Divide the equilateral hexagon $$A B C D E F$$ into isosceles triangles $$A B F, C B D$$, and $$E D F$$ and triangle $$B D F$$. 
The three isosceles triangles are congruent by SAS congruence. Вy CPCTC, $$B F=B D=D F$$,
so triangle $$B D F$$ is equilateral.
Let the side length of the hexagon be $$s$$. 
The area of each isosceles triangle is
$$\frac{1}{2} a b \sin \angle C=\frac{1}{2} \times s \times s \times \sin 30^{\circ}$$
$$\quad \quad \quad =\frac{1}{4} s^{2} $$
By the Law of Cosines on triangle $$A B F$$,
$$B F^{2}=s^{2}+s^{2}-2 s^{2} \cos 30^{\circ}$$
$$\quad=2 s^{2}-\sqrt{3} s^{2} $$
Hence, the area of the equilateral triangle $$B D F$$ is
$$\frac{\sqrt{3}}{4} B F^{2}=\frac{\sqrt{3}}{4}(2 s^{2}-\sqrt{3} s^{2})=\frac{\sqrt{3}}{2} s^{2}-\frac{3}{4} s^{2}$$
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or
$$3(\frac{1}{4} s^{2})+(\frac{\sqrt{3}}{2} s^{2}-\frac{3}{4} s^{2})=\frac{\sqrt{3}}{2} s^{2}=6 \sqrt{3} \text {. }$$
Hence, $$s=2 \sqrt{3}$$, and 
The perimeter of the hexagon is $$6 s= 12 \sqrt{3}$$.

Divide the equilateral hexagon $A B C D E F$ into isosceles triangles $A B F, C B D$ , and $E D F$ and triangle $B D F$ .
The three isosceles triangles are congruent by SAS congruence. Вy CPCTC, $B F=B D=D F$ ,
so triangle $B D F$ is equilateral.
Let the side length of the hexagon be $s$ .
The area of each isosceles triangle is
$\frac{1}{2} a b \sin \angle C=\frac{1}{2} \times s \times s \times \sin 30^{\circ}$
$\quad \quad \quad =\frac{1}{4} s^{2}$
By the Law of Cosines on triangle $A B F$ ,
$B F^{2}=s^{2}+s^{2}-2 s^{2} \cos 30^{\circ}$
$\quad=2 s^{2}-\sqrt{3} s^{2}$
Hence, the area of the equilateral triangle $B D F$ is
$\frac{\sqrt{3}}{4} B F^{2}=\frac{\sqrt{3}}{4}(2 s^{2}-\sqrt{3} s^{2})=\frac{\sqrt{3}}{2} s^{2}-\frac{3}{4} s^{2}$
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or
$3(\frac{1}{4} s^{2})+(\frac{\sqrt{3}}{2} s^{2}-\frac{3}{4} s^{2})=\frac{\sqrt{3}}{2} s^{2}=6 \sqrt{3} \text {. }$
Hence, $s=2 \sqrt{3}$ , and
The perimeter of the hexagon is $6 s= 12 \sqrt{3}$ .

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