12.033. A rectangular trapezoid with an acute angle $\alpha$ is circumscribed around a circle. Find the height of the trapezoid if its perimeter is $P$.
## Solution.

Let $\angle C D A=\alpha, \angle A B C=90^{\circ}, P-$ be the perimeter of the trapezoid (Fig. 12.36) $(B C \| A D)$. Since $B C+A D=A B+C D$, and $P=B C+A D+A B+C D$, then $B C+A D=\frac{P}{2}=D C+A B$. Denote $A B=C E=x$. From $\triangle C E D$,

$$
\angle C E D=90^{\circ}, C D=\frac{x}{\sin \alpha}, \text { then } \frac{P}{2}=x+\frac{x}{\sin \alpha} \text { and from this }
$$

$$
x=\frac{P \sin \alpha}{2(1+\sin \alpha)}=\frac{P}{2} \frac{\sin \alpha}{\left(\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}\right)^{2}}=\frac{P \sin \alpha}{4 \cdot \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}
$$

Answer: $\frac{P \sin \alpha}{4 \cdot \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$.