4. As shown in Figure 3, in the convex hexagon $A B C D E F$, $A D$, $B E$, and $C F$ intersect at point $O$. The area of each triangle formed by every three consecutive vertices is equal to 1. Then $S_{\text {hexagon } A B C D E F}=$ $\qquad$
4.6.

As shown in Figure 5, connect $B D$ and $C E$. Since
$$
S_{\triangle B C D}=S_{\triangle E C D}=1,
$$

therefore, $B E \parallel C D$.
Since $S_{\triangle M F}=S_{\triangle S A F}$
$$
=1 \text {, therefore, } B E \parallel A F \text {. }
$$

Thus, $B E \parallel A F \parallel C D$.
Similarly, $C F \parallel D E \parallel B A, A D \parallel F E \parallel B C$.
Since $A D$, $B E$, and $C F$ intersect at point $O$, it follows that quadrilaterals $O C D E$, $O E F A$, and $O A B C$ are all parallelograms. It is easy to see that the area of each parallelogram is 2.