6. As shown in Figure 2, points $R, T$ are on the side $CD$ of $\square ABCD$, such that $DR = RT = TC$. Lines $AR, AT$ intersect the extension of $BC$ at points $M, L$, respectively, and lines $BT, BR$ intersect the extension of $AD$ at points $S, P$, respectively. Given that the area of $\square ABCD$ is 48, then the area of the shaded part is $\qquad$
6. 66.

Let the intersection points of $AL$ and $BP$, and $BS$ and $AM$ be $O$ and $E$, respectively.
Given $RT \in \mathbb{E} \frac{1}{3} AB$, we know that $T$ and $R$ are the trisection points of $EB$ and $EA$, respectively, and also the trisection points of $LA$ and $PB$, respectively.

Thus, $P$ and $L$ are the midpoints of $AS$ and $BM$, and $E$ is the midpoint of $BS$ and $AM$.
Let the area of $\triangle OAB$ be $x$. Then,
$S_{\triangle EAO} = S_{\triangle EBO} = \frac{x}{2}, S_{\triangle BOT} = \frac{x}{3}$,
$S_{\text{quadrilateral } ROTE} = \frac{x}{2} + \frac{x}{2} - \frac{x}{3} \times 2 = \frac{x}{3}$.
Also, $S_{\triangle ABT} = \frac{4x}{3}$, and since $T$ is the centroid of $\triangle ABM$, we have
$S_{\triangle TBM} = \frac{4x}{3}, S_{\triangle TML} = \frac{2x}{3}, S_{\text{quadrilateral } ETL M} = \frac{4x}{3}$.
Noting that, $S_{\triangle ABCD} = 2 S_{\triangle ABT} = \frac{8x}{3}$.

Thus, $x = 48 \times \frac{3}{8} = 18$.
Therefore, the area of the shaded part is
$$
\frac{4x}{3} \times 2 + \frac{x}{3} + \frac{x}{3} \times 2 = \frac{11x}{3} = 66.
$$