[ Pythagorean Theorem (direct and inverse).
[ Law of Sines
[ Isosceles, inscribed, and circumscribed trapezoids ]
[ $\underline{\text { Chords and secants (other). }}$ ]

The lengths of two parallel chords of a circle are 40 and 48, and the distance between them is 22. Find the radius of the circle.
Draw the diagonal of the resulting trapezoid and find the radius of the circumscribed circle of one of the resulting triangles.

## Solution

Let $AB=48$ and $CD=40$ be the parallel chords. Find the desired radius $R$.

First method. Drop a perpendicular $CM$ to $AB$. Then $AM=44, BM=4, AC^2=AM^2+CM^2=44^2+22^2=$ $5 \cdot 22^2, BC^2=CM^2+BM^2=500, S_{ABC}=11 AB$,

$R=AC \cdot BC \cdot AB / 4 S_{ABC}=AC \cdot BC / 44=25$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-03.jpg?height=572&width=606&top_left_y=183&top_left_x=729)

Second method. According to the problem, $\sqrt{R^2-20^2}+\sqrt{R^2-24^2}=22$. Solving this equation, we get $R=25$.

## Answer

25.