Solution:

Since $AB \parallel CD,$ the distance $18$ from $A$ to $CD$ is the height. Put $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18),$ and let $u = AB = c - 2a \gt 0.$ The point-to-line distance formulas give $$\frac{18u}{\sqrt{324 + a^2}} = 15 \quad (\text{to } BC), \qquad \frac{18u}{\sqrt{324 + (u + a)^2}} = 10 \quad (\text{to } BD),$$ so $324 + a^2 = \left(\frac{6u}{5}\right)^2$ and $324 + (u + a)^2 = \left(\frac{9u}{5}\right)^2.$

Subtracting, $u(u + 2a) = \frac{81 - 36}{25}u^2 = \frac{9}{5}u^2,$ hence $u + 2a = \frac{9}{5}u$ and $a = \frac{2}{5}u.$ Substituting back, $324 = \frac{36u^2}{25} - \frac{4u^2}{25} = \frac{32u^2}{25},$ so $u^2 = \frac{2025}{8}$ and $u = \frac{45\sqrt{2}}{4}.$

Then $CD = c = u + 2a = \frac{9}{5}u = \frac{81\sqrt{2}}{4},$ and $$K = \frac{AB + CD}{2} \cdot 18 = 9\left(\frac{45\sqrt{2}}{4} + \frac{81\sqrt{2}}{4}\right) = \frac{567\sqrt{2}}{2},$$ so $\sqrt{2} \cdot K = 567.$