In a circle of radius 10 cm one chord measures 16 cm and another measures 12 cm. Which chord lies nearer the centre, and by how much?

The perpendicular from the centre bisects a chord, so for the 16 cm chord the distance is √$(10^2-8^2)=$ √36 = 6 cm, and for the 12 cm chord it is √$(10^2-6^2)=$ √64 = 8 cm. So the 16 cm chord is nearer, by 8 − 6 = 2 cm. The general rule matches: in one circle the LONGER chord is always the nearer to the centre, and only a diameter passes through it. Choosing the 12 cm chord inverts that rule, and 4 cm comes from differencing the half-chords 8 and 6 against the chords themselves rather than the two distances.