Let the n-gon be $A_1A_2...A_n$ with side length 1, then clearly exactly one edge will have exactly 2 vertices of the n+1-gon, and WLOG let that side be $A_1A_2$. Note that all other edges have exactly one vertex. Suppose the vertices $C,D$ are on $A_1A_2$ with $C$ closer to $A_1$ and let $a=DA_2$. Finding the min or max of the area is equivalent to finding the max or min of the set of triangles outside the n+1-gon but inside the n-gon, and the area of those set of triangles is exactly \[(\frac12 \sin\alpha)(a(\frac{n}{n+1}-a)+(a-\frac{1}{n+1})(\frac{n-1}{n+1}-a)+\]\[(a-\frac{2}{n+1})(\frac{n-2}{n+1}-a)+...+(a-\frac{n-1}{n+1})(\frac{1}{n+1}-a))) \]where $\alpha$ is the angle of the n-gon. Now we remove any constant terms or multiples since we are only looking for min and max, so this is equivalent to \[\sum_{i=0}^{n-1}{(a-\frac{i}{n+1})(\frac{n-i}{n+1}-a)} \]\[-na^2+\frac{a}{n+1}\sum_{i=0}^{n-1}{\frac{n-2i}{n+1}} \]\[-na^2+\frac{n}{n+1}a \]\[-a^2+\frac{1}{n+1}a \]and this attains max when $a=\frac{1}{2(n+1)}$ and min when $a=0\text{ or }\frac{1}{n+1}$ since $0\leq a\leq \frac{1}{n+1}$. Thus the minimum area is when $CA_1=DA_2$ (i.e. One of the edge lies on the center of the other edge) and maximum is when one of the vertex coincides with the other polygon's vertex.