6. In the Cartesian coordinate system $x O y$, it is known that points $A(3,5)$, $B(1,3)$, and $C(4,3)$. Now, $\triangle A B C$ is translated to the position of $\triangle A_{1} B_{1} C_{1}$, and the translation direction forms an angle of $30^{\circ}$ with the $x$-axis. If the area of the common part of $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ is equal to $m$ times the area of $\triangle A B C$ $(0<m<1)$, then the translation distance $d$ is equal to ( ).
(A) $3(\sqrt{3}-1)(1-\sqrt{m})$
(B) $3 \sqrt{3}(1-\sqrt{m})$
(C) $2(\sqrt{3}-1)(1-\sqrt{m})$
(D) $2 \sqrt{3}(1-\sqrt{m})$
6. A.

As shown in Figure 2, let $A_{1} B_{1}$ intersect $B C$ at point $K$, then
$$
\begin{array}{l}
\frac{C K}{C B}=\sqrt{\frac{S_{\text {common }}}{S_{\triangle U C}}}=\sqrt{m}, \\
\frac{B K}{C B}=1-\frac{C K}{C B}=1-\sqrt{m} .
\end{array}
$$

Let the line $l$ passing through points $C$ and $C_{1}$ intersect $A B$ and $A_{1} B_{1}$ at $T$ and $T_{1}$, respectively. Then
$$
\begin{array}{l}
\frac{T T_{1}}{T C}=\frac{B K}{B C}=1-\sqrt{m}, \\
d=T T_{1}=(1-\sqrt{m}) T C .
\end{array}
$$

Notice that the angle between $A B$ and $B C$ is $45^{\circ}$, and the angle between $l$ and $B C$ is $30^{\circ}$. In $\triangle B C T$, by the Law of Sines, we have
$$
T C=B C \cdot \frac{\sin 45^{\circ}}{\sin \left(45^{\circ}+30^{\circ}\right)}=3(\sqrt{3}-1) \text {. }
$$

Therefore, $d=A A_{1}=T T_{1}=3(\sqrt{3}-1)(1-\sqrt{m})$.