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Question

Two chords AB, CD of length 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm. find the radius of the circle.Assume that the chords are on the same side of the center.

A

5.02cm

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B

6.02cm

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C

7.02cm

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D

8.02cm

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Solution

The correct option is B 6.02cm Given: Chord AB=5 cm, chord CD = 11 cm and AB ∥ CD. Perpendicular distance ML between AB and CD = 3 cm. To find: Radius of the circle. Construction:Join OB, OD and draw perpendicular bisectors OL of AB and OM of CD. ln right angled triangle OMD, OD2 = MD2 + OM2 [By PythagorasTheorem] Let OM = x cm r2 = (112)2 + x2 ..........(i) And In right triangle OLB, OB2 = BL2 + OL2 [By Pythagorastheorem] r2 = (52)2 + 3+x2 .........(ii) From (i) and (ii), we get r2 = (112)2 +x2 - (52)2 + 3+x2 1214 + x2 - 254 + 9 + x2 + 6x x2 - x2 + 6x = 1214 - 254 - 9 6x = 121−25−364 - 604 - 15 x = 156 cm - 52 cm From (1), r2 = (112)2 + (52)2 - 121+254 - 1464 r = (√146)2 cm

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