5. (6 points) As shown in the figure, a square with a side length of $12 \, \text{cm}$ partially overlaps with a circle of diameter $16 \, \text{cm}$ (the center of the circle is one vertex of the square). Let $S_{1}, S_{2}$ represent the areas of the two blank regions, respectively. Then $S_{1}-S_{2}=$ $\qquad$ $\text{cm}^{2}$ (take $\pi$ as 3).
5. (6 points) As shown in the figure, a square with a side length of $12 \mathrm{~cm}$ partially overlaps with a circle with a diameter of $16 \mathrm{~cm}$ (the center of the circle is one of the vertices of the square).
3).
【Solution】Solution: $3 \times(16 \div 2)^{2}-12^{2}$
$=192$ - 144,
$=48$ (square centimeters);
Answer: $S_{1}-S_{2}=48 \mathrm{~cm}^{2}$.
Therefore, the answer is: 48.