7. $(3 \sqrt{3}-3)(1-\sqrt{m})$ Detailed Explanation: Let $A_{1} B_{1}$ intersect $B C$ at $K$, then $\frac{C K}{C B}=\sqrt{\frac{S_{\text {out }}}{S_{\triangle A B C}}}=\sqrt{m}, \frac{B K}{C B}=1-\frac{C K}{C B}=1-\sqrt{m}$. Suppose the line $l$ passing through points $C$ and $C_{1}$ intersects $A B$ and $A_{1} B_{1}$ at $T$ and $T_{1}$, respectively. Then $\frac{T T_{1}}{T C}=\frac{B K}{B C}=1-\sqrt{m}, d=T T_{1}=(1-\sqrt{m}) T C$. Noting 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 get $T C=B C \cdot \frac{\sin 45^{\circ}}{\sin \left(45^{\circ}+30^{\circ}\right)}=3(\sqrt{3}-1)$, so $d=(3 \sqrt{3}-3)(1-\sqrt{m})$.
7. In the Cartesian coordinate system $x O y$, there are three points $A(3,5), B(1,3), C(4,3)$. Now, $\triangle A B C$ is translated to the position of $\triangle A_{1} B_{1} C_{1}$, and the translation direction forms a $30^{\circ}$ angle 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 $d$ equals $\qquad$.