6. In a convex quadrilateral $ABCD$, $\angle BAD + \angle ADC = 240^{\circ}$, $E$ and $F$ are the midpoints of sides $AD$ and $BC$, respectively, and $EF = \sqrt{7}$. If two squares $A_1$ and $A_2$ are drawn with sides $AB$ and $CD$, respectively, and a rectangle $A_3$ is drawn with length $AB$ and width $CD$, find the sum of the areas of the three figures $A_1$, $A_2$, and $A_3$.
6.28 .

As shown in Figure 6, extend $BA$ and $CD$ to intersect at point $P$.
Given $\angle BAD + \angle ADC = 240^{\circ}$, we have
$$
\angle BPC = 60^{\circ}.
$$

Connect $BD$, and take the midpoint $G$ of $BD$, then connect $EG$ and $FG$.
By the Midline Theorem of a triangle, we know
$$
EG \perp \frac{1}{2} AB, FG \perp \frac{1}{2} CD, \angle EGF = 120^{\circ}.
$$

In $\triangle EGF$, by the Law of Cosines, we get
$$
\begin{array}{l}
7 = EF^2 = EG^2 + FG^2 + EG \cdot FG \\
= \left(\frac{AB}{2}\right)^2 + \left(\frac{CD}{2}\right)^2 + \left(\frac{AB}{2}\right)\left(\frac{CD}{2}\right),
\end{array}
$$

which simplifies to $AB^2 + CD^2 + AB \times CD = 28$.
Therefore, the sum of the areas of the three figures $A_1, A_2$, and $A_3$ is 28.