Inscribe hexagon in an equilateral triangle by making intersection of $AB,CD$, $CD,EF$ and $EF, AB$. Then $AB+BC=DE+EF$ and so on. Hence $BC=4,\ DE=2,\ FA=3$ and side length of triangle is $10$. Hence the answer $$\left(10^2-4^2-3^2-2^2\right)\cdot\frac{\sqrt{3}}4=71\cdot\frac{\sqrt{3}}4.$$