$O$ is the centre of the circle with radius $5\ cm$ . Chords $AB$ and $CD$ are parallel. $AB = 6\ cm$ and $CD = 8\ cm$. If $PQ$ is distance between $AB$ and $CD$, then $PQ$ is equal to

Using the property that a perpendicular from the center bisects a chord, the distance from center O to AB is the square root of (5 squared minus 3 squared), which equals 4 cm. Similarly, the distance from center O to CD is the square root of (5 squared minus 4 squared), which equals 3 cm. Since the two parallel chords are on opposite sides of the center, the total distance between them is the sum of these two distances. Adding 4 cm and 3 cm gives the distance PQ equal to 7 cm.

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