Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=30$, and $EF=6+8\sqrt3$. What is the side-length of the square?

1. Following the same steps as the main problem, with $a=AB=30$ and $b=EF=6+8\sqrt3$: $\triangle ABG$ is a $30$-$60$-$90$ triangle (with $G$ the foot of the perpendicular from $A$ to $\overline{BC}$), so $BG=\frac a2=15$, $AG=\frac{\sqrt3a}2=15\sqrt3$.
2. As before, $\triangle AGX\cong\triangle XHY$ and $\triangle ABX\cong\triangle YEZ$, $\triangle YWX\cong\triangle AFZ$, giving (with $x=BX$) the relation $x=\dfrac{\sqrt3\,b-a}{\sqrt3-1}$.
3. Substituting: $\sqrt3\,b=\sqrt3(6+8\sqrt3)=6\sqrt3+24$, so $x=\dfrac{6\sqrt3+24-30}{\sqrt3-1}=\dfrac{6\sqrt3-6}{\sqrt3-1}=6$.
4. Then $XG=x+\frac a2=6+15=21$.
5. $AX^2=AG^2+XG^2=(15\sqrt3)^2+21^2=675+441=1116$, so $AX=\sqrt{1116}=6\sqrt{31}$.
6. The answer is $\boxed{6\sqrt{31}}$.