Problem 3. A circle with radius $r=12 \mathrm{~cm}$ has an isosceles triangle $A B C$ inscribed in it, with the angle at the vertex equal to $30^{\circ}$. Determine the perimeter and area of triangle $A B C$.
Solution. Since the angle at the vertex of the isosceles triangle $ABC$ is $30^{\circ}$, it follows that the central angle $\measuredangle AOB$ is $60^{\circ}$. Therefore, triangle $AOB$ is equilateral, so $\overline{AB}=12 \mathrm{~cm}$. The height of triangle $ABC$ is equal to the sum of the height of triangle $AOB$ and the radius of the circumscribed circle. Thus,

![](https://cdn.mathpix.com/cropped/2024_06_05_8aa76aa53b232df96dffg-11.jpg?height=447&width=407&top_left_y=1130&top_left_x=1075)

$$
h_{C}=\overline{OC}+\frac{\overline{OB} \sqrt{3}}{2}=12+6 \sqrt{3}=6(2+\sqrt{3}) \mathrm{cm}
$$

From the Pythagorean theorem, it follows that $\overline{AC}=\overline{BC}=12 \sqrt{2+\sqrt{3}} \mathrm{~cm}$. Therefore, the perimeter of the triangle is $24 \sqrt{2+\sqrt{3}}+12 \mathrm{~cm}$, and the area is $36(2+\sqrt{3}) \mathrm{cm}^{2}$.