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Question

AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.

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Solution

let O be the centre of the given circle and let its radius be r cm. Draw OPAB and OQCD. Since OPAB,OQCD and AB||CD, therefore, points P, O and Q are collinear. So, PQ = 17 cm.
Let OP = x cm. Then, OQ = (17 – x) cm.
Join OA and OC. Then, OA = OC = r.
Since the perpendicular from the centre to a chord of teh circle bisets the chord.
AP=PB=5 cm andCQ=QD=12 cm
In right triangle OAP and OCQ , we have
OA2=OP2+AP2 and OC2=OQ2+CQ2
r2=x2+52 ....(i)
and, r2=(17x2)+122 ....(ii)
x2+52=(17x)2+122
[On equating the valyes of r2]
x2+25=28934x+x2+144
34x=408x=12 cm
Putting x = 12 cm in equation (i), we get
r2=122+52=169
r=13 cm
Hence, the radius of the circle is 13 cm.

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