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The problem involves a circle with center \(O\) and a radius of 15 cm. There are two chords, \(AB\) and \(XY\), which are parallel to each other, and you are given the lengths \(OM = 9\) cm and \(XN = 9\) cm. The objective is to find the lengths of \(AB\) and \(MN\).

Let's break down the information and approach the problem step by step:

### Given Information:
- Radius of the circle, \(OQ = 15\) cm.
- The distances from the center to the chords \(OM = XN = 9\) cm.
- Chords \(AB\) and \(XY\) are parallel.

We can solve the problem geometrically using the properties of circles and right triangles.

### (a) To find the length of \(AB\):
- \(OM\) is perpendicular to the chord \(AB\) because the perpendicular from the center of a circle to a chord bisects the chord.
- Let \(AM = MB = \frac{AB}{2}\).
- In the right triangle \(OAM\), by the Pythagorean theorem:
  \[
  OQ^2 = OM^2 + AM^2
  \]
  Substituting the values:
  \[
  15^2 = 9^2 + AM^2
  \]
  \[
  225 = 81 + AM^2
  \]
  \[
  AM^2 = 225 - 81 = 144
  \]
  \[
  AM = \sqrt{144} = 12 \, \text{cm}
  \]
  Therefore, \(AB = 2 \times AM = 2 \times 12 = 24\) cm.

### (b) To find the length of \(MN\):
- Since \(OM\) is perpendicular to \(AB\) and \(XN\) is perpendicular to \(XY\), and the chords are parallel, \(MN\) is the distance between these two parallel chords.
- Thus, \(MN = OX - OM\), where \(OX\) is the distance from the center to the second chord \(XY\).
- To find \(OX\), use the same Pythagorean theorem method for the triangle \(OXN\). We are given \(XN = 9\) cm:
  \[
  OQ^2 = OX^2 + XN^2
  \]
  Substituting the values:
  \[
  15^2 = OX^2 + 9^2
  \]
  \[
  225 = OX^2 + 81
  \]
  \[
  OX^2 = 225 - 81 = 144
  \]
  \[
  OX = \sqrt{144} = 12 \, \text{cm}
  \]
  Therefore, \(MN = OX - OM = 12 - 9 = 3\) cm.

### Final Answers:
- \(AB = 24\) cm
- \(MN = 3\) cm

Do you want any further details or explanations?

Here are 5 related questions to extend this problem:
1. What would the length of \(AB\) be if \(OM = 6\) cm instead of 9 cm?
2. How would the problem change if the radius of the circle were 20 cm instead of 15 cm?
3. If \(OM = 9\) cm and \(XN = 12\) cm, how would you find the new length of \(MN\)?
4. What happens to the length of a chord as it moves closer to the center of the circle?
5. How would you prove that the chords \(AB\) and \(XY\) are indeed parallel using geometric properties?

**Tip:** When dealing with chords in circles, remember that the perpendicular from the center to a chord always bisects the chord.

The figure below is a circle center O and a radius of 15 cm. The two chords, AB and XY, are parallel to each other. If OM = XN = 9 cm, find: (a) AB (b) MN

Solve a geometry problem involving a circle with a radius of 15 cm, where two parallel chords AB and XY are given. Learn how to find the lengths of chord AB and the distance MN using the Pythagorean theorem and circle properties. Suitable for students in Grades 9-12.

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