In a circle of radius 10 cm, two parallel chords have lengths 12 cm and 16 cm respectively. What is the distance between these two chords inside the circle (considering both possible positions of the chords)?

Introduction / Context:
This problem involves two parallel chords in a circle and asks for the distance between them. When chords are parallel, their distances from the centre of the circle determine how far apart they are. Because chords can lie on the same side of the centre or on opposite sides of the centre, two possible distances often arise. This question tests understanding of the relationship between chord length, radius, and perpendicular distance from the centre of the circle to the chord.

Given Data / Assumptions:
- Radius of the circle r = 10 cm. - Length of one chord is 12 cm. - Length of the other chord is 16 cm. - Chords are parallel and lie inside the same circle. - Perpendicular distance from the centre to a chord is used to find how far chords are from the centre and from each other.

Concept / Approach:
For a circle of radius r, if a chord of length L is at a perpendicular distance d from the centre, we can use the right triangle formed by the radius, half of the chord, and the distance d. The relation is r^2 = d^2 + (L / 2)^2. From this we get d = √(r^2 - (L / 2)^2). Once we find the distances of both chords from the centre, the distance between the chords is either the difference of these distances (if chords are on the same side of the centre) or the sum of these distances (if chords lie on opposite sides of the centre).

Step-by-Step Solution:
Step 1: For the chord of length 12 cm, half the chord is L1 / 2 = 6 cm. Step 2: Use r^2 = d1^2 + (L1 / 2)^2 with r = 10. So 10^2 = d1^2 + 6^2 which gives 100 = d1^2 + 36. Step 3: Therefore d1^2 = 100 - 36 = 64, so d1 = √64 = 8 cm. Step 4: For the chord of length 16 cm, half the chord is L2 / 2 = 8 cm. Step 5: Use r^2 = d2^2 + (L2 / 2)^2 again. So 10^2 = d2^2 + 8^2 which gives 100 = d2^2 + 64. Step 6: Therefore d2^2 = 100 - 64 = 36, so d2 = √36 = 6 cm. Step 7: If both chords lie on the same side of the centre, the distance between them is the difference of their distances from the centre: |d1 - d2| = |8 - 6| = 2 cm. Step 8: If the chords lie on opposite sides of the centre, the distance between them is the sum of their distances from the centre: d1 + d2 = 8 + 6 = 14 cm. Step 9: Hence there are two possible distances between the two parallel chords: 2 cm or 14 cm.

Verification / Alternative check:
You can sketch a circle with centre O, place one chord closer to the centre and the other further out, and visually check that the computed distances 8 cm and 6 cm from the centre are reasonable for chords of lengths 12 cm and 16 cm. Since both distances are less than the radius 10 cm, both chords indeed lie inside the circle. Considering them on the same side or on opposite sides of the centre naturally gives the two possible distances, confirming our reasoning.

Why Other Options Are Wrong:
Option 1 cm or 7 cm does not match the correct distances derived from the radius and chord relations. Option 3 cm or 21 cm and Option 4 cm or 28 cm both give values that are inconsistent with the maximum possible separation within a circle of radius 10 cm. In particular, a distance of 21 cm or 28 cm would exceed the diameter, which is impossible for two chords inside the circle.

Common Pitfalls:
A common mistake is to forget to divide the chord length by 2 before applying the Pythagoras relation with the radius and the perpendicular distance. Another error is to think that there is only one possible distance between parallel chords, ignoring the possibility that chords can be on opposite sides of the centre. Learners may also incorrectly add or subtract distances without considering the geometry of where each chord lies relative to the centre.

Final Answer:
The distance between the two parallel chords can be either 2 cm or 14 cm depending on their positions relative to the centre of the circle.