9. [10] Let $A D, B E$, and $C F$ be segments sharing a common midpoint, with $A B<A E$ and $B C<B F$. Suppose that each pair of segments forms a $60^{\circ}$ angle, and that $A D=7, B E=10$, and $C F=18$. Let $K$ denote the sum of the areas of the six triangles $\triangle A B C, \triangle B C D, \triangle C D E, \triangle D E F, \triangle E F A$, and $\triangle F A B$. Compute $K \sqrt{3}$.
Answer: 141
Solution: Let $M$ be the common midpoint, and let $x=7, y=10, z=18$. One can verify that hexagon $A B C D E F$ is convex. We have
$$
[A B C]=[A B M]+[B C M]-[A C M]=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{y}{2}+\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{y}{2} \cdot \frac{z}{2}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{z}{2}=\frac{\sqrt{3}(x y+y z-z x)}{16}
$$

Summing similar expressions for all 6 triangles, we have
$$
K=\frac{\sqrt{3}(2 x y+2 y z+2 z x)}{16} .
$$

Substituting $x, y, z$ gives $K=47 \sqrt{3}$, for an answer of 141 .
Remark: As long as hexagon $A B C D E F$ is convex, $K$ is the area of this hexagon.