A circle has a chord AB of length 24 cm. OC is the perpendicular bisector of AB, with C between O and B. If OC = x cm and CE = 8 cm, where E is a point on the circle, find the value of x.

AB is a chord of length 24 cm, and C is its midpoint.

O is the center of the circle, and E is a point on the circle, so OE is the radius.
We are given $CO = x$ and $CE = 8$ cm.
From the diagram, O, C, and E are collinear, with C between O and E.
Therefore, the radius $r = OE = OC + CE = x + 8$ cm.

Since CE $\perp$ AB, and O, C, E are collinear, OC $\perp$ AB. This means $\triangle OAC$ is a right-angled triangle with the right angle at C.
Using the Pythagorean theorem ($hypotenuse^2 = side1^2 + side2^2$):

Substitute $OA = r = x + 8$, $OC = x$, and $AC = 12$:

Expand the left side and simplify:

Subtract $x^2$ from both sides:

Subtract 64 from both sides:

Divide by 16:

The value of x is 5 cm.

The final answer is $\text{5 cm}$.