9.64 Two identical chessboards $(8 \times 8$ squares $)$ overlap each other. Now, one of them is rotated $45^{\circ}$ around the center. If the area of each square is 1, find the total area of the intersection of all the black squares of the two chessboards.
[Solution] The intersecting figure of the two chessboards is a regular octagon, with an area of $128(\sqrt{2}-1)$. This regular octagon can be divided into 4 parts: (white, white), (white, black), (black, white), (black, black), denoted as $S_{1}, S_{2}, S_{3}, S_{4}$ respectively. Since rotating one chessboard $90^{\circ}$ around its center swaps the two colors of the squares, when both chessboards are rotated $90^{\circ}$ around their centers, the order of the 4 parts is exactly reversed, hence we get $S_{1}=S_{4}, S_{2}=S_{3}$. Furthermore, when the two chessboards are rotated in opposite directions by $45^{\circ}$ and their positions are swapped, it is equivalent to one chessboard being stationary while the other rotates $90^{\circ}$, which can prove that $S_{1}=S_{3}$, thus the areas of the 4 parts are all equal, hence we get $S_{4}=$ $32(\sqrt{2}-1)$.