Two circles of radii 13 cm and 15 cm intersect at two points. The distance between their centres is 14 cm. What is the length of their common chord?

The line of centres is the perpendicular bisector of the common chord. Let the distances from the two centres to the chord be $d$₁ and $d$₂ with $d$₁ + $d$₂ = 14. Then $13^2-d$₁² = $15^2-d$₂², so $d$₂² − $d$₁² = 56, and since $d$₂ + $d$₁ = 14 we get $d$₂ − $d$₁ = 4. Solving, $d$₁ = 5 and $d$₂ = 9. Half the chord is √$(169-25)=12$, so the chord is 24 cm. (12 cm is only the half-chord.)