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Let's break down the problem step by step:

We are given a circle with:
- Diameter \( CD = 2r \), where \( r \) is the radius of the circle.
- Chord \( EF \) is \( \frac{1}{5} \) of the diameter, so the length of chord \( EF \) is:
  \[
  EF = \frac{1}{5} \times 2r = \frac{2r}{5}
  \]
  
We are tasked with finding the distance between the chord \( EF \) and the diameter \( CD \) in terms of \( r \).

### Approach:
1. **Center and symmetry**: Since \( EF \) is a chord and the circle is symmetric about the diameter \( CD \), the perpendicular distance from the center \( A \) of the circle to the chord \( EF \) will give the required distance.
  
2. **Right triangle setup**: Let \( M \) be the midpoint of the chord \( EF \), and let \( AM \) be the perpendicular distance from the center \( A \) to the chord. By symmetry, \( M \) lies on \( AM \), and \( AM \) is the distance we are looking for.

3. **Using Pythagoras' Theorem**: In the right triangle \( AME \), where:
   - \( AE = r \) (radius of the circle),
   - \( EM = \frac{EF}{2} = \frac{r}{5} \) (half of the chord length),
   - \( AM \) is the perpendicular distance we need to find.

   By applying the Pythagorean theorem:
   \[
   AE^2 = AM^2 + EM^2
   \]
   Substituting the values:
   \[
   r^2 = AM^2 + \left(\frac{r}{5}\right)^2
   \]
   Simplifying:
   \[
   r^2 = AM^2 + \frac{r^2}{25}
   \]
   Multiply through by 25 to clear the fraction:
   \[
   25r^2 = 25AM^2 + r^2
   \]
   \[
   24r^2 = 25AM^2
   \]
   \[
   AM^2 = \frac{24r^2}{25}
   \]
   \[
   AM = \frac{r\sqrt{24}}{5} = \frac{2r\sqrt{6}}{5}
   \]

Thus, the distance between the chord and the diameter is \( \frac{2r\sqrt{6}}{5} \).

This corresponds to the answer choice:
\[
\boxed{(c) \frac{\sqrt{24}}{5}r}
\]

Would you like more details on any of the steps? Here are some related questions to explore further:

1. How do you determine the length of a chord in a circle?
2. What is the relationship between a chord and the perpendicular from the center of a circle?
3. How does the Pythagorean theorem apply to solving problems in circles?
4. What other methods can be used to find the distance between a chord and the center of a circle?
5. How does the symmetry of a circle simplify geometric problems?

**Tip**: Always look for symmetry in geometric figures—it can significantly simplify problem-solving!

In the circle shown, if chord EF is 1/5 of diameter CD, what is the distance between the chord and the diameter in terms of r?

Solve a geometry problem involving a circle where a chord EF is 1/5 of the diameter CD. Learn how to use the Pythagorean Theorem to find the distance between the chord and the diameter in terms of the circle's radius.

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