## Task B-3.3.

Let $A B C$ be a right triangle with a right angle at vertex $C$. The longer leg $\overline{A C}$ is the diameter of a semicircle that intersects the hypotenuse at point $D$. Determine the length of the semicircle if the length of the shorter leg is $30 \mathrm{~cm}$, and the length of the chord $\overline{C D}$ is $24 \mathrm{~cm}$.
## First solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-21.jpg?height=525&width=714&top_left_y=297&top_left_x=591)

Angle $CDA$ is a right angle because it subtends the diameter of the circle, so triangles $ADC$ and $CDB$ are right triangles.

By the Pythagorean theorem, we have $|BD|^2 = |CB|^2 - |CD|^2 = 30^2 - 24^2 = 324$, so

$$
|BD| = 18 \text{ cm}
$$

If we denote the angle at vertex $B$ as $\beta$, from the right triangle $CDB$ it follows that

$$
\tan \beta = \frac{24}{18} = \frac{4}{3}
$$

In the right triangle $ABC$, $\tan \beta = \frac{|AC|}{30}$.

Thus, $\frac{|AC|}{30} = \frac{4}{3}$, so the longer leg $|AC| = 40 \text{ cm}$.

Finally, the length of the semicircle is $l = r \pi = 20 \pi \text{ cm}$.