A regular tetrahedron may be formed such that each of its edges is a diagonal of a face of a cube; then the tetrahedron has been inscribed in the cube.

It's apparent that a plane passing through the midpoints of three parallel edges of the cube cuts the regular tetrahedron into two congruent pentahedrons and that the intersection figure is a square, the midpoint $M$ of which is the centroid of the tetrahedron.

The angles between the four half-lines from the centroid $M$ of the regular tetrahedron to the vertices are $2\arctan\!{\sqrt{2}}$ ($\approx 109^\circ$ ), which is equal the angle between the four covalent bonds of a carbon atom. A half of this angle, $\alpha$ , can be found from the right triangle in the below figure, where the catheti are $\frac{s}{\sqrt{2}}$ and $\frac{s}{2}$ .

One can consider the regular tetrahedron as a cone. Let its edge be $a$ and its height $h$ . Because of symmetry, a height line intersects the corresponding base triangle in the centroid of this equilateral triangle. Thus we have (see the below diagram) the rectangular triangle with hypotenuse $a$ , one cathetus $h$ and the other cathetus $\frac{2}{3}\!\cdot\!\frac{a\sqrt{3}}{2} = \frac{a}{\sqrt{3}}$ (i.e. $\frac{2}{3}$ of the median $\frac{a\sqrt{3}}{2}$ of the equilateral triangle -- see the common point of triangle medians). The Pythagorean theorem then gives $$h \;=\; \sqrt{a^2-\left(\frac{a}{\sqrt{3}}\right)^2} \;=\; \frac{a\sqrt{6}}{3}.$$

Consequently, the height of the regular tetrahedron is $\displaystyle\frac{a\sqrt{6}}{3}$ .

Since the area of the base triangle is $\frac{a^2\sqrt{3}}{4}$ , the volume (one third of the product of the base and the height) of the regular tetrahedron is $\displaystyle\frac{a^3\sqrt{2}}{12}$ .