Ex 10.6, 2 (Optional)
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, find the radius of the circle
Given: Two Chords AB & CD of a circle
where AB = 5 cm & CD = 11 cm
and AB ∥ CD
Also, Distance between AB & CD = 6 cm
Construction:
Draw Perpendiculars from O to AB and CD.
Also, Join OB and OD
To find: Radius of the circle
Finding:
We know that
perpendicular drawn from the center of a circle bisects the chord.
Here, distance between AB & CD will be perpendicular distance and the perpendicular PQ will pass through center O.
So, PQ = OP + OQ
Let O bisect AB at P and CD at Q.
∴ AP = BP = 𝐴𝐵/2 = 5/2 cm
& CQ = DQ = 𝐶𝐷/2 = 11/2 cm
Let r be the radius of the circle
Let distance OP = x cm
So, distance OQ = (6 − x) cm.
Now,
In ∆ OPB,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
OB2 = OP2 + PB2
r2 = x2 + (5/2)^2
…(1)
In ∆ OQD,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
OD2 = OQ2 + QD2
r2 = (6 − x)2 + (11/2)^2
…(2)
From (1) and (2)
x2 + (5/2)^2=(6−𝑥)^2+(11/2)^2
x2 + 25/4 = 36 + x2 − 12x + 121/4
x2 − x2 + 12x = 36 + 121/4 − 25/4
12x = 36 + 96/4
12x = 36 + 24
12x = 60
x = 60/12
x = 5 cm
Putting x = 5cm in (1),
r2 = x2 + (5/2)^2
r2 = (5)2 + (5/2)^2
r2 = 25 + 25/4
r2 = (100 +25)/4
r2 = 125/4
r = ± √(125/4)
r = ± √125/√4
r = ± √(25 × 5)/√4
r = ± (5√5)/2
Since Radius cannot be negative
∴ Radius = (𝟓√𝟓)/𝟐 cm

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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