3. Two circles with equal radii touch at point $C$ and touch externally a third circle with radius $r=5$ cm at points $A$ and $B$. Calculate the area of triangle $ABC$, if $\overline{AB}=6$ cm.
Solution. Let

$$
\overline{A O}_{3}=5 \text{ cm}, \overline{A B}=6 \text{ cm}, \overline{C D}=x
$$

From $\triangle O_{1} C O_{3} \sim \triangle A_{3}$ we get

$$
\begin{aligned}
& \overline{O_{1} C}: \overline{A D}={\overline{O_{1} O_{3}}: \overline{A O}}_{3} \Rightarrow \\
& \overline{O_{1} C}: \overline{A B}=\overline{O_{1} A}: \overline{A O_{3}} \Rightarrow \\
& \overline{O_{1} C}: 3=\left(\overline{O_{1} C}+5\right): 5 \Rightarrow \\
& 5 \overline{O_{1} C}=3 \overline{O_{1} C}+15 \Rightarrow \\
& \overline{O_{1} C}=\frac{15}{2}=7.5 \text{ cm}
\end{aligned}
$$

According to this,

![](https://cdn.mathpix.com/cropped/2024_06_05_b31b6b11300135c24fa8g-5.jpg?height=560&width=495&top_left_y=1027&top_left_x=954)

$$
\overline{O_{3} D}=\sqrt{\overline{A D}_{3}^{2}-\overline{A D}^{2}}=\sqrt{25-7}=4
$$

From

$$
\overline{O_{1} C}: \overline{A D}=\overline{O_{3} C}: \overline{C D}
$$

we get $\frac{15}{2}: 3=(x+4): 4$, i.e., $x=6 \text{ cm}$. Finally, $P=\frac{\overline{A B} x}{2}=18 \text{ cm}^{2}$.