What are the conditions for a polygon to be tessellated?

Upon one of my mathematical journey's (clicking through wikipedia), I encountered one of the most beautiful geometrical concept that I have ever encountered in my 16 and a half years on this oblate spheroid that we call a planet.

I'm most interested in the tessellation of regular polygons and their 3D counterparts. I've noticed that simple shapes like squares or cubes can be tessellated but not circles or spheres.

Somewhere after hexagons, shapes lose that ability to be tessellated by only themselves. Although it is intuitively clear to me when shape can be tessellated, I cant put it into words. Please describe to me, in a fair amount of detail, what the lesser sided shapes had that the greater sided shapes did not inorder to be tessellated.

A regular polygon can only tessellate the plane when its interior angle (in degrees) divides $360$ (this is because an integral number of them must meet at a vertex). This condition is met for equilateral triangles, squares, and regular hexagons.

You can create irregular polygons that tessellate the plane easily, by cutting out and adding symmetrically.

First, let's see the case that we use only one polygon and its copies to tessellate the plane.

(1) We can easily prove that there are only three regular polygons $(n=3,4,6)$ which tessellate the plane with one polygon.

(2) You'll see that any parallelogram can tessellate the plane. (I found some figures here though the language is Japanese.)

(3) The fact (2) means that any triangle can tessellate the plane because you can make a parallelogram using two copies of a triangle.

(4) A hexagon with three pairs of parallel edges can tessellate the plane.

(5) The fact (4) means that any quadrilateral can tessellate the plane because you can make a hexagon with three pairs of parallel edges using two copies of a quadrilatelral which is not a parallelogram.

(6) Some pentagons with a special condition can tessellate the plane. For example, you can divide a hexagon of (4) into two congruent pentagons.

Second, let's see the case we can use more than two distinct polygons and its copies to tessellate the plane.

You can find helpful comments in other's answer. Also, you'll find some figures in the same page as above. For example, (3,3,3,3,6) means there exist four equilateral triangles and one hexagon at every vertex.

I hope you like this answer.

Edit 1 : This is a question which I asked at mathoverflow. You may be interested in the question.

Edit 2 : Let us consider $3D$ version. Fedorov found that there are exactly five 3-dimensional parallelohedra. You can see beautiful figures here. You'll be interested in these figures.