AB and CD are two equal chords of a circle with centre O. The chord AB is at a distance of 12 cm from O, and CD \(=18\) cm. What is the radius of the circle?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15 cm
Equal chords of a circle are equidistant from the centre, and the perpendicular from the centre to a chord bisects the chord. So the radius is the hypotenuse of a right triangle with legs “half the chord” and “distance from O”.
AB \(=\) CD \(=18\) cm, so half of AB \(=9\) cm
Distance of AB from O \(=12\) cm (and CD is also 12 cm from O)
Radius \(=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\) cm
