jmc

Any equiangular octagon has all its interior angles equal to $135^{\circ }$ and can thus be inscribed in a square or rectangle. We draw the octagon and extend four of its sides to form a rectangle $ABCD$:

Notice that the area of the octagon is equal to the area of $ABCD$ minus the area of the four triangles. All four triangles are isosceles right triangles, so we can find their leg lengths and areas. The triangle with $A$ as a vertex has leg length $4/\sqrt{2}=2\sqrt{2}$ and area $(1/2)(2\sqrt{2}{)}^2=4$. Similarly, the triangles with $B$, $C$, and $D$ as a vertex have leg lengths $\sqrt{2}$, $2\sqrt{2}$, and $\sqrt{2}$ respectively, and areas $1$, $4$, and $1$ respectively.

Now we can compute the sides of rectangle $ABCD$. $AB=2\sqrt{2}+1+\sqrt{2}=1+3\sqrt{2}$ and $CB=\sqrt{2}+2+2\sqrt{2}=2+3\sqrt{2}$. It follows that the area of $ABCD$ is $$(1+3\sqrt{2})(2+3\sqrt{2})=20+9\sqrt{2}.$$Finally, the area of the octagon is $20+9\sqrt{2}-1-4-1-4=\boxed{10+9\sqrt{2}}$.