What is the distance (in cm) between two parallel chords of lengths 48 cm and 14 cm in a circle of radius 25 cm?

A perpendicular drawn from the centre to a chord bisects the chord, creating two right triangles. Using the Pythagorean theorem, the distance from the centre to the 48 cm chord is the square root of (25^2 - 24^2), which is the square root of 49 or 7 cm. The distance from the centre to the 14 cm chord is the square root of (25^2 - 7^2), which is the square root of 576 or 24 cm. Since the chords must be on opposite sides of the centre for the answer choices to match, the total distance between them is 24 + 7 = 31 cm.

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