## Task B-2.4.

The lengths $\overline{A D}$ and $\overline{E H}$, both 12 units long, are divided by points $B$ and $C$, and $F$ and $G$, respectively, into three equal parts. Over each of these segments, semicircles are constructed that touch each other at points $R, S$, and $T$ as shown in the figure. The circle $k$ with center at point $S$ has a radius of 4 units. What is the probability that a randomly chosen point inside circle $k$ will not be inside the semicircles but in the shaded area?

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-11.jpg?height=314&width=489&top_left_y=588&top_left_x=795)
## Solution.

The radius of each of the semicircles is $12: 6=2$.

First, let's determine the area of the shaded circular segment shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-11.jpg?height=432&width=688&top_left_y=1206&top_left_x=227)

The distance from the chord of the circle $k$ (which includes points $B$ and $C$) to the center $S$ is equal to the radius of the semicircle and is 2. Let's denote the angle $\alpha$ of the corresponding circular sector as shown in the figure. Since the radius of the circle $k$ is 4, we have $\cos \frac{\alpha}{2}=\frac{2}{4}=\frac{1}{2}$, which means $\alpha=120^{\circ}$.

Then, the area $P_{1}$ of the marked circular segment is:

$P_{1}=\frac{4^{2} \pi \cdot 120^{\circ}}{360^{\circ}}-\frac{4 \cdot 4 \cdot \sin 120^{\circ}}{2}=\frac{16 \pi}{3}-4 \sqrt{3}$.

Next, let's determine the area of the shaded part between the semicircles.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-11.jpg?height=348&width=574&top_left_y=2142&top_left_x=204)

The area of the two shaded parts in the figure is $P_{2}=2\left(4^{2}-2^{2} \pi\right)=32-8 \pi$.

The total area of all the shaded parts of the given circle $k$ is:

$P_{O}=2 P_{1}+P_{2}=2\left(\frac{16 \pi}{3}-4 \sqrt{3}\right)+32-8 \pi=\frac{8}{3} \pi+32-8 \sqrt{3}$.

The desired probability is the ratio of the shaded area to the area of the circle $k$, that is:

$p=\frac{P_{O}}{P_{k}}=\frac{\frac{8}{3} \pi+32-8 \sqrt{3}}{16 \pi}$.