We could find triangles area
$\triangle (ABC) = \frac{8^2\sqrt{3}}{4}= 16\sqrt{3}$
$\triangle (DEF) = \frac{3^2\sqrt{3}}{4}= \frac{16\sqrt{3}}{4}$

Now 3 triangles no equilaterals area is $\triangle (ABC)-\triangle (DEF) = \frac{55\sqrt{3}}{12}$

By Heron's Theorem:
$\frac{55\sqrt{3}}{4} =\sqrt{\frac{11}{2}+\frac{5}{2}+\frac{2AE-5}{2}+\frac{2A}{11}}$
$\frac{55}{3}=4AE^2+12AE-55$
$0=4AE^2+12AE-55-\frac{55}{3}$
$0=4AE^2+12AE-\frac{220}{3}$

By quadratic formula
$AE= \frac{\sqrt{741}-9}{6}$
$AE=FC=BD \therefore AE+FC+BD=\frac{\sqrt{741}-9}{2}= \approx 9.11$