10.224. Given an isosceles triangle with a base of 12 cm and a lateral side of $18 \mathrm{~cm}$. What segments need to be laid off from the vertex of the triangle on its lateral sides, so that by connecting their ends, a trapezoid with a perimeter of 40 cm is obtained?
## Solution.

By the condition in $\triangle A B C$ (Fig. 10.33) we have: $A B=B C=18, A C=12$. Let $B D=B E=x$. Then $A D=E C=18-x$. The perimeter of trapezoid $A D E C$ is: $P=A C+D E+2 A D=40$, from which $D E=40-(2 A D+A C)=2 x-8$. $\triangle A B C \sim$ $\sim \triangle D B E$, therefore, $\frac{B D}{D E}=\frac{A B}{A C}$ or $\frac{x}{2 x-8}=\frac{18}{12}, x=6$.

Answer: 6 cm.