8,9

A circle with a diameter of 15 is circumscribed by an isosceles trapezoid with a side length of 17. Find the bases of the trapezoid.
The height of the given trapezoid is equal to the diameter of the inscribed circle, and the midline is equal to the lateral side.

## Solution

Let $M$ be the base of the height drawn from vertex $C$ of the smaller base $B C$ of trapezoid $A B C D$ to the larger base $A D$. Then

$$
M D=\sqrt{C D^{2}-C M^{2}}=8
$$

$A M$ is equal to the midline of the trapezoid: $A M=17$. Therefore,

$$
A D=17+8=25
$$

Since $B C+A D=A B+B C=34$, then

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-05.jpg?height=452&width=604&top_left_y=2358&top_left_x=727)

## Problem