} \cdot \frac{AB \cdot C'C}{2} = 48 \text{ volume units, }
$$

(i.e., 48 units of a cube with unit edge length).

Remarks. 1. The right angle of the $ABD$ face can also be used by folding the faces $ABC$ and $ADC$ into the plane of this face (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_4493c756a7f64913d899g-2.jpg?height=789&width=681&top_left_y=126&top_left_x=711)

Figure 2

Based on the above, $AC_1' = 3$ units, similarly $AC_2' = 2$, $C'A^2 = C_1'A^2 + C_2'A^2 = 13$ and $m_{ABD} = 6$ units, the base area is 24 square units.

2. The edge system of the tetrahedron under consideration is a good example of a so-called rational tetrahedron. A tetrahedron is called rational if all 6 edges and the volume are rational (in our case, integers). The sequential nature of the measurements reminds us of the 3, 4, 5 unit right triangle (the so-called Egyptian triangle). However, we note that the spatial arrangement of the edges is also important; by placing the opposite pairs of edges in a different way, the volume can be different. For example, by swapping the 6 and 8 unit edges, instead of the 6 unit height obtained in Remark 1, 6.97 is obtained, while the base area remains unchanged.