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properties of regular tetrahedron

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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 \PMlinkname{vertices}{Polyhedron} are $2\arctan\!{\sqrt{2}}$ ($\approx 109^\circ$), which is equal the angle between the four covalent bonds of a carbon \PMlinkescapetext{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 


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 \PMlinkescapetext{diagram}) the rectangular triangle with hypotenuse $a$, one cathetus $h$ and the other \PMlinkname{cathetus}{Cathetus} \,$\frac{2}{3}\!\cdot\!\frac{a\sqrt{3}}{2} = \frac{a}{\sqrt{3}}$\, (i.e. $\frac{2}{3}$ of the \PMlinkname{median}{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 \PMlinkname{base triangle}{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 $\displaystyle\frac{a^3\sqrt{2}}{12}$.\\