AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that `AB=10 cm`, `CD=24cm` and the distance between AB and CD is `17 cm`. Find the radius of the circle.

Explanation

To find the radius of the circle given two parallel chords AB and CD, we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Circle and Chords**:
- Draw a circle with center O.
- Draw two parallel chords AB and CD such that AB = 10 cm and CD = 24 cm. Place AB above CD, with a distance of 17 cm between them.

2. **Label Distances**:
- Let the distance from the center O to chord AB be \( OP = x \).
- The distance from the center O to chord CD will then be \( OQ = 17 - x \) (since the total distance between the two chords is 17 cm).

3. **Bisect the Chords**:
- Since the perpendicular from the center of the circle to a chord bisects the chord:
- For chord AB, let \( AP = PB = \frac{AB}{2} = \frac{10}{2} = 5 \) cm.
- For chord CD, let \( CQ = QD = \frac{CD}{2} = \frac{24}{2} = 12 \) cm.

4. **Apply the Pythagorean Theorem**:
- For triangle OPB (right triangle):
\[ OB^2 = OP^2 + PB^2 \implies r^2 = x^2 + 5^2 \implies r^2 = x^2 + 25 \quad \text{(Equation 1)} \]
- For triangle OQD (right triangle):
\[ OD^2 = OQ^2 + QD^2 \implies r^2 = (17 - x)^2 + 12^2 \implies r^2 = (17 - x)^2 + 144 \quad \text{(Equation 2)} \]

5. **Set Equations Equal**:
- Since both expressions equal \( r^2 \):
\[ x^2 + 25 = (17 - x)^2 + 144 \]

6. **Expand and Simplify**:
- Expand \( (17 - x)^2 \):
\[ (17 - x)^2 = 289 - 34x + x^2 \]
- Substitute back into the equation:
\[ x^2 + 25 = 289 - 34x + x^2 + 144 \]
- Cancel \( x^2 \) from both sides:
\[ 25 = 433 - 34x \]
- Rearranging gives:
\[ 34x = 433 - 25 \implies 34x = 408 \implies x = \frac{408}{34} = 12 \]

7. **Find the Radius**:
- Substitute \( x = 12 \) back into Equation 1:
\[ r^2 = 12^2 + 25 = 144 + 25 = 169 \implies r = \sqrt{169} = 13 \text{ cm} \]

### Final Answer:
The radius of the circle is **13 cm**.

---