10.042. One of the angles of the trapezoid is $30^{\circ}$, and the lines containing the lateral sides of the trapezoid intersect at a right angle (Fig. 10.42). Find the length of the shorter lateral side of the trapezoid if its midline is 10 cm, and one of the bases is $8 \mathrm{~cm}$.
## Solution.

Given $\angle B C A=30^{\circ}, \angle A B C=90^{\circ}, K M=10$ cm, $D E=8$ cm, $\frac{8+A C}{2}=10, A C=12$ cm (since $K M$ is the midline). Since $\triangle D B E \sim \triangle A B C$, then $\frac{A B}{D B}=\frac{A C}{D E} ; \frac{x+D B}{D B}=\frac{12}{8}$ (where $A D=x$ ), $x=D B \cdot \frac{1}{2}$. In $\triangle D B E, \angle D B E=90^{\circ}, \angle B E D=\angle B C A=30^{\circ}, D B=D E \cdot \sin 30^{\circ}=8 / 2=4$, so, $x=4 \cdot \frac{1}{2}=2$ cm.

Answer: 2 cm.