10.291. The center of the circle inscribed in a right trapezoid is 1 cm and 2 cm away from the ends of the lateral side. Find the area of the trapezoid.
Solution.

Let in trapezoid $ABCD$ (Fig. 10.84) $BC \| AD, AB \perp AD, O$ - the center

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0772.jpg?height=50&width=1169&top_left_y=1388&top_left_x=94)
of the angle bisectors of $\angle BCD$ and $\angle ADC$. Since $\angle BCD + \angle ADC = 180^\circ$, then $\angle OCD + \angle ODC = \frac{1}{2} \angle BCD + \frac{1}{2} \angle ADC = 90^\circ$ and $\angle COD = 90^\circ$, $CD = \sqrt{OC^2 + OD^2} = \sqrt{5}$ cm. $K$ is the point of tangency of the inscribed circle with side $CD$ of the trapezoid. Then $OK$ is the radius of this circle and the height of the right $\triangle COD$. Therefore, $OK = \frac{OC \cdot OD}{CD} = \frac{2}{\sqrt{5}}$ cm.
$AB = 2 \cdot OK = \frac{4}{\sqrt{5}}$ cm. Since a circle is inscribed in the given trapezoid,

then $BC + AD = AB + CD = \frac{4}{\sqrt{5}} + \sqrt{5} = \frac{9}{\sqrt{5}}$ cm.

The area of the trapezoid $S = \frac{BC + AD}{2} \cdot AB = \frac{9}{2 \sqrt{5}} \cdot \frac{4}{\sqrt{5}} = 3.6 \text{ cm}^2$.

Answer: $3.6 \text{ cm}^2$.