12.024. Find the ratio of the perimeter of a trapezoid circumscribed about a circle to the length of this circle, if the angles at the larger base of the trapezoid are $\alpha$ and $\beta$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-732.jpg?height=328&width=504&top_left_y=1028&top_left_x=730)

Fig. 12.27
## Solution.

Given that $ABCD$ is a trapezoid circumscribed around a circle, $\angle BAD = \alpha, \angle CDA = \beta$. Let $R$ be the radius of the inscribed circle. The circumference of the circle is $l = 2\pi R$ (Fig. 12.27). $P = BC + AD + AB + CD$, $BC + AD = AB + CD$, $2(AB + CD) = P$. From $\triangle BFA$, $\angle BFA = 90^\circ$, $AB = \frac{2R}{\sin \alpha}$.

From $\triangle CED$, $\angle CED = 90^\circ$, $CD = \frac{2R}{\sin \beta}$. Then $P = 4R\left(\frac{1}{\sin \alpha} + \frac{1}{\sin \beta}\right)$; $\frac{P}{l} = \frac{2(\sin \alpha + \sin \beta)}{\pi \sin \alpha \cdot \sin \beta} = \frac{4 \cdot \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}}{\pi \cdot \sin \alpha \sin \beta}$.

Answer: $\frac{4 \cdot \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}}{\pi \cdot \sin \alpha \sin \beta}$.