Ex 10.6, 2 (Optional) - Two chords AB and CD of lengths 5 cm and 11 cm

Ex 10.6, 2 (Optional) - Chapter 10 Class 9 Circles - Part 2
Ex 10.6, 2 (Optional) - Chapter 10 Class 9 Circles - Part 3
Ex 10.6, 2 (Optional) - Chapter 10 Class 9 Circles - Part 4
Ex 10.6, 2 (Optional) - Chapter 10 Class 9 Circles - Part 5 Ex 10.6, 2 (Optional) - Chapter 10 Class 9 Circles - Part 6

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Question 2 Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, find the radius of the circle Given: Two Chords AB & CD of a circle where AB = 5 cm & CD = 11 cm and AB βˆ₯ CD Also, Distance between AB & CD = 6 cm Construction: Draw Perpendiculars from O to AB and CD. Also, Join OB and OD To find: Radius of the circle 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 = 𝐴𝐡/2 = 5/2 cm & CQ = DQ = 𝐢𝐷/2 = 11/2 cm Let r be the radius of the circle Let distance OP = x cm So, distance OQ = (6 βˆ’ x) cm. Now, In βˆ† OPB, (Hypotenuse)2 = (Perpendicular)2 + (Base)2 OB2 = OP2 + PB2 r2 = x2 + (5/2)^2 …(1) In βˆ† OQD, (Hypotenuse)2 = (Perpendicular)2 + (Base)2 OD2 = OQ2 + QD2 r2 = (6 βˆ’ x)2 + (11/2)^2 …(2) From (1) and (2) x2 + (5/2)^2=(6βˆ’π‘₯)^2+(11/2)^2 x2 + 25/4 = 36 + x2 βˆ’ 12x + 121/4 x2 βˆ’ x2 + 12x = 36 + 121/4 βˆ’ 25/4 12x = 36 + 96/4 12x = 36 + 24 12x = 60 x = 60/12 x = 5 cm Putting x = 5cm in (1), r2 = x2 + (5/2)^2 r2 = (5)2 + (5/2)^2 r2 = 25 + 25/4 r2 = (100 +25)/4 r2 = 125/4 r = Β± √(125/4) r = Β± √125/√4 r = Β± √(25 Γ— 5)/√4 r = Β± (5√5)/2 Since Radius cannot be negative ∴ Radius = (πŸ“βˆšπŸ“)/𝟐 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