$37 \cdot 7$ Place a circle with a radius of 4 in a square chessboard composed of $8 \times 8$ squares with a side length of 8. If the sum of the areas inside the circle of all the small squares that the circumference passes through is denoted as $S_{1}$, and the sum of the areas outside the circle of all the small squares that the circumference passes through is denoted as $S_{2}$, then the integer part of $\frac{S_{1}}{S_{2}}$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(China Junior High School Mathematics League, 1992)
[Solution] According to the symmetry of the square chessboard, we only need to examine one quarter of it (as shown in the figure). Let the sum of the areas inside the circle of all the small squares that the circumference passes through be $S_{1}^{\prime}$, and the sum of the areas outside the circle of all the small squares that the circumference passes through be $S_{2}^{\prime}$.
Then we have $S_{1}^{\prime}=4 \pi-8$, and $S_{2}^{\prime}=15-4 \pi$.

Thus, $\quad \frac{S_{1}}{S_{2}}=\frac{4 S_{1}^{\prime}}{4 S_{2}^{\prime}}=\frac{4 \pi-8}{15-4 \pi} \approx \frac{4.56}{2.44}$.
Also,
$1<\frac{4.56}{2.44}<2$.

Therefore, the answer is $(B)$.