12.032. In an isosceles triangle, the angle between the lateral sides is $\alpha$, and the radius of the inscribed circle is $r$. Find the area of the triangle.
## Solution.

Given in $\triangle ABC \quad AB=BC, \angle ABC=\alpha, r$ - the radius of the inscribed circle. From $\triangle OEC \quad \angle OEC=90^{\circ}$ (Fig. 12.35), $EC=r \cdot \operatorname{ctg} \angle ECO=$ $=r \cdot \operatorname{ctg}\left(\pi / 4-\frac{\alpha}{4}\right) \cdot$ From $\triangle BEC, \angle BEC=90^{\circ}$, we have $BE=EC \cdot \operatorname{ctg} \frac{\alpha}{2}=$ $=r \cdot \operatorname{ctg}\left(\pi / 4-\frac{\alpha}{4}\right) \cdot \operatorname{ctg} \frac{\alpha}{2}$. Then $S=\frac{2}{2} r^{2} \cdot \operatorname{ctg}^{2}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right) \cdot \operatorname{ctg} \frac{\alpha}{2}=$ $=r^{2} \cdot \operatorname{ctg}^{2}\left(\frac{\pi-\alpha}{4}\right) \cdot \operatorname{ctg} \frac{\alpha}{2}$.

Answer: $\quad r^{2} \cdot \operatorname{ctg}^{2}\left(\frac{\pi-\alpha}{4}\right) \cdot \operatorname{ctg} \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-739.jpg?height=362&width=482&top_left_y=97&top_left_x=143)

Fig. 12.36

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-739.jpg?height=368&width=356&top_left_y=94&top_left_x=780)

Fig. 12.37