4. In a convex pentagon $A B C D E$, $\angle A=\angle E$ $=120^{\circ}, E A=A B=B C$ $=2, C D=D E=4$. Then its area is ( ).
(A) $6 \sqrt{3}$
(B) $7 \sqrt{3}$
(C) $8 \sqrt{3}$
(D) $9 \sqrt{3}$
4. (B).

Extend $E A$ and $C B$ to intersect at point $F$.
$$
\because \angle E A B=\angle C B A=
$$
$120^{\circ}$,
$$
\therefore \angle F A B=\angle F B A=60^{\circ} \text {, }
$$

Therefore, $\triangle A B F$ is an equilateral triangle.
$$
\begin{array}{c}
A F=F B=A B=2 . \\
\therefore C D=D E=E F=F C=4 .
\end{array}
$$

Thus, quadrilateral $C D E F$ is a rhombus.
$$
\begin{aligned}
\therefore S_{A B C D E} & =S_{C G F}-S_{\triangle A B F} \\
& =E F \cdot F C \sin 60^{\circ}-\frac{1}{2} F A \cdot F B \sin 60^{\circ} \\
& =8 \sqrt{3}-\sqrt{3}=7 \sqrt{3} .
\end{aligned}
$$