Let a circle with center $O$ and radius $r$, inscribed in an isosceles trapezoid $A B C D$, touch the bases $A D$ and $B C$ at points $K$ and $L$ respectively, and the lateral side $C D$ at point $M$. Given that $C M=4, D M=9$.

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,

$$
r=O M=\sqrt{C M \cdot D M}=\sqrt{4 \cdot 9}=6
$$

Then, 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, $K L=$ $2 r=12$. Therefore,

$$
S_{A B C D}=\frac{A D+B C}{2} \cdot K L=\frac{18+8}{2} \cdot 12=156
$$

## Answer

156.
$3-$ Area of a trapezoid $\quad$ K [ Mean proportionals in a right triangle ]

A circle inscribed in an isosceles trapezoid divides its lateral side into segments equal to 4 and 9. Find the area of the trapezoid.