Given a circle with radius 5cm and chord AB, measuring 6cm, find the perpendicular distance of "d" between AB and the center of the circle. Ans: Remember that: The perpendicular "d" from the center of the circle to the chord, bisects the chord. Let the centre of the circle be O, and the point where d cuts AB be called C. Now: d, CB (half the chord) and OB (radius) form a right triangle with OB as the hypotenuse. OB = 5 CB = half of AB = 3 Applying Pythagoras theorem d^2 = OB^2 - CB^2 = 25 - 9 = 16 Therefore d = 4 cm. Hope you got it :)