[ Chords and secants (etc.). $]$

A chord subtends an arc of $90^{\circ}$ and is equal to 16. Find its distance from the center.
The foot of the perpendicular dropped from the center of the circle to the given chord bisects it.

## Solution

Let $M$ be the foot of the perpendicular dropped from the center $O$ of the circle to the chord $A B$. Then $M$ is the midpoint of $A B$ and the right triangle $O M A$ is isosceles. Therefore, $O M = M A = 8$.

## Answer

8.