4. A circle is given with center $S$ and diameter $\overline{A B}$ of length $2 r$. The angle between the radii $\overline{S B}$ and $\overline{S C}$ is $30^{\circ}$. At point $C$, a tangent is constructed which intersects the line $A B$ at point $D$. Calculate the area and perimeter of the figure bounded by the circular arc $B C$ and the segments $\overline{C D}$ and $\overline{B D}$.
4. 

![](https://cdn.mathpix.com/cropped/2024_05_30_b01eff6f13fa46d9f24bg-14.jpg?height=711&width=831&top_left_y=333&top_left_x=313)

Let $|SD| = d$.

Triangle $\triangle SDC$ is a right triangle with acute angles of $30^{\circ}$ and $60^{\circ}$, i.e., half of an equilateral triangle with side length $d$.

The length of $\overline{SC}$ is the height of that equilateral triangle, so $\frac{d \sqrt{3}}{2} = r$, from which it follows that $d = \frac{2 r \sqrt{3}}{3}$.

The hypotenuse of the right triangle $\triangle SCD$ has a length of $\frac{2 r \sqrt{3}}{3}$, and the leg $\overline{DC}$ is half as long, so it has a length of $\frac{r \sqrt{3}}{3}$.

The area of triangle $\triangle SDC$ is $P_{\Delta} = \frac{r \cdot \frac{r \sqrt{3}}{3}}{2} = \frac{r^2 \sqrt{3}}{6}$.

The area of the circular sector with a central angle of $30^{\circ}$ is $P_{i} = \frac{r^2 \cdot \pi \cdot 30^{\circ}}{360^{\circ}} = \frac{r^2 \cdot \pi}{12}$.

The area of the given shape is $P = \frac{r^2 \sqrt{3}}{6} - \frac{r^2 \cdot \pi}{12} = \frac{r^2}{12}(2 \sqrt{3} - \pi)$.

The length of the circular arc $BC$ is $/ = \frac{r \cdot \pi \cdot 30^{\circ}}{180^{\circ}} = \frac{r \cdot \pi}{6}$,

and the perimeter of the given shape is $O = \frac{r \pi}{6} + \frac{r \sqrt{3}}{3} + \frac{2 r \sqrt{3}}{3} - r = \frac{r \pi}{6} + r \sqrt{3} - r = r\left(\frac{\pi}{6} + \sqrt{3} - 1\right)$.