Solution. Since $\left|S_{1} S_{2}\right|=12,\left|C S_{2}\right|=6$ and triangle $S_{1} C S_{2}$ is a right triangle, it is equal to half of an equilateral triangle, so $\angle S_{1} S_{2} C=60^{\circ}$.

(4 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_9c56249806d755c5106ag-07.jpg?height=471&width=899&top_left_y=364&top_left_x=584)

From the area $P_{1}$ of trapezoid $A B S_{2} S_{1}$, we need to subtract the area $P_{2}$ of the circular segment on the smaller circle with a central angle of $120^{\circ}$ and the area of the circular segment $P_{3}$ on the larger circle with a central angle of $60^{\circ}$.

Thus, we have

$$
\begin{gathered}
P_{1}=\frac{\left|A S_{1}\right|+\left|B S_{2}\right|}{2} \cdot\left|S_{1} C\right|=\frac{3+9}{2} \cdot \sqrt{12^{2}-6^{2}} \mathrm{~cm}^{2}=36 \sqrt{3} \mathrm{~cm}^{2} \\
P_{2}=\frac{3^{2} \pi \cdot 120^{\circ}}{360^{\circ}} \mathrm{cm}^{2}=3 \pi \mathrm{cm}^{2} \\
P_{3}=\frac{9^{2} \pi \cdot 60^{\circ}}{360^{\circ}} \mathrm{cm}^{2}=\frac{27 \pi}{2} \mathrm{~cm}^{2}
\end{gathered}
$$

From this, we get

$$
P=P_{1}-P_{2}-P_{3}=\left(36 \sqrt{3}-\frac{33 \pi}{2}\right) \mathrm{cm}^{2}
$$

(4 points)
Task B-2.4. (20 points) Two circles with diameters of $6 \mathrm{~cm}$ and $18 \mathrm{~cm}$ touch each other externally. Calculate the area of the figure bounded by the circles and their common external tangent.