Answer: 35

Solution:

A segment cuts the rectangle into a triangle and a trapezoid only if it runs from a vertex to a point on a nonadjacent side, so $U_2$ is a right triangle whose legs lie along two sides of the rectangle, one leg being a full side ($6$ or $7$). Since $U_1 \sim U_2,$ the cut in the $3$-$4$-$5$ right triangle $ABC$ must also produce a right triangle, so it is parallel to a leg, and then $U_1 \sim ABC.$ Hence $U_2$ is a $3$-$4$-$5$ triangle too: its legs are $6$ and $\frac{9}{2}$ (full side $6$) or $7$ and $\frac{21}{4}$ (full side $7$); the other orientations need legs $8$ or $\frac{28}{3},$ which do not fit.

In both cases the trapezoid $V_2$ has two right angles and an acute angle between the cut and its longer base with tangent $\frac{6}{9/2} = \frac{7}{21/4} = \frac{4}{3}.$ In triangle $ABC,$ a cut parallel to the leg of length $3$ gives $V_1$ an acute angle with tangent $\frac{4}{3},$ matching, while a cut parallel to the leg of length $4$ gives tangent $\frac{3}{4},$ which cannot match. So the cut is parallel to the side of length $3,$ and the parallel bases of $V_1$ are the cut segment $s$ and the side of length $3.$

Similarity of the trapezoids forces $s : 3$ to equal the ratio of the bases of $V_2,$ which is $\frac{7 - 9/2}{7} = \frac{5}{14}$ in the first case and $\frac{6 - 21/4}{6} = \frac{1}{8}$ in the second. Then $[U_1] = \left(\frac{s}{3}\right)^2 [ABC],$ giving $\left(\frac{5}{14}\right)^2 \cdot 6 = \frac{75}{98}$ or $\left(\frac{1}{8}\right)^2 \cdot 6 = \frac{3}{32}.$ The minimum is $\frac{3}{32},$ so $m + n = 3 + 32 = 35.$