Rhombuses

The central point of this partially constructed figure is at the far left. For any beginning angle $\theta$ , you can see the next set of rhombuses will have angle $2\theta$ , then $3\theta$ and so on. There will be $\lfloor\frac{n}{2}\rfloor$ sets of rhombuses (angles up to $180$ degrees.)

The area of each rhombus with angle $\alpha$ is simply $\sin(\alpha)$ so we just need to add up all the rhombuses. The general formula for any $n$ is

$n*\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor} \sin{(i\frac{360}{n})}$

Using $n=360$ gives the solution: $\boxed{41251.91405}$

Which, not coincidentally is quite close to $\frac{360^{2}}{\pi}$

For even $n$ , the final polygon is a regular $n$ -polygon of side $2$ . The area is, for side $a=2$

$A = \frac{1}{4}na^2Cot(\frac{\pi}{n}) = 360Cot(\frac{\pi}{360}) \approx 41251.914$

Jeremy's solution shows how and why the added rhombuses reach that final regular $n$ -polygon.

Here's the $179 \cdot 360 = 64440$ rhombuses put together. The Moire patterns is an artifact of having such a dense mat of lines in a rasterized display. But you can barely see the rhombuses "open up" in the middle zone between center and perimeter.