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A Chord Cd of a Circle Whose Centre is O, is Bisected at P by a Diameter Ab. Given Oa = Ob = 15 Cm and Op = 9 Cm. Calculate the Length Of: (I) Cd (Ii) Ad (Iii) Cb - ICSE Class 10 - Mathematics

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Question

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

Given OA = OB = 15 cm and OP = 9 cm. calculate the length of:
(i) CD (ii) AD (iii) CB

 

Solution

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 =(15)2 - (9)2 = 144
∴ CP =12 cm
∴ CD =12×2 = 24 cm

(ii) Join BD
∴ BP = OB - OP = 15 - 9 = 6 cm
In right ΔBPD,
BD2 = BP2 + PD2

= (6)2 + ( 12)2 = 180

In ΔADB, ∠ADB = 90°

(Angle in a semicircle is a right angle)

∴  AB2 = AD+ BD2

⇒  AD2 = AB2 + BD2 = (30)2 - 180 =720

∴ `"AD" =sqrt 720 = 26.83 "cm"`

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

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Solution A Chord Cd of a Circle Whose Centre is O, is Bisected at P by a Diameter Ab. Given Oa = Ob = 15 Cm and Op = 9 Cm. Calculate the Length Of: (I) Cd (Ii) Ad (Iii) Cb Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).
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