Understanding the Circle Geometry Problem

The question asks us to find the distance between two parallel chords, XY and PQ, in a circle. We are given the radius of the circle, the lengths of the two chords, and that the chords are located on opposite sides of the circle's centre.

Let's break down the given information:

• Radius of the circle (R) = 10 cm.
• Length of chord XY = 12 cm.
• Length of chord PQ = 16 cm.
• The chords are parallel and on opposite sides of the centre.

We need to find the distance between these two parallel chords.

Key Concepts for Solving Circle Problems

To solve this problem, we need to recall some important properties of circles and chords:

• A perpendicular drawn from the centre of a circle to a chord bisects the chord.
• The distance of a chord from the centre is the length of the perpendicular segment from the centre to the chord.
• We can use the Pythagorean theorem in the right triangle formed by the radius, half of the chord length, and the distance of the chord from the centre.

Step-by-Step Solution to Find Distance Between Chords

Let O be the centre of the circle. Let M be the midpoint of chord XY, and N be the midpoint of chord PQ. Since the perpendicular from the centre bisects the chord, OM is perpendicular to XY and ON is perpendicular to PQ.

The distance of chord XY from the centre is OM. The distance of chord PQ from the centre is ON.

Since the chords are parallel and on opposite sides of the centre, the distance between the chords XY and PQ is the sum of their distances from the centre, i.e., Distance = OM + ON.

Calculating Distance of Chord XY from Centre

Chord XY has length 12 cm. The perpendicular from the centre bisects it, so MY = XY/2 = 12/2 = 6 cm.

In the right-angled triangle OMY, we have:

• Hypotenuse = Radius (OY) = 10 cm
• One leg = Half chord length (MY) = 6 cm
• Other leg = Distance from centre (OM)

Using the Pythagorean theorem: $\text{OY}^2 = \text{OM}^2 + \text{MY}^2$

Substituting the values:

$10^2 = \text{OM}^2 + 6^2$

$100 = \text{OM}^2 + 36$

$\text{OM}^2 = 100 - 36$

$\text{OM}^2 = 64$

$\text{OM} = \sqrt{64}$

$\text{OM} = 8 \text{ cm}$

So, the distance of chord XY from the centre is 8 cm.

Calculating Distance of Chord PQ from Centre

Chord PQ has length 16 cm. The perpendicular from the centre bisects it, so NQ = PQ/2 = 16/2 = 8 cm.

In the right-angled triangle ONQ, we have:

• Hypotenuse = Radius (OQ) = 10 cm
• One leg = Half chord length (NQ) = 8 cm
• Other leg = Distance from centre (ON)

Using the Pythagorean theorem: $\text{OQ}^2 = \text{ON}^2 + \text{NQ}^2$

Substituting the values:

$10^2 = \text{ON}^2 + 8^2$

$100 = \text{ON}^2 + 64$

$\text{ON}^2 = 100 - 64$

$\text{ON}^2 = 36$

$\text{ON} = \sqrt{36}$

$\text{ON} = 6 \text{ cm}$

So, the distance of chord PQ from the centre is 6 cm.

Finding the Total Distance Between the Chords

Since the two parallel chords XY and PQ are on opposite sides of the centre, the total distance between them is the sum of their distances from the centre.

Distance between chords = Distance of XY from centre + Distance of PQ from centre

Distance between chords = OM + ON

Distance between chords = 8 cm + 6 cm

Distance between chords = 14 cm

Conclusion

The distance between the two parallel chords XY and PQ, which are on opposite sides of the centre in a circle with a radius of 10 cm and lengths 12 cm and 16 cm respectively, is 14 cm.

Revision Table: Circle Chords and Distance

Additional Information: Circle Geometry Concepts

Understanding the properties of circles and chords is fundamental in geometry. Here are a few additional points:

• The longest chord in a circle is its diameter, which passes through the centre.
• Equal chords are equidistant from the centre. Conversely, chords equidistant from the centre are equal in length.
• A chord closer to the centre is longer than a chord farther from the centre.
• Parallel chords cut off equal arcs.

Problems involving chords and their distances from the centre often require constructing right triangles and applying the Pythagorean theorem, as demonstrated in this solution.