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$
Solution
Each side of the octagon subtends an inscribed angle of $\frac{180^\circ}8=22.5^\circ$. So $\angle GDA=45^\circ$ (arc $GHA$) and $\angle DAG=\angle DGA=67.5^\circ$ (arcs $DEFG$ and $ABCD$); hence $DA=DG$.
Draw $EK\bot DG$ and $FL\bot DG$. Since $\angle EDG=45^\circ$, $DK=GL=2\cos45^\circ=\sqrt2$ and $KL=EF=2$, so $DG=2+2\sqrt2$.
