Geometry quickie: volume of a tetrahedron

I saw this formula on Wikipedia for the volume of tetrahedron and was intrigued.

Any two opposite edges of a tetrahedron lie on two skew lines, and the distance between the edges is defined as the distance between the two skew lines. Let $d$ be the distance between the skew lines formed by opposite edges... 

then it gave a volume formula. I want to derive it. This post gives my thought process, where I keep using the trick of reducing a general problem to a good example. This is a general strategy for solving problems, often written as "wlog".
(I write it as "wolog" and pronounce as "volog", as if it's a German adverb. So that I can write "We assume wolog that...", "... and thus wolog we have...")

How would a true geometer approach this problem? She would first do a dimension analysis. The volume has dimension $meter^3$, and the lengths have dimension $meter$, so it ought to look like
$$Volume = k \cdot AB \cdot CD \cdot h$$
where $h$ is the distance between the two skew lines of $AB$ and $CD$, and $k$ is some dimensionless number yet to be determined.

The binomial numbers, the Rule 60 celluar automata, and the Sierpinski triangle

Consider the pattern of the parity of the binomial numbers. Color the even numbers white and odd numbers black. You get a Sierpinski Triangle.

We will prove this using spacetime-geometric reasoning. It would help a lot if you have done some geometry of special relativity.

Define the parity function $f(t, n) = C(t, n) \mod 2$, defined on $\mathbb{N}\times\mathbb{Z}$.

We can consider $f(t, n)$ as encoding a 1-d celluar automata, with $t \ge 0, n \in \mathbb{Z}$. Then we have the evolution rule:

$$ f(t+1, n) = f(t, n) + f(t, n-1) \mod 2$$

This is just the Rule 60 celluar automata. 

Then it's easy to prove the following:

The heart and soul of elliptic, flat (Euclidean), and hyperbolic spaces

... is constant, polynomial, and exponential.

Think about building cellphone towers in a Euclidean plane. First, we choose the central station arbitrarily at point $O$, and call it the "origin". To cover all space that's within a distance $R$ to the origin, we need $O(R^2)$ stations.

This is polynomial growth.

Then think about building cellphone towers on a spherical surface, on a little asteroid. Then it's clear that there's only a constant number of towers needed no matter how big $R$ is chosen, since the whole surface has finite area. So we need $O(1)$ stations.

This is constant "growth".

Then, if you know how to play in hyperbolic space (if not, go play Hyperrogue!), you'd know that the number of stations grows as $O(\exp{(CR)})$, where $C=(-K)^{-1/2}$, and $K$ is the Gaussian curvature of the plane. This can be derived by using the formula of the area of a hyperbolic circle of radius $R$:
$$\frac{4\pi}{-K} \sinh^2 \frac{r}{2\sqrt{-K}} $$

We normalize $K = -1$ to get the growth rate $O(\exp{R})$.