I have got the following task here:
Prove, that you can't cover the "Plane" with convex polygons, which have more than $\,6\,$ vertices!
The answer is pretty obvious for $\,n=3\,$ vertices, because $ 6\cdot 60^\circ = 360^\circ$.
For $\,n=4\,$ it works too, because $4\cdot 90^\circ = 360^\circ$.
I think that $\,n=6\,$ is good too, but how do I prove, that other than that, it isn't possible to do that?
On
It is possible to cover the plane with convex polygons with more than $6$ vetices, if you do not put a bound on their size. Take a tiling of the hyperbolic plane by heptagons in the disc model and apply a map $(r,\theta)\mapsto(\frac{r}{1-r},\theta)$ to yield a tiling of the whole plane (if we straighten out the edges after applying the transformation).
If you do bound the area and perimeter of such heptagons, it can't be done - see this /r/mathriddles thread for a proof.
I'm going to assume you mean tiling the plane with copies of the same regular $n$-gon; so we can't mix squares and hexagons, for example.
Why are $60^\circ, 90^\circ,$ and $120^\circ$ (for a hexagon) important, feature of the polygon do they measure?
What the corresponding angles be, if we consider a regular $n$-gon, where $n > 6$? Would angles like that make sense, in a tiling situation?