In a circle of radius 10 cm the chord AB is 12 cm long and lies 8 cm from the centre. A second chord CD of the same circle is also 12 cm long. How far is CD from the centre?

Chords of equal length in one circle are equidistant from the centre, so CD is 8 cm from the centre as well — where it sits on the circle makes no difference. The arithmetic confirms it: the perpendicular from the centre bisects the chord, so the distance is √$(10^2-6^2)=$ √64 = 8 cm. The trap 6 cm is the half-chord rather than the distance, and the last option overlooks the equal-chords theorem altogether.