Cover a polygon with polygons Besides right angled triangles, is there any polygon I could use to cover any given (regular or not) polygon? It's clear that given a triangle, square, hexagon or rectangle you would other options. Can this be proven, if right? That is, that the right angled triangle is the only polygon able to $100\%$ cover any other polygon. 
 A: You'll want to look at Laczkovich, M. "Tilings of Polygons with Similar Triangles." Combinatorica 10, 281-306, 1990.  
46 triangles of type $45^\circ-60^\circ-75^\circ$ can perfectly cover a square.

A: First, let me try to formulate your question properly. Let $C$ denote the set of congruence classes of all planar polygons; let $C'\subset C$ be a subset. For instance, $C'$  can consist of all right-angles triangles. Or, fix an angle $\alpha\in (0,\pi)$ and let $C'=T_{\alpha}$ consist of all triangles where one of the angles is $\alpha$. Another example is the class of all isosceles triangles. Or $C'$ can consist of all triangles similar to the given one (as in the Laczkovich's paper in the answer by Ed Pegg). Or you can take $C'=C$ (this is not very interesting, of course).  
Definition. Fix a subset $C'\subset C$. We say that $C'$ is universal if every planar polygon $P$ can be subdivided into a finite collection of subpolygons which all belong to $C'$. The latter means that 
there is a finite collection of polygons $T_i\in C'$, such that the union of $T_i$'s equals $P$ and the intersection $T_i \cap T_j$ has empty interior whenever i≠j. 
Theorem. The classes $T_{\pi/2}$ (all right  angled triangles) and $T_{2\pi/3}$ are universal. 
Proof. First of all, every planar polygon admits a triangulation, see 
here. Thus, to prove universality of some $C'$, it suffices to show that every triangle can be subdivided into a finite collection of polygons in $C'$. 
Each triangle $ABC$ can be subdivided in two right-angled triangles: One of the altitudes of $ABC$ (say, the one from $A$) connects $A$ to a point $D\in BC$. (This is the case if the angle $\angle BAC$ is the largest angle of $ABC$.) Now, cut $ABC$ along $AD$. 
The proof of universality of triangles of the angle $2\pi/3$ is a bit more complex. First, each triangle can be subdivided in two or three triangles $T$ such that each $T$ has at least one angle in the interval $(\pi/3, \pi/6)$. Then you show that such $T=ABC$ admits an interior point $D$ such that all three angles $\angle ADB, \angle BDC, \angle CDA$ are the same, equal to $2\pi/3$. qed
It is an interesting question which other classes $T_\alpha$ are universal. Or if the class of isosceles triangles is universal. 
Another remark is that the class of all convex quadrilaterals is universal. To prove this, given a triangle $T$, pick its interior point $D$ and interior points $P, Q, R$ of the three edges of $T$. Now, connect $D$ to $P, Q, R$ by segments. 
A: In FEA (Finite Element Analysis), meshing (specially triangle meshing) is required to perform the analysis.
Perhaps some insights of your proof can be found in Wikipedia, Types of Mesh: https://en.wikipedia.org/wiki/Types_of_mesh
