In a circle of radius 5 cm, PQ and RS are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the distance between the chords if they are on:
opposite sides of the centre.

Answer

To solve the problem, we need to find the distance between two parallel chords, PQ and RS, in a circle of radius 5 cm, where the lengths of the chords are 8 cm and 6 cm respectively, and they are on opposite sides of the center.

### Step-by-step Solution:

1. **Identify the Circle and Chords**:
- We have a circle with a radius of 5 cm.
- Chord PQ has a length of 8 cm.
- Chord RS has a length of 6 cm.
- Both chords are parallel and on opposite sides of the center of the circle.

2. **Find the Midpoints of the Chords**:
- Let A be the midpoint of chord PQ and B be the midpoint of chord RS.
- Since PQ is 8 cm long, the length from the center to A (denoted as AQ) is half of 8 cm:
\[ AQ = \frac{8}{2} = 4 \text{ cm} \]
- Similarly, for chord RS which is 6 cm long, the length from the center to B (denoted as BS) is half of 6 cm:
\[ BS = \frac{6}{2} = 3 \text{ cm} \]

3. **Apply the Pythagorean Theorem**:
- In triangle OAQ (where O is the center of the circle), we can apply the Pythagorean theorem:
\[ OQ^2 = OA^2 + AQ^2 \]
where \( OQ \) is the radius (5 cm):
\[ 5^2 = OA^2 + 4^2 \]
\[ 25 = OA^2 + 16 \]
\[ OA^2 = 25 - 16 = 9 \]
\[ OA = \sqrt{9} = 3 \text{ cm} \]

- Now, for triangle OBS (where B is the midpoint of chord RS):
\[ OS^2 = OB^2 + BS^2 \]
where \( OS \) is also the radius (5 cm):
\[ 5^2 = OB^2 + 3^2 \]
\[ 25 = OB^2 + 9 \]
\[ OB^2 = 25 - 9 = 16 \]
\[ OB = \sqrt{16} = 4 \text{ cm} \]

4. **Calculate the Distance Between the Chords**:
- Since the chords are on opposite sides of the center, the total distance between the chords PQ and RS is:
\[ AB = OA + OB = 3 \text{ cm} + 4 \text{ cm} = 7 \text{ cm} \]

### Final Answer:
The distance between the chords PQ and RS is **7 cm**.