Solution:

Any chord in a unit circle has length at most $2$ (if it is a diameter), so all side lengths must be $1$, $\sqrt{2}$, $\sqrt{3}$, or $2$. We can do casework based on the number of sides:

- Triangles: The possible combinations of side lengths are $(\sqrt{2}, \sqrt{2}, 2)$, $(1,1, \sqrt{3})$, $(1, \sqrt{3}, 2)$, and $(\sqrt{3}, \sqrt{3}, \sqrt{3})$. This is $4$ possibilities.

- Quadrilaterals: The possible ordered combinations of side lengths are $(1,1,1,2)$, $(1, \sqrt{3}, \sqrt{2}, \sqrt{2})$, $(1, \sqrt{3}, 1, \sqrt{3})$, $(1, \sqrt{3}, \sqrt{3}, 1)$, $(\sqrt{2}, \sqrt{2}, \sqrt{2}, \sqrt{2})$, and $(1, \sqrt{2}, \sqrt{3}, \sqrt{2})$. This is $6$ possibilities.

- Pentagons: The possible combinations of side lengths are $(1,1,1,1,1, \sqrt{3})$, $(1,1,1, \sqrt{2}, \sqrt{2})$, and $(1,1, \sqrt{2}, 1, \sqrt{2})$. This is $3$ possibilities.

- Hexagons: The only option is that all sides are $1$, so there is $1$ possibility.

Putting these together, there are $4+6+3+1=14$ polygons total.