An equiangular octagon has four sides of length $2$ and four sides of length $3$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?
All angles are $135^\circ$. Extending the four sides of length $a$ gives a square; each side of length $b$ cuts off an isosceles right triangle with legs $\frac b{\sqrt2}$ at a corner. The square has side $a+2\cdot\frac b{\sqrt2}=a+\sqrt2\,b$, so the octagon has area $(a+\sqrt2\,b)^2-4\cdot\frac12\cdot\frac{b^2}2=a^2+2\sqrt2\,ab+b^2$.
With $a=2,b=3$: area $=4+12\sqrt2+9=\boxed{13+12\sqrt2}$.
