AD is the diameter of a circle of radius 17cm and AB is of length 30 cm. Find the distance of AB 
from the centre.

To find the distance of line segment \( AB \) from the center of the circle, we can use the following steps:

1. **Identify key components**:
   - The radius of the circle \( r = 17 \) cm.
   - The diameter \( AD = 2 \times r = 34 \) cm.
   - The length of segment \( AB = 30 \) cm.

2. **Use the right triangle relationship**:
   - Let \( O \) be the center of the circle.
   - The distance from \( O \) to \( AB \) is the perpendicular distance we want to find, denoted as \( d \).
   - The midpoint \( M \) of \( AB \) is where the perpendicular from \( O \) meets \( AB \).

3. **Apply the Pythagorean theorem**:
   - The distance \( OA \) from the center \( O \) to point \( A \) is equal to the radius, which is \( 17 \) cm.
   - The length \( AM \) (half of \( AB \)) is \( \frac{30}{2} = 15 \) cm.

Using the Pythagorean theorem in triangle \( OAM \):
\[
OA^2 = OM^2 + AM^2
\]
\[
17^2 = d^2 + 15^2
\]
\[
289 = d^2 + 225
\]
\[
d^2 = 289 - 225 = 64
\]
\[
d = \sqrt{64} = 8 \text{ cm}
\]

Thus, the distance of \( AB \) from the center of the circle is **8 cm**.

### Do you want details or have any questions?

Here are some related questions you might find interesting:
1. What would happen to the distance if the length of \( AB \) were greater than the diameter?
2. How do you find the area of the circle given the radius?
3. Can you explain how to find the angle subtended by chord \( AB \) at the center?
4. What are other methods to find the distance from a point to a line?
5. How would the situation change if \( AB \) were a secant line instead? 

**Tip**: Always remember to check whether a line segment lies entirely within the circle or intersects it to apply the right geometric principles.

AD is the diameter of a circle of radius 17cm and AB is of length 30 cm. Find the distance of AB from the centre.

Explore how to calculate the distance from a line segment AB (30 cm) to the center of a circle with a radius of 17 cm. This detailed solution utilizes the Pythagorean theorem and explores essential geometry concepts. Suitable for students in Grades 7-9.

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