A chord 6 cm long is at a distance of 4 cm from the centre of a circle. Find the length of a chord which is at a distance of 3 cm from the centre of the circle.

Options

A. 10 cm
B. 6 cm
C. 8 cm
D. 12 cm

Answer

To find the length of a chord that is at a distance of 3 cm from the center of the circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Length of the first chord (AB) = 6 cm
- Distance from the center (C) to the first chord = 4 cm

2. **Determine the Half-Length of the First Chord:**
- Since the chord is bisected by the perpendicular from the center, we can find the half-length of the chord:
\[ AP = PB = \frac{AB}{2} = \frac{6}{2} = 3 \text{ cm} \]

3. **Apply the Pythagorean Theorem in Triangle ACP:**
- In triangle ACP, where AC is the radius, CP is the distance from the center to the chord, and AP is half the chord length:
\[ AC^2 = AP^2 + CP^2 \]
- Plugging in the values:
\[ AC^2 = 3^2 + 4^2 \]
\[ AC^2 = 9 + 16 = 25 \]
\[ AC = \sqrt{25} = 5 \text{ cm} \]

4. **Identify the Distance for the Second Chord:**
- Now we need to find the length of the second chord (DE) which is at a distance of 3 cm from the center.

5. **Apply the Pythagorean Theorem in Triangle COD:**
- Again, using the Pythagorean theorem for triangle COD:
\[ CD^2 = OD^2 + OC^2 \]
- Here, OC is the distance from the center to the second chord (3 cm), and CD is the radius (5 cm):
\[ 5^2 = OD^2 + 3^2 \]
\[ 25 = OD^2 + 9 \]
\[ OD^2 = 25 - 9 = 16 \]
\[ OD = \sqrt{16} = 4 \text{ cm} \]

6. **Calculate the Length of the Second Chord:**
- Since the perpendicular from the center bisects the chord, DE can be calculated as:
\[ DE = OD + OE = 4 + 4 = 8 \text{ cm} \]

### Final Answer:
The length of the chord which is at a distance of 3 cm from the center of the circle is **8 cm**.

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