Answer: 23

Solution:

Since the interior angles are all $135^\circ$ and the $\sqrt{2}$ sides are diagonals of unit squares, the octagon fits on a lattice: $C = (0, 0),$ $A = (1, 1),$ $R = (2, 1),$ $O = (3, 0),$ $L = (3, -1),$ $I = (2, -2),$ $N = (1, -2),$ $E = (0, -1).$ The path $CORNELIA$ is carried to itself by the $180^\circ$ rotation about $(\tfrac{3}{2}, -\tfrac{1}{2}).$ Let $Y$ and $Z$ be the points where $AI$ and $RN$ cross $CO,$ and let $W = AI \cap RN.$ Segment $AI$ has slope $-3,$ so $Y = (\tfrac{4}{3}, 0),$ and by symmetry $Z = (\tfrac{5}{3}, 0)$ and $W = (\tfrac{3}{2}, -\tfrac{1}{2}).$

The six enclosed regions are the four congruent corner triangles like $CAY$ and the two small congruent triangles like $YZW.$ Triangle $CAY$ has base $CY = \tfrac{4}{3}$ and height $1,$ so its area is $\tfrac{2}{3}.$ Triangle $YZW$ has base $YZ = \tfrac{1}{3}$ and height $\tfrac{1}{2},$ so its area is $\tfrac{1}{12}.$ Therefore $$\begin{aligned} K &= 4 \cdot \frac{2}{3} + 2 \cdot \frac{1}{12} \\ &= \frac{8}{3} + \frac{1}{6} = \frac{17}{6}, \end{aligned}$$ and $a + b = 17 + 6 = 23.$