The convex pentagon $ABCDE$ has $\angle A = \angle B = 120^\circ$, $EA = AB = BC = 2$ and $CD = DE = 4$.  What is the area of $ABCDE$?

[asy]

unitsize(1 cm);

pair A, B, C, D, E;

A = (0,0);

B = (1,0);

C = B + dir(60);

D = C + 2*dir(120);

E = dir(120);

draw(A--B--C--D--E--cycle);

label("$A$", A, SW);

label("$B$", B, SE);

label("$C$", C, dir(0));

label("$D$", D, N);

label("$E$", E, W);

[/asy]
We can divide the pentagon into 7 equilateral triangles of side length 2.

[asy]
unitsize(1 cm);

pair A, B, C, D, E;

A = (0,0);
B = (1,0);
C = B + dir(60);
D = C + 2*dir(120);
E = dir(120);

draw(A--B--C--D--E--cycle);
draw(C--E);
draw((C + D)/2--(D + E)/2);
draw(A--(C + D)/2);
draw(B--(D + E)/2);

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, dir(0));
label("$D$", D, N);
label("$E$", E, W);
[/asy]

The area of each equilateral triangle is \[\frac{\sqrt{3}}{4} \cdot 2^2 = \sqrt{3},\]so the area of pentagon $ABCDE$ is $\boxed{7 \sqrt{3}}$.