3.17. A circle of radius $R$ is circumscribed around an isosceles trapezoid with an acute angle $\alpha$ at the base. Find the perimeter of the trapezoid.
3.17. Given that $ABCD$ is a trapezoid, $AB = CD$, $O$ is the center of the circle inscribed in the trapezoid, $OE \perp AD$, $OE = R$, $\angle BAD = \alpha$, $\alpha < \frac{\pi}{2}$ (Fig. 3.19), we need to find $P_{ABCD} = AD + 2AB + BC$. According to the property of a circumscribed quadrilateral, we have $AD + BC = 2AB$; therefore, $P_{ABCD} = 4AB$. Draw $BK \| OE$; then $BK = 2OE = 2R$. Finally, from $\triangle BKA$ we find $AB = \frac{2R}{\sin \alpha}$ and $P_{ABCD} = \frac{8R}{\sin \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-510.jpg?height=414&width=544&top_left_y=416&top_left_x=609)

Fig. 3.19