Task B-2.2. On a circle with a radius of $9 \mathrm{~cm}$, the center of another circle with a radius of $6 \mathrm{~cm}$ is located. Determine the length of the common chord of these two circles.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9c6c142e457d4763ad17g-06.jpg?height=697&width=900&top_left_y=1885&top_left_x=583)

First, let's calculate the area of triangle $ABC$ using Heron's formula.

$$
\begin{aligned}
s & =\frac{9+9+6}{2}=12 \\
P & =\sqrt{12 \cdot 3 \cdot 3 \cdot 6}=18 \sqrt{2} \text{ cm}^2
\end{aligned}
$$

The area of triangle $ABC$ can also be calculated as

$$
P=\frac{|AB| \cdot h}{2}
$$

so the height of triangle $ABC$ is

$$
h=\frac{2 P}{|AB|}=\frac{2 \cdot 18 \sqrt{2}}{9}=4 \sqrt{2} \text{ cm}
$$

Then the length of the common chord $\overline{CD}$ is $2 h=8 \sqrt{2} \text{ cm}$.