A regular octagon $ABCDEFGH$ has sides of length two.  Find the area of $\triangle ADG$.
$\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2$
The area of the triangle $ADG$ can be computed as $\frac{DG \cdot AP}2$. We will now find $DG$ and $AP$.
Clearly, $PFG$ is a right isosceles triangle with hypotenuse of length $2$, hence $PG=\sqrt 2$. 
The same holds for triangle $QED$ and its leg $QD$. The length of $PQ$ is equal to $FE=2$.
Hence $GD = 2 + 2\sqrt 2$, and $AP = PD = 2 + \sqrt 2$.
Then the area of $ADG$ equals $\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}$.