[tex]\bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad
\begin{cases}
s=\textit{length of a side}\\
----------\\
perimeter=s+s+s\\
perimiter=3s\\
36=3s\\
\frac{36}{3}=s\\
12=s
\end{cases}
\\\\\\
A=\cfrac{12^2\sqrt{3}}{4}\implies A=36\sqrt{3}[/tex]
ok.. .based on a side of 12, that's the area of the equilateral triangle, now, the hexagon has the same area... so... let's use the area of a polygon to see what's the length of a side
[tex]\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}ns^2\ cot\left( \frac{180}{n} \right)\qquad
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of one side}\\
----------\\
n=6\\
A=36\sqrt{3}
\end{cases}
\\\\\\
36\sqrt{3}=\cfrac{1}{4}\cdot 6\cdot s^2\cdot cot\left( \frac{180}{6} \right)\implies 36\sqrt{3}=\cfrac{1}{4}\cdot 6\cdot s^2\cdot \sqrt{3}
\\\\\\
36\sqrt{3}=\cfrac{6s^2\sqrt{3}}{4}\implies \cfrac{4\cdot 36\sqrt{3}}{6\sqrt{3}}=s^2
\\\\\\
24=s^2\implies \sqrt{24}=s\implies 2\sqrt{6}=s[/tex]
now, in case you want to check how much is the cot(30°), check your Unit Circle, recall, cotangent is cosine/sine
