[

A circle inscribed in a right trapezoid divides its larger lateral side into segments of 1 and 4. Find the area of the trapezoid.

#
Let a circle with center $O$ and radius $r$, inscribed in a right trapezoid $A B C D$ with bases $A D$ and $B C$ and right angles at vertices $A$ and $B$, touch the bases $A D$ and $B C$ at points $K$ and $L$ respectively, and the larger lateral side $C D$ at point $M$. Given that $C M=1, D M=4$.

Since $C O$ and $D O$ are the angle bisectors of angles $B C D$ and $A D C$, the sum of which is $180^{\circ}$, angle $C O D$ is a right angle, so $O M$ is the altitude of the right triangle $C O D$ drawn from the vertex of the right angle. Therefore,

$$
r=O M=\sqrt{C M \cdot D M}=\sqrt{1 \cdot 4}=2
$$

Then

$$
A D=A K+D K=r+C D=2+4=6, B C=B L+L C=r+C M=2+1=3
$$

Since points $K, O$, and $L$ lie on a line perpendicular to the bases of the trapezoid, $K L$ is the height of the trapezoid, $K L=$ $2 r=4$. Therefore,

$$
S_{A B C D}=\frac{A D+B C}{2} \cdot K L=\frac{6+3}{2} \cdot 4=18
$$

## Answer

18.