A chord CD of a circle, whose center is O, is bisected at P by a diameter AB.

Given OA = OB = 15 cm and OP = 9 cm. Calculate the lengths of :

(i) CD

(ii) AD

(iii) CB.

We know that,

A straight line drawn from the center of a circle to bisect a chord is at right angles to the chord.

∴ OP ⊥ CD.

Join OC, AD and BC.

Given,

OA = OB = 15 cm

∴ Radius of circle = 15 cm.

∴ OC = 15 cm.

(i) In right-angled triangle OCP,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OP² + CP²

⇒ 15² = 9² + CP²

⇒ 225 = 81 + CP²

⇒ CP² = 225 - 81

⇒ CP² = 144

⇒ CP = $\sqrt{144}$ = 12 cm.

Since, chord CD is bisected at point P.

∴ CD = 2 × CP = 2 × 12 = 24 cm.

Hence, CD = 24 cm.

(ii) Since, chord CD is bisected at point P.

∴ PD = CP = 12 cm.

From figure,

⇒ AP = OA + OP = 15 + 9 = 24 cm.

In right-angled triangle APD,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AD² = AP² + PD²

⇒ AD² = 24² + 12²

⇒ AD² = 576 + 144

⇒ AD² = 720

⇒ AD = $\sqrt{720}$ = 26.83 cm.

Hence, AD = 26.83 cm.

(iii) From figure,

AB is the diameter of the circle.

∴ AB = 2 × OA = 2 × 15 = 30 cm.

⇒ PB = AB - AP = 30 - 24 = 6 cm.

In right-angled triangle CPB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ CB² = CP² + PB²

⇒ CB² = 12² + 6²

⇒ CB² = 144 + 36

⇒ CB² = 180

⇒ CB = $\sqrt{180}$ = 13.42 cm.

Hence, CB = 13.42 cm.