A chord AB is drawn in a circle with centre O and radius 5 cm. If the shortest distance between centre and chord is 4 cm, find the length of chord AB?

Answer

To find the length of the chord AB in a circle with a center O and radius 5 cm, where the shortest distance from the center to the chord is 4 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Radius of the circle (OA) = 5 cm
- Shortest distance from the center to the chord (OK) = 4 cm

2. **Understand the Geometry:**
- The shortest distance from the center O to the chord AB is perpendicular to the chord and meets the chord at point K.
- Since OK is perpendicular to AB, it bisects the chord. Therefore, AK = KB.

3. **Apply the Pythagorean Theorem:**
- In triangle OAK, we can use the Pythagorean theorem:
\[ OA^2 = AK^2 + OK^2 \]
- Substituting the known values:
\[ 5^2 = AK^2 + 4^2 \]
- This simplifies to:
\[ 25 = AK^2 + 16 \]

4. **Solve for AK:**
- Rearranging the equation gives:
\[ AK^2 = 25 - 16 \]
\[ AK^2 = 9 \]
- Taking the square root of both sides:
\[ AK = 3 \text{ cm} \]

5. **Calculate the Length of the Chord AB:**
- Since AK = KB, the total length of the chord AB is:
\[ AB = AK + KB = AK + AK = 2 \times AK = 2 \times 3 = 6 \text{ cm} \]

### Final Answer:
The length of chord AB is **6 cm**.

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