In a circle of radius 25 cm, one chord lies 7 cm from the centre and a second chord lies 20 cm from the centre. By how much is the first chord longer than the second?

The perpendicular from the centre bisects a chord. Half the first chord = √$(25^2-7^2)=$ √576 = 24, so that chord is 48 cm. Half the second = √$(25^2-20^2)=$ √225 = 15, so that chord is 30 cm. The difference is 48 − 30 = 18 cm. Stopping at the half-chords gives 24 − 15 = 9 cm, which is only half the required difference.