Given a rectangular trapezoid. A circle constructed on the smaller lateral side as a diameter touches the other lateral side and divides it into segments equal to $a$ and $b$. Find the radius of the circle.

#
Segments connecting the center of the circle with the ends of the larger lateral side of the trapezoid are mutually perpendicular

## Solution

The radius drawn from the center $O$ of the circle to the point $C$ of tangency of the circle with the lateral side $AB$ is the height of the triangle $AOB$, dropped from the vertex of the right angle to the hypotenuse. Therefore,

$$
OC^2 = AC \cdot CB = ab
$$

The radius drawn from the center $O$ of the circle to the point $C$ of tangency of the circle with the lateral side $AB$ is the height of the triangle $AOB$, dropped from the vertex of the right angle to the hypotenuse. Therefore,

$$
OC^2 = AC \cdot CB = ab
$$

The radius drawn from the center $O$ of the circle to the point $C$ of tangency of the circle with the lateral side $AB$ is the height of the triangle $AOB$, dropped from the vertex of the right angle to the hypotenuse. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-11.jpg?height=437&width=420&top_left_y=456&top_left_x=819)

Answer

$\sqrt{ab}$.