Solution:

Place $B = (0, 0),$ $C = (11, 0),$ $A = (0, 3),$ $D = (11, 3).$ Solving $AF = 7$ and $CF = 9$ (consistent since $AC^2 = 11^2 + 3^2 = 130 = 7^2 + 9^2$) gives $F = \left(\frac{28}{5}, \frac{36}{5}\right),$ and since the diagonals of rectangle $AECF$ bisect each other, $E = A + C - F = \left(\frac{27}{5}, -\frac{21}{5}\right).$

Sides $AE$ and $FC$ have direction $(3, -4),$ lying on the lines $4x + 3y = 9$ and $4x + 3y = 44;$ sides $AF$ and $EC$ lie on $3x - 4y = -12$ and $3x - 4y = 33.$ Every point of $ABCD$ satisfies $-12 \le 3x - 4y \le 33,$ so the common region is just the part of the strip $0 \le y \le 3$ between the lines $4x + 3y = 9$ and $4x + 3y = 44:$ a parallelogram with vertices $A = (0, 3),$ $\left(\frac{9}{4}, 0\right),$ $C = (11, 0),$ and $\left(\frac{35}{4}, 3\right).$

Its horizontal sides have length $11 - \frac{9}{4} = \frac{35}{4}$ and the height between them is $3,$ so the area is $\frac{35}{4} \cdot 3 = \frac{105}{4},$ and $m + n = 105 + 4 = 109.$