$37 \cdot 8$ A disk with diameter $D$ is placed on an $8 \times 8$ chessboard with a width of $D$, and their centers coincide. The number of squares on the chessboard that are completely covered by the disk is
(A) 48.
(B) 44
(C) 40.
(D) 36.
(E) 32.
(23rd American High School Mathematics Examination, 1972)
[Solution] All the edge squares of the chessboard are not fully covered by the disk, there are $8+8+$ $6+6=28$ of them. Among the remaining 36 internal squares, only the four corner squares of the $6 \times 6$ chessboard are not fully covered. Because the distance from these corners to the center of the circle is
$$
\sqrt{3^{2}+3^{2}} \cdot \frac{D}{8}=\frac{3 \sqrt{2}}{8} D,
$$

Since $\frac{3 \sqrt{2}}{8}>\frac{1}{2}$, i.e., $\frac{3 \sqrt{2}}{8} D$ is greater than the radius of the circle $\frac{D}{2}$.
Therefore, there are $36-4=32$ squares that are fully covered by the disk.
Hence, the answer is $(E)$.