AD is a diameter of a circle and AB is a chord. If AD = 34cm, AB = 30cm, the distance of AB form the centre of the circle is

Options

A. 17 cm
B. 15 cm
C. 4 cm
D. 8 cm

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Identify the Given Information
- Diameter AD = 34 cm
- Chord AB = 30 cm

### Step 2: Calculate the Radius of the Circle
The radius (OA) can be calculated as half of the diameter (AD):
\[ OA = \frac{AD}{2} = \frac{34 \text{ cm}}{2} = 17 \text{ cm} \]

### Step 3: Bisect the Chord
Since OL is perpendicular to AB and bisects it, we can find the lengths AL and BL:
\[ AL = BL = \frac{AB}{2} = \frac{30 \text{ cm}}{2} = 15 \text{ cm} \]

### Step 4: Apply the Pythagorean Theorem
In the right triangle AOL, we can apply the Pythagorean theorem:
\[ OA^2 = AL^2 + OL^2 \]
Substituting the known values:
\[ 17^2 = 15^2 + OL^2 \]

### Step 5: Calculate the Squares
Calculating the squares:
\[ 289 = 225 + OL^2 \]

### Step 6: Solve for OL^2
Rearranging the equation to solve for OL^2:
\[ OL^2 = 289 - 225 \]
\[ OL^2 = 64 \]

### Step 7: Find OL
Taking the square root of both sides:
\[ OL = \sqrt{64} = 8 \text{ cm} \]

### Conclusion
The distance of chord AB from the center of the circle (OL) is 8 cm.

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