Math, asked by Rohitsirohi4300, 10 months ago

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

Answers

Answered by amirgraveiens
23

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,

OD^2=OQ^2+QD^2

r^2 = x^2 + 24^2    [By Pythagoras theorem]

r^2 = x^2 + 576                       (1)

In right ΔOPB,  

OB^2=OP^2+PB^2             [given distance between the chords, QP = 42 cm ]

r^2 = (x + 42)^2 + 18^2       [By Pythagoras theorem]

r^2 = x^2 + 84x+1764 + 324

r^2 = x^2 + 84x+2088           (2)

From (1) and (2) we get  

x^2 + 576 = x^2 + 84x + 2088

⇒ 576 = 84x + 2088

⇒ 84x = 576 - 2088

⇒ 84 x = -1512

x = -\frac{1512}{84}

⇒ x = - 18

Put x = - 18 in (1), we get

r^2 = (-18)^2 + 576

r^2 = 324 + 576

r^2 = 900

r = \sqrt{900}

r = 30 cm

Therefore radius = 30 cm.

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