The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is

Options

A. `sqrt(5) c m`
B. `2sqrt(5) c m`
C. `2 sqrt(7) c m`
D. `sqrt(7) c m`

Answer

To find the perpendicular distance from the center of the circle to the chord, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Radius of the circle (r) = 6 cm
- Length of the chord (AB) = 8 cm

2. **Determine the Midpoint of the Chord:**
- Let O be the center of the circle, and let D be the midpoint of the chord AB. Since the perpendicular from the center to the chord bisects the chord, we have:
\[ AD = DB = \frac{AB}{2} = \frac{8 \text{ cm}}{2} = 4 \text{ cm} \]

3. **Apply the Pythagorean Theorem:**
- In the right triangle OBD, we can apply the Pythagorean theorem:
\[ OB^2 = OD^2 + BD^2 \]
- Here, OB is the radius (6 cm), OD is the perpendicular distance we want to find, and BD is half the chord length (4 cm).

4. **Substitute the Known Values:**
- Substitute the values into the equation:
\[ 6^2 = OD^2 + 4^2 \]
- This simplifies to:
\[ 36 = OD^2 + 16 \]

5. **Solve for OD^2:**
- Rearranging the equation gives:
\[ OD^2 = 36 - 16 = 20 \]

6. **Find OD:**
- Taking the square root of both sides:
\[ OD = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \text{ cm} \]

### Final Answer:
The perpendicular distance from the center of the circle to the chord is \( 2\sqrt{5} \) cm.

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