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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.
Solution
(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.
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