Two concentric circles having common centre O and chord AB of the outer circle intersect the inner circle at points C and D. If distance of chord from the centre is 3 cm, outer radius is 13 cm inner radius is 7 cm, then length of AC (in cm) is

Options

A. `8sqrt 10`
B. `6 sqrt(10)`
C. `4 sqrt 10`
D. `2 sqrt10`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have two concentric circles with a common center O. The outer circle has a radius of 13 cm, and the inner circle has a radius of 7 cm. The chord AB of the outer circle is at a distance of 3 cm from the center O. This means that if we draw a perpendicular line from O to the chord AB, it will intersect the chord at point K, and the length OK will be 3 cm.

### Step 2: Set Up the Right Triangle
Since OK is perpendicular to AB, we can form two right triangles: triangle OCK (with OC as the hypotenuse) and triangle OAK (with OA as the hypotenuse).

### Step 3: Apply the Pythagorean Theorem to Triangle OCK
In triangle OCK:
- OC = 7 cm (inner radius)
- OK = 3 cm (distance from center to chord)
- CK = ? (half the length of the chord)

Using the Pythagorean theorem:
\[ OC^2 = OK^2 + CK^2 \]
Substituting the known values:
\[ 7^2 = 3^2 + CK^2 \]
\[ 49 = 9 + CK^2 \]
\[ CK^2 = 49 - 9 = 40 \]
\[ CK = \sqrt{40} = 2\sqrt{10} \text{ cm} \]

### Step 4: Apply the Pythagorean Theorem to Triangle OAK
In triangle OAK:
- OA = 13 cm (outer radius)
- OK = 3 cm (distance from center to chord)
- AK = ? (half the length of the chord)

Using the Pythagorean theorem:
\[ OA^2 = OK^2 + AK^2 \]
Substituting the known values:
\[ 13^2 = 3^2 + AK^2 \]
\[ 169 = 9 + AK^2 \]
\[ AK^2 = 169 - 9 = 160 \]
\[ AK = \sqrt{160} = 4\sqrt{10} \text{ cm} \]

### Step 5: Find Length AC
Since AC is the segment from point A to point C, we can find it by subtracting CK from AK:
\[ AC = AK - CK \]
Substituting the values we found:
\[ AC = 4\sqrt{10} - 2\sqrt{10} = 2\sqrt{10} \text{ cm} \]

### Final Answer
The length of AC is \( 2\sqrt{10} \) cm.

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