12.014. The area of a rectangular trapezoid is $S$, the acute angle is $\alpha$. Find the height of the trapezoid if its smaller diagonal is equal to the larger base.
Solution.

Given $\angle C B A=90^{\circ}, \angle A D C=\alpha, A C=A D . S_{A B C D}=S$, then $\quad S=\frac{B C+A D}{2} \cdot h$, from which $\quad h=\frac{2 S}{B C+A D}=\frac{2 S}{A D-E D+A D}$ (Fig. 12.17). From $\triangle C E D, \angle C E D=90^{\circ}, E D=h \cdot \operatorname{ctg} \alpha$. From $\triangle A E C$ $\angle A E C=90^{\circ}, \angle C A E=180^{\circ}-2 \alpha \quad(A C=A D), \quad A C=\frac{h}{\sin 2 \alpha}$, then $h=\frac{2 S}{\frac{2 \cdot h}{2 \cdot \sin \alpha \cdot \cos \alpha}-h \frac{\cos \alpha}{\sin \alpha}}, h^{2}=\frac{2 S \cdot \sin \alpha \cdot \cos \alpha}{\sin ^{2} \alpha}, h=\sqrt{2 S \operatorname{ctg} \alpha}$.

Answer: $\sqrt{2 S \operatorname{ctg} \alpha}$.