Area Between Outer Hull and Inner Hull

Given a set $S$ of $N$ points on the plane, namely S = \{(x_0, y_0), (x_1, y_1), \ldots, (x_{N-1}, y_{N-1})\}, the outer hull $Ho$ of $S$ is simply just the convex hull of $S$. It is a subset of points of $S$ which when joined by line segments will form a convex polygon $P$ of minimum area that includes all points of $S$ inside $P$ or on $P$. Points of $S$ falling on $P$ but are not corner vertices of $P$ are not considered part of the convex hull. The inner hull $Hi$ of $S$ is obtained by getting the convex hull of S - Ho. That is, remove from set $S$ all points that belong to the outer hull $Ho$, to get the set (S - Ho). Then compute the inner hull $Hi$ as the convex hull of (S - Ho).

For example given the set $S$ of $8$ points $A(2.0,5.0)$, B(2.0, 4.0), C(2.0, 2.0), D(1.0, 1.0), E(4.0, 1.0), F(0.0, 0.0), $G(3.0,0.0)$, and H(5.0, 0.0), the outer hull $Ho$ of $S$ can be computed as the set Ho = \{F, H, A, F\}. If we remove $Ho$ from $S$, we get S - Ho = \{B, C, D, E, G\}. The inner hull $Hi$ of $S$ is computed as the convex hull of S - Ho, namely Hi = \{D, G, E, B, D\}. Now the area inside $Ho$ is \text{area}(Ho) = 12.5. The area inside $Hi$ is \text{area}(Hi) = 6.0. Thus the area between the outer hull $Ho$ and the inner hull $Hi$ is 12.5 - 6.0 = 6.5.

Your problem is, given a set $S$ of $N$ points, to write a program that computes the area between the outer hull $Ho$ of $S$, and the inner hull $Hi$ of $S$.