A regular octagon $ABCDEFGH$ has sides of length four. Find the area of $\triangle ACG$.

1. $C$ and $G$ are opposite vertices, so $CG$ is a diameter of the circumcircle. Since $A$ lies on this circle, $\angle CAG=90^\circ$.
2. The central angles subtending $AC$ and $AG$ are each $90^\circ$, so $AC=AG=R\sqrt2$. Therefore $[ACG]=\frac12(AC)(AG)=\frac12(R\sqrt2)^2=R^2$.
3. The side length is $4=2R\sin 22.5^\circ$, hence
   $$[ACG]=R^2=\frac{16}{2-\sqrt2}=\boxed{16+8\sqrt2}.$$