Inside the square

Considering overlaying four congruent right triangles with legs 3 and 5 (shown in light blue) to form a $5 \times 5$ square $ABCD$ with side length of 5 and a square hole $MNOP$ in the center. The four dark blue congruent right triangles are the areas where two blue triangles overlap. Therefore, the area of $MNOP$ $=$ the area of square $ABCD$ $-$ the area of 4 blue triangles $+$ the area of 4 dark blue triangles. In formula:

$\begin{aligned} [MNOP] & = [ABCD] - 4[ABG] + 4 \color{#3D99F6} [AFO] & \small \color{#3D99F6} \triangle ABG \text{ and } \triangle AFO \text{ are similar.} \\ & = 5 \times 5 - 4 \times \frac 12 \times 3 \times 5 - 4 \color{#3D99F6} \times \frac 9{34} \times \frac 12 \times 3 \times 5 & \small \color{#3D99F6} \implies [AFO] = \frac {AF^2}{AG^2} [ABG] \\ & = \frac {50}{17} \end{aligned}$

Therefore, $m+n = 50+17 = \boxed{67}$ .

The quadrilateral MNOP is a square with side length 5/√(34). So it's area is 50/17, and the required sum is 67