[Math] The Area of an Irregular Hexagon

Question

So I have a hexagon with $3$ side lengths of length $2$, and $3$ side lengths of length $1$. All side length of length $1$ are next to each other, and all sides of length $2$ are as well. The two angles between the sides of length $2$ are the same, as are the two angles shared by sides of length $1$. The two angles shared by sides $1$ and $2$ at opposite ends of the hexagon are the same as well. I would like to find out how to calculate out the area. I've tried splitting it into triangles, but that hasn't worked, and of course it's not regular, so I can't use the normal formula for a hexagon. Any help would be appreciated.

Answer 1

Edit: I mis-read the lengths. The same analysis should still work.

I don't think that your question has a well defined answer.

Suppose that the lengths are arranged 2,1,2,1,2,1 around the hexagon. One can inscribe such a hexagon in an equilateral triangle of side length four and then remove the corners of the equilateral triangle to get that: $$ A = \frac{\sqrt{3}}{4} \left(4^2 - 3 \cdot 1^2 \right) \approx 5.62 $$

One can also arrange the lengths 1,1,1,2,2,2 around the hexagon. You can arrange so that there is a square of side length one with a trapezoid of sidelengths 2,2,2,1 and altitude $\sqrt{3}/2$ on top of it. This figure has area: $$ A = 1^2 + 2 \left( \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right) + 1 \cdot \frac{\sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{2} \approx 2.29 $$