Given a circle with center O and radius r = 13 cm. A chord AB is located at a distance d = 12 cm from the center. A perpendicular is drawn from O to chord AB, intersecting at point M. Thus, OM = 12 cm and AM = MB = AB/2. In the right triangle OAM, applying the Pythagorean theorem: \[OA^2 = OM^2 + AM^2\] \[13^2 = 12^2 + AM^2\] \[169 = 144 + AM^2\] \[AM^2 = 169 - 144 = 25\] \[AM = 5 cm\] The length of chord AB is 2 × AM = 2 × 5 = 10 cm.