PQ and RS are two parallel chords of a circle with centre C such that PQ = 8 cm and RS = 16 cm. If the chords are on the same side of the centre and the distance between them is 4 cm, then the radius of the circle is :

Options

A. `3 sqrt(2)` cm
B. `3 sqrt(5)` cm
C. `4 sqrt(5)` cm
D. `5 sqrt(5)` cm

Answer

To find the radius of the circle with the given parallel chords PQ and RS, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Length of chord PQ = 8 cm
- Length of chord RS = 16 cm
- Distance between the chords = 4 cm

2. **Determine the Midpoints of the Chords:**
- Let A be the midpoint of chord PQ and B be the midpoint of chord RS.
- Since PQ = 8 cm, PA = AQ = 8/2 = 4 cm.
- Since RS = 16 cm, RB = BS = 16/2 = 8 cm.

3. **Set Up the Coordinate System:**
- Place the center of the circle C at the origin (0, 0).
- Let the distance from the center C to chord PQ be 'x' cm. Therefore, the distance from C to chord RS will be (x + 4) cm (since the distance between the two chords is 4 cm).

4. **Use the Pythagorean Theorem:**
- For triangle QAC (where Q is a point on chord PQ):
\[ QC^2 = AC^2 + AQ^2 \]
\[ r^2 = x^2 + 4^2 \quad \text{(1)} \]
- For triangle SBC (where S is a point on chord RS):
\[ SC^2 = BC^2 + BS^2 \]
\[ r^2 = (x + 4)^2 + 8^2 \quad \text{(2)} \]

5. **Set the Two Equations Equal:**
- Since both equations equal \( r^2 \):
\[ x^2 + 16 = (x + 4)^2 + 64 \]

6. **Expand and Simplify:**
- Expand the right side:
\[ (x + 4)^2 = x^2 + 8x + 16 \]
Therefore:
\[ r^2 = x^2 + 8x + 16 + 64 = x^2 + 8x + 80 \]
- Set the equations equal:
\[ x^2 + 16 = x^2 + 8x + 80 \]
- Cancel \( x^2 \) from both sides:
\[ 16 = 8x + 80 \]
- Rearranging gives:
\[ 8x = 16 - 80 = -64 \]
\[ x = -8 \]

7. **Calculate the Radius:**
- Substitute \( x \) back into either equation for \( r^2 \):
\[ r^2 = (-8)^2 + 16 = 64 + 16 = 80 \]
- Therefore, \( r = \sqrt{80} = 4\sqrt{5} \) cm.

### Final Answer:
The radius of the circle is \( 4\sqrt{5} \) cm.