What is the formula for calculating volume of tetrahedron in $m$ dimension ?
I am looking for equation dependent on radius of described circle.
I know equations only for $\text{2D}$ and $\text{3D}$ cases.
Tetrahedron:
For $\text{2D}$ it is: $$ V(r) = \frac{3 \sqrt{3}}{4} * r^2 $$
For $\text{3D}$ it is:
$$
V(r) = \frac{8 \sqrt{3}}{27} * r^3
$$
Dodecahedron :
For $\text{2D}$ it is: $$ V(r) = \frac{5}{4} \sqrt{ \frac{1}{2} (5 + \sqrt{5}) } * r^2 $$
For $\text{3D}$ it is:
$$
V(r) = \frac{2}{9} \sqrt{3} (5 + \sqrt{5}) * r^3
$$
I was looking fo the fromula for at least few hours without any effects.
Only what I can find is some info about count of verticles, etc.
First of all, you must know that this field has been studied through and through, about as thoroughly as the geography of a particular patch of Earth in London, UK, between Downing street and Westminster Bridge. Therefore, whenever you look or ask for it online, you'll get a ton of search results and answers, each containing a whole lot of words that you don't need in front of the words that you do need. Skipping the unnecessary part can be tiresome (I can attest to that myself), so have my answer instead.
Second, all terms in math have precise meaning (unless they have another meaning), and if you use them in a different meaning, you are bound to get garbled search results.
With that in mind, now to the point.
Tetrahedron...
...is a figure in 3 dimensions. A plausible analog in all dimensions is called a simplex. The volume is $\dfrac1{n!}\sqrt{(n+1)^{(n+1)}\over n^n}r^n$, which coincides with your findings for $n=2$ and $3$.
Dodecahedron...
...is a figure in 3 dimensions. A plausible analog in all dimensions does not exist. As to how, turn your face to the nearest load-bearing wall and try to walk through it. That's how much it is impossible. As to why, it is because of math.
For the few values of $n$ that leave a possibility of an analog, the results are:
That will be all.