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.