jmc

We will calculate the area of the hexagon in two different ways. Let the interior point in the figure be called $P$, and let $s$ be the side length of the hexagon. The areas of the triangles $APB$, $BPC$, $CPD$, $DPE$, $EPF$, and $FPA$ are $\frac{1}{2}(s)(4)$, $\frac{1}{2}(s)(6)$, $\frac{1}{2}(s)(9)$, $\frac{1}{2}(s)(10)$, $\frac{1}{2}(s)(8)$, and $\frac{1}{2}(s)(5)$, respectively. Also, the area of a regular hexagon with side length $s$ is $3s^2\sqrt{3}/2$. Setting the sum of the triangles' areas equal to $3s^2\sqrt{3}/2$ gives $$\begin{array}{rl}\frac{1}{2}s(4+6+9+10+8+5)& =\frac{3s^2\sqrt{3}}{2}⟹\\ 21s& =\frac{3s^2\sqrt{3}}{2}⟹\\ 14s& =s^2\sqrt{3}⟹\\ s=0\text{or}s& =\frac{14}{\sqrt{3}}\\ & =\frac{14}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\boxed{\frac{14\sqrt{3}}{3}}\text{ cm}.\end{array}$$