properties of regular tetrahedron

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 (http://planetmath.org/Polyhedron) are $2\arctan \sqrt{2}$ ($\approx {109}^{\circ }$), which is equal the angle between the four covalent bonds of a carbon .  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 ) the rectangular triangle with hypotenuse $a$, one cathetus $h$ and the other cathetus (http://planetmath.org/Cathetus)  $\frac{2}{3}\cdot \frac{a\sqrt{3}}{2}=\frac{a}{\sqrt{3}}$  (i.e. $\frac{2}{3}$ of the median (http://planetmath.org/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 (http://planetmath.org/EquilateralTriangle) 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 ${\displaystyle \frac{a^3\sqrt{2}}{12}}$.