imc

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 $\frac{DG\cdot AP}{2}=\frac{(2+2\sqrt{2})(2+\sqrt{2})}{2}=\frac{8+6\sqrt{2}}{2}=\boxed{4+3\sqrt{2}}$.