Question

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.

Answer 1

(i) OP ⊥ CD
∴ OP bisects CD.   ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
⇒ CP = `"CD"/2`

In right ΔOPC,
OC² = OP² + CP²
⇒ CP² = OC² - OP² 
⇒ 15² - 9² = 144
∴ CP = 12 cm
∴ CD = 12 x 2 = 24 cm

(ii) Join BD,
∴ BP = OB - OP = 15 - 9 = 6 cm.
In right ΔBPD,
BD² = BP² + PD² 
       = 6² + 12² = 180
In ΔADB,
∠ADB = 90°     ...( Angle in a semi-circle is a right angle )
∴ AB² = AD² + BD²
⇒ AD² = AB² - BD²
= 30² - 180 = 720
∴ AD = `sqrt(720)` = 26.83 cm

(iii) Also, BC = BD = `sqrt(180)` = 13.42 cm.