Find the length of a chord in a circle with radius 8 cm, if the angle subtended by the chord at the center is 90 degrees.

The radius of the circle is $r = 8 \text{ cm}$.
The angle subtended by the chord at the center is $\theta = 90^\circ$.

Consider the triangle formed by the two radii connecting the center to the endpoints of the chord, and the chord itself. Let the center be $O$ and the endpoints of the chord be $A$ and $B$.
Then $OA = OB = r = 8 \text{ cm}$.
The angle $\angle AOB = 90^\circ$.
This means $\triangle OAB$ is a right-angled isosceles triangle, with the chord $AB$ as its hypotenuse.

According to the Pythagorean theorem, for a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let the length of the chord be $c$.

The length of the chord is $8\sqrt{2} \text{ cm}$.