In reply to Ian Wakeling.

Because the area of the inner polygon is always less than pi and the area of the outer polygon is always greater than pi, the average (A_in + A_out)/2 will be a better approximation than either. From the plots, you can see that A_out converges faster than A_in, so a weighted sum that puts more weight on A_out is preferable.

Here's a challenge: Given a parameter p in (0,1), consider the approximation A = p*A_in + (1-p)*A_out. By the previous reasoning, A is always a better approximation to pi than A_in or A_out. For what value of p is the approximation the best? The answer presumably depends on n, the number of sides. What is p as a function of n? Enjoy!