304. Two identical chessboards $(8 \times 8$ cells) have a common center, with one of them rotated relative to the other by $45^{\circ}$ about the center. Find the total area of all intersections of the black cells of these two boards, if the area of one cell is 1.
304. Answer: $32(\sqrt{2}-1)$. You need to use the fact that when one of the boards is rotated by $90^{\circ}$, the black cells are replaced by white ones and vice versa, and when both boards are rotated, the black cells are replaced by white ones on both boards. Therefore, the desired area is a quarter of the area of the octagon formed by the intersection of the boards.