Solution:

Opposite sides are parallel, equal in length, and traversed in opposite directions, so $\overrightarrow{AB} = \overrightarrow{ED},$ $\overrightarrow{BC} = \overrightarrow{FE},$ $\overrightarrow{CD} = \overrightarrow{AF}:$ the hexagon is centrally symmetric, and opposite vertices' $y$-coordinates share a common sum, namely $\frac{0 + 2 + \cdots + 10}{3} = 10.$ From $y_A = 0$ and $y_B = 2$ we get $y_D = 10,$ $y_E = 8,$ and convexity puts $y_C = 6,$ $y_F = 4.$ Write $\overrightarrow{AB} = (b, 2),$ $\overrightarrow{BC} = (p, 4),$ $\overrightarrow{CD} = (q, 4).$ Equal side lengths give $s^2 = b^2 + 4 = p^2 + 16 = q^2 + 16,$ so $p = \pm q;$ since $p = q$ would make $B,$ $C,$ $D$ collinear, $p = -q.$

Since $\overrightarrow{AF} = \overrightarrow{CD},$ we have $F = (q, 4),$ and $\angle FAB = 120^\circ$ gives $$\overrightarrow{AB} \cdot \overrightarrow{AF} = bq + 8 = -\frac{s^2}{2} = -\frac{b^2 + 4}{2}.$$ Taking $b \gt 0$ forces $q \lt 0,$ so $q = -\sqrt{b^2 - 12},$ and the equation becomes $b\sqrt{b^2 - 12} = \frac{b^2 + 20}{2}.$ Squaring yields $3b^4 - 88b^2 - 400 = 0,$ so $b^2 = \frac{100}{3},$ giving $b = \frac{10}{\sqrt{3}},$ $q = -\frac{8}{\sqrt{3}},$ $p = \frac{8}{\sqrt{3}}.$

The vertices are $A = (0, 0),$ $B = \left(\frac{10}{\sqrt{3}}, 2\right),$ $C = (6\sqrt{3}, 6),$ $D = \left(\frac{10}{\sqrt{3}}, 10\right),$ $E = (0, 8),$ $F = \left(-\frac{8}{\sqrt{3}}, 4\right).$ The hexagon splits into the parallelogram $ABDE,$ with vertical side $AE = 8$ and horizontal offset $b$ (area $8b$), plus the two congruent triangles $BCD$ and $EFA,$ each with vertical base $8$ and horizontal height $\frac{8}{\sqrt{3}}.$ The total area is $$8 \cdot \frac{10}{\sqrt{3}} + 2 \cdot \frac{1}{2} \cdot 8 \cdot \frac{8}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3},$$ so $m + n = 48 + 3 = 51.$