A radius of circle is 2.5cm AB and CD two parallel chords are at distance of 2.7cm. It AB=4.8cm, then value of CD is

Options

A. 4.8cm
B. 2.4cm
C. 3cm
D. 4cm

Answer

To solve the problem, we need to find the length of the chord CD in a circle with a radius of 2.5 cm, where the distance between two parallel chords AB and CD is 2.7 cm, and the length of chord AB is 4.8 cm.

### Step-by-Step Solution:

1. **Draw the Circle and Chords**:
- Draw a circle with a radius of 2.5 cm.
- Mark two parallel chords AB and CD such that the distance between them is 2.7 cm.

2. **Identify the Midpoints**:
- Let M be the midpoint of chord AB and N be the midpoint of chord CD.
- Since AB = 4.8 cm, the length of AM (or MB) will be half of AB:
\[ AM = MB = \frac{4.8}{2} = 2.4 \text{ cm} \]

3. **Use the Radius and Distance**:
- The distance from the center O of the circle to chord AB is denoted as OM, and the distance from O to chord CD is ON.
- Given that the distance between the two chords is 2.7 cm, we can express this as:
\[ ON = OM + 2.7 \]

4. **Apply the Pythagorean Theorem**:
- For triangle OMB (where OB is the radius):
\[ OB^2 = OM^2 + MB^2 \]
- Substituting the known values:
\[ (2.5)^2 = OM^2 + (2.4)^2 \]
- This simplifies to:
\[ 6.25 = OM^2 + 5.76 \]

5. **Solve for OM**:
- Rearranging gives:
\[ OM^2 = 6.25 - 5.76 = 0.49 \]
- Taking the square root:
\[ OM = \sqrt{0.49} = 0.7 \text{ cm} \]

6. **Find ON**:
- Using the distance relationship:
\[ ON = OM + 2.7 = 0.7 + 2.7 = 3.4 \text{ cm} \]

7. **Apply Pythagorean Theorem for CD**:
- For triangle OND (where OD is the radius):
\[ OD^2 = ON^2 + ND^2 \]
- Substituting the known values:
\[ (2.5)^2 = (3.4)^2 + ND^2 \]
- This simplifies to:
\[ 6.25 = 11.56 + ND^2 \]

8. **Solve for ND**:
- Rearranging gives:
\[ ND^2 = 6.25 - 11.56 = -5.31 \]
- Since this is not possible, we need to correct our earlier calculations.

9. **Recalculate**:
- The correct distance from the center to chord CD should be:
\[ ON = 2.5 - 0.7 = 1.8 \text{ cm} \]
- Now apply the Pythagorean theorem again:
\[ OD^2 = ON^2 + ND^2 \]
- Substituting the known values:
\[ (2.5)^2 = (1.8)^2 + ND^2 \]
- This simplifies to:
\[ 6.25 = 3.24 + ND^2 \]
- Rearranging gives:
\[ ND^2 = 6.25 - 3.24 = 3.01 \]
- Taking the square root:
\[ ND = \sqrt{3.01} \approx 1.73 \text{ cm} \]

10. **Find Length of CD**:
- Since N is the midpoint of CD:
\[ CD = 2 \times ND = 2 \times 1.73 \approx 3.46 \text{ cm} \]

### Final Answer:
The length of chord CD is approximately **3.46 cm**.