What is the distance in cm between two parallel chords of length 32 cm and 24…

For a chord of a circle, the perpendicular drawn from the center to the chord bisects it. If a chord of half-length a lies at perpendicular distance d from the center of a circle of radius r, then r2 = d2 + a2, so d = √(r2 − a2). For two parallel chords, the distance between them equals the SUM of their perpendicular distances from the center when the chords lie on opposite sides of the center, or the DIFFERENCE when they lie on the same side.

1. Chord 1 has length 32 cm, so its half-length is 32/2 = 16 cm.

2. Using r2 = d2 + a2 with r = 20 cm, the perpendicular distance of this chord from the center is d1 = √(202 − 162) = √(400 − 256) = √144 = 12 cm.

3. Chord 2 has length 24 cm, so its half-length is 24/2 = 12 cm.

4. Using the same relation, the perpendicular distance of this chord from the center is d2 = √(202 − 122) = √(400 − 144) = √256 = 16 cm.

5. If the two chords lie on opposite sides of the center, the distance between them is d1 + d2 = 12 + 16 = 28 cm.

6. If the two chords lie on the same side of the center, the distance between them is d2 − d1 = 16 − 12 = 4 cm.

Both distances check out: 122 + 162 = 144 + 256 = 400 = 202, confirming they are consistent with the given radius. Since the problem does not specify which side of the center the chords lie on, both configurations are valid, giving the distance as 4 cm or 28 cm.