Question 1.
\nTwo circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
\nSolution:
\nR = 5 cm, r = 3 cm. d = 4 cm.
\nlength of the common chord = ?
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-1.png\")
\nWhen measured Chord CD = 6 cm.
\nCircles with centre A and B, intersect at points C and D.
\nCD is common chord.
\nPerpendicular bisector of CD is AB.
\nAC = 5cm. BC = 3cm.
\n\u2234 Chord, CD= 2 \u00d7 OC = 2 \u00d7 3
\n\u2234 Chord, CD = 6 cm.<\/p>\n
Question 2.
\nIf two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
\nSolution:
\nData : Circle with centre ‘O’, AB and CD chords are equal and intersect at E.
\nTo Prove: Segment EAD = Segment ACB.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-2.png\")
\nConstruction: Draw OP\u22a5CD and OQ\u22a5AB.
\nJoin OE.
\nProof: AQ = QB (\u2235 Chord drawn perpendicular CP = PD ,to centre bisects chord.).
\nIn \u2206OPE and \u2206OQE,
\n\u2220OPE = \u2220OQE = 90\u00b0 (Construction)
\nOP = OQ
\nOE is common.
\n\u2234 \u2206OPE \u2245 \u2206OQE (RHS Theorem)
\nQE = PE ……….. (i)
\nBut QB = PD
\n\u2234 QB – QE = PD – PE
\nEB = ED ……….. (ii)
\nFrom (i) and (ii),
\nAE = CE.
\nCBD is minor segment formed by Chord CD.
\nADB is minor segment formed by Chord AB.
\n\u2234 Minor segment CBD = Minor segment ADB
\nMajor segment CAD = Major segment ACB.<\/p>\n
Question 3.
\nIf two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-3.png\")
\nSolution:
\nData: O is the centre of Circle. Chord AB = Chord CD. These chords intersects at E. OE is joined.
\nTo Prove: \u2220OEP = \u2220OEQ Construction:
\nDraw OP\u22a5CD and OQ\u22a5AB.
\nProof: In \u2206OPE and \u2206OQE,
\n\u2220OPE = \u2220OQE = 90\u00b0 (construction)
\nOE is common.
\nOP = OQ (Equal chords are equidistant)
\n\u2234 \u2206OPE \u2245 \u2206OEQ (RHS postulate)
\n\u2234 \u2220OEP = \u2220OEQ.<\/p>\n
Question 4.
\nIf a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-4.png\")
\nSolution:
\nData: A line intersects two concentric circles with centre O at A, B, C, and D.
\nTo Prove: AB = CD.
\nConstruction: Draw OM\u22a5AD.
\nProof: AD is the chord of a bigger circle.
\nOM\u22a5AD
\n\u2234 AM = MD ………(i) (\u2235 Perpendicular bisects the chord from centre)
\nSimilarly, BC is the chord of the smaller circle.
\nOM\u22a5BC.
\n\u2234 BM = MC ………. (ii)
\nSubtracting (ii) from (i),
\nAM – BM = MD – MC
\n\u2234 AB = CD.<\/p>\n
Question 5. Question 6. <\/p>\n","protected":false},"excerpt":{"rendered":" Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.4 Question 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. Solution: R = 5 cm, r = 3 cm. d = 4 cm. length …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2445"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2445"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2445\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2445"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2445"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2445"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nThree girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
\nSolution:
\nData: Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 cm drawn in a park. Distance between Reshma and Salma and between Salma and Mandip is 6 cm. each.
\nTo Prove: Distance between Reshma and Mandip = ?, RM = ?
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-5.png\")
\nConstruction: Draw OX\u22a5RS and OY\u22a5SM. OR, OS and OM are joined.
\nProof: OR = OS = OM = 5 m.
\nRX = XS = 3 m.
\nNow, in \u22a5\u2206ORX,
\nOR2<\/sup> = OX2<\/sup> + RX2<\/sup>
\n(5)2<\/sup> = (OX)2<\/sup> + (3)2<\/sup>
\n25 = OX2<\/sup> + 9
\n25 – 9 = OX2<\/sup>
\n16 = OX2<\/sup>
\nOX2<\/sup> = 16
\n\u2234 OX = 4 WI …………… (i)
\nIn quadrilateral ORSM,
\nOR = OM and SR = SM
\n\u2234 ORSM is a kite.
\nDiagonals of a kite bisect at the right angle.
\n\u2220RAS = 90\u00b0
\n\u2234 RA = AM.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-6.png\")
\nNow, RM = 2RC = 2 \u00d7 4.8 = 9.8 m
\n\u2234 RM = 9.8 m.
\n\u2234 Distance between Reshma and Mandip is 9.8 m.<\/p>\n
\nA circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-7.png\")
\nSolution:
\nData: A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed, and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other.
\nTo Prove: Length of wire of telephone =?
\nProof: In this figure, AS = SD = DA
\n\u2234 ASD is an equilateral triangle.
\n\u2234 OA = 20 m. (\u2235 radius)
\nMedian passed through centre of circle.
\n\u2234 AB is the median.
\nAD \u22a5 SD
\nMedian of an equilateral triangle divides in the ratio 2 : 1 in the centre?
\n\u2234 AO : OB = 2 : 1
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-8.png\")
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.4-9.png\")
<\/p>\nKSEEB Solutions for Class 9 Maths<\/a><\/h5>\n