Chords AB and CD of a circle are parallel to each other and lie on opp sides of the centre of the circle.if AB=36cm,CD=48cm and the distance between the chords is 42cm find the radius
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Radius = 30 cm
Step-by-step explanation:
Given:
Here, chords AB = 6 cm, CD = 12 cm and AB║CD
Draw OP⊥ AB. Let it intersect CD at Q and AB at P
∴ AP = PB = 18 cm and CQ = DQ = 24 cm [Since perpendicular draw from the centre of the chord bisects the chord]
Let OD = OB = r
In right Δ OQD,
[By Pythagoras theorem]
(1)
In right ΔOPB,
[given distance between the chords, QP = 42 cm ]
[By Pythagoras theorem]
(2)
From (1) and (2) we get
⇒ 576 = 84x + 2088
⇒ 84x = 576 - 2088
⇒ 84 x = -1512
⇒
⇒ x = - 18
Put x = - 18 in (1), we get
r = 30 cm
Therefore radius = 30 cm.
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