78.35. Let us position the polygon so that one of its vertices $A_{1}$ is at the origin, and the side $A_{1} A_{2}$ lies along the x-axis (see Fig. 52). Since all the lengths of the diagonals and sides are rational, the cosines of the angles $A_{k} A_{1} A_{l}$ are also rational. Therefore, since $\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$, all the sines of the angles $A_{k} A_{1} A_{2}$ are rational multiples of some fixed number $t$. Consequently, we can assume that all points $A_{1}, A_{2}, \ldots, A_{n}$ lie at the nodes of a rectangular grid with a horizontal step of 1 and a vertical step of $h$, as we can transition from rational coordinates to integer coordinates using homothety. Next, consider all polygons obtained from $M$ by parallel translation $\overrightarrow{A_{1} P}$, where $P$ is any node of the grid, as well as all polygons obtained from the polygon $M^{\prime}$, which is centrally symmetric to $M$ with respect to the origin, by the same translations. Then the set of such polygons intersecting the square $K$ will provide the required covering, as it includes, along with any polygon $N$ with a side $X Y$ intersecting $K$, the polygon $N$ centrally symmetric to $N$ with respect to the midpoint of the segment $X Y$. Therefore, the multiplicity of the covering does not change when crossing the side $X Y$, which is what we needed to prove. (The solution was provided by $\Phi . L$ L. Nazarov.)