Ex 10.6, 3 (Optional) - The lengths of two parallel chords of a circle

Ex 10.6, 3 (Optional) - Chapter 10 Class 9 Circles - Part 2
Ex 10.6, 3 (Optional) - Chapter 10 Class 9 Circles - Part 3
Ex 10.6, 3 (Optional) - Chapter 10 Class 9 Circles - Part 4

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Question 3 The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the center? Given: Two chords AB & CD of a circle where AB = 6 cm & CD = 8 cm and AB ∥ CD Also, Distance between O and AB = 4 cm Construction: Draw Perpendiculars from O to AB and CD. Also, Join OB and OD To find: Distance of O from CD i.e. OQ 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 = 6/2 = 3 cm & CQ = DQ = 𝐶𝐷/2 = 8/2 = 4 cm Let r be the radius of the circle Let distance OQ = x cm In ∆ OPB, (Hypotenuse)2 = (Perpendicular)2 + (Base)2 OB2 = OP2 + PB2 r2 = 42 + 32 …(1) In ∆ OQD, (Hypotenuse)2 = (Perpendicular)2 + (Base)2 OD2 = OQ2 + QD2 r2 = x2 + 42…(2) From (1) and (2) 42 + 32 = x2 + 42 32 = x2 x2 = 32 x = ± 3 cm Since distance cannot be negative x = 3 cm ∴ The distance of chord CD from the center is 3 cm.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo