Set up coordinates; the tilt of the square makes $3$-$4$-$5$ right triangles at the corners

The whole square lies inside the rectangle except one triangle sticking out past the top edge; subtract that triangle from $25$

Place the rectangle as $[0,8] \times [0,4].$ The tilted square, using the $3$-$4$-$5$ right triangles, has vertices $(4, 0),$ $(0, 3),$ $(3, 7),$ and $(7, 4).$

The entire square lies inside the rectangle except for the triangle poking above the top edge $y = 4.$ That triangle has vertices $(0.75, 4),$ $(3, 7),$ and $(7, 4),$ with area $$\dfrac12 \cdot (7 - 0.75) \cdot (7 - 4) = \dfrac{75}{8}.$$

The region inside both is $25 - \dfrac{75}{8} = \dfrac{125}{8} = 15\dfrac58.$

Thus, the correct answer is D.