On the circle of radius $30$ cm are given $2$ points A,B with $AB = 16$ cm and $C$ is a midpoint of $AB$. What is the perpendicular distance from $C$ to the circle?


1. Given a circle with radius \( R = 30 \) cm, and two points \( A \) and \( B \) on the circle such that \( AB = 16 \) cm. Let \( C \) be the midpoint of \( AB \).

2. Since \( C \) is the midpoint of \( AB \), we have \( AC = CB = \frac{AB}{2} = \frac{16}{2} = 8 \) cm.

3. The perpendicular distance from \( C \) to the circle is the distance from \( C \) to the center of the circle, denoted as \( O \).

4. Consider the triangle \( OAC \). Since \( O \) is the center of the circle and \( A \) is on the circle, \( OA = R = 30 \) cm. 

5. In triangle \( OAC \), we have:
   - \( OA = 30 \) cm (radius)
   - \( AC = 8 \) cm (half of \( AB \))

6. We need to find the distance \( OC \). Using the Pythagorean theorem in triangle \( OAC \):
   \[
   OC^2 + AC^2 = OA^2
   \]
   \[
   OC^2 + 8^2 = 30^2
   \]
   \[
   OC^2 + 64 = 900
   \]
   \[
   OC^2 = 900 - 64
   \]
   \[
   OC^2 = 836
   \]
   \[
   OC = \sqrt{836} = 2\sqrt{209}
   \]

Therefore, the perpendicular distance from \( C \) to the center of the circle is \( 2\sqrt{209} \) cm.

The final answer is \( \boxed{ 2\sqrt{209} } \) cm.