2<\/sup> = 9 \n\u21d2 AC = 3 cm \nHence CD passes through the centre of a smaller circle and equal to the diameter of the circle. \nCD = 2AC = 2 \u00d7 3 = 6 cm<\/p>\n <\/p>\n
Question 2. \nIf two equal chords of a circle intersect within the circle, prove that the segments of one chord are corresponding segments of the other chord. \nSolution: \nGiven: AB and CD are two equal chords of the circle which intersect each other at P. \n \nTo prove that: Segment ACB \u2245 Segment CBD. \nProof: We have given, \nchord AB = chord CD \n\u2234 \\(\\widetilde{\\mathrm{AB}} \\cong \\widetilde{\\mathrm{CD}}\\) \nWe know that, the segment made between congruent arcs and equal chords are congruent. \n\u2234 Segment ACB \u2245 Segment CBD.<\/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. \nSolution: \nGiven: AB and CD are two equal chords of a circle whose centre is O. Chord AB and CD intersect each other at E. \n \nTo prove that: \u22201 = \u22202 \nConstruction: Draw OM \u22a5 AB and ON \u22a5 CD, and join OE. \nProof: In \u2206OME and \u2206ONE \nOM = ON (Equal chords are equidistance from the centre) \n\u2220OME = \u2220ONE (Each 90\u00b0) \nOE = OE (Common) \nSo, by R-H-S congruency condition \n\u2206OME \u2245 \u2206ONE \n\u22201 = \u22202 (By CPCT)<\/p>\n
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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 (see Fig. 10.25) \n \nSolution: \nLet OM be perpendicular from O on line l. \nWe know that the perpendicular from the centre of a circle to a chord; bisect the chord. Since BC is a chord of the smaller circle and OM \u22a5 BC. \n\u2234 BM = CM ……(i) \nAgain, AD is a chord of the larger circle and OM \u22a5 AD. \n\u2234 AM = DM ……(ii) \nSubtracting equation (i) from (ii), we get \nAM – BM = DM – CM \nor, AB = CD.<\/p>\n
Question 5. \nThree girls Reshma, Salma, and Mandeep are playing a game by standing on & circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Resma. 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: \nLet Reshma, Salma and Mandip are standing on the point R, S, and M respectively where O is the centre of the circle. \n \nRS = SM = 6m \nOR = 5m \n\u2206RSM is Issosclus Triangle \n\u2234 SP \u22a5 RM and also P is mid-point of RM \nRP = PM \nar (\u2206ORS) = \\(\\frac {1}{2}\\) \u00d7 OS \u00d7 PR \n= \\(\\frac {1}{2}\\) \u00d7 5 \u00d7 x …….(i) \nDraw OT \u22a5 RS here OR = OS (Radii of circle) \nSo, \u2206ORS is Issosceles Triangle \n\u2234 RT = TS = 3 cm \nIn right \u2206OTS, \nOS = 5 cm, TS = 3 cm \nUsing Pythagoras theorem, \nOT2<\/sup> = OS2<\/sup> – TS2<\/sup> \n\u21d2 OT2<\/sup> = (5)2<\/sup> – (3)2<\/sup> \n\u21d2 OT2<\/sup> = 16 \n\u21d2 OT = \u221a16 = 4 cm \nAgain, ar (\u2206ORS) = \\(\\frac {1}{2}\\) \u00d7 RS \u00d7 OT \n= \\(\\frac {1}{2}\\) \u00d7 6 \u00d7 4 ……(ii) \nFrom equation (i) and (ii) \n\\(\\frac {1}{2}\\) \u00d7 5 \u00d7 x = \\(\\frac {1}{2}\\) \u00d7 6 \u00d7 4 \nor \\(\\frac {5x}{2}\\) = 12 \nx = \\(\\frac{12 \\times 2}{5}=\\frac{24}{5}\\) \nWe have to calculate \nRM = 2RP = 2 \u00d7 \\(\\frac{24}{5}\\) = \\(\\frac{48}{5}\\) \n\u2234 RM = 9.6 m \nDistance between Reshma and Mandip is equal to 9.6 m.<\/p>\n <\/p>\n
Question 6. \nA circular park of a radius of 20 m is situated in a colony. Three boys Ankur, Syed, and David are sitting at an equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone. \nSolution: \nLet three boys Ankur, Syed and David are sitting on the point A, B, and C respectively. \nO is the centre of the circle. \nAccording to question AB = BC = CA = x (Let) \n \nSo, \u2206ABC is an equilateral triangle \nOA = OB = OC = Radius of die circle = 20 meter \nJoin A, O and extand to D. \nAs \u2206ABC is an equilateral triangle \n\\(\\frac{\\mathrm{OA}}{\\mathrm{OD}}=\\frac{2}{1}\\) \nor \\(\\frac{\\mathrm{20}}{\\mathrm{OD}}=\\frac{2}{1}\\) (\u2235 OA = 20 m) \nOD = 10 meter \nIn right angle triangle ACD \n(AD)2<\/sup> = (CD)2<\/sup> + (AD)2<\/sup> \n\u21d2 (x)2 = \\(\\left(\\frac{x}{2}\\right)^{2}\\) + (30)2<\/sup> (\u2235 AD \u22a5 BC and D is the mid point) \n\u21d2 x2<\/sup> = \\(\\frac{x^{2}}{4}\\) + 900 \n\u21d2 x2<\/sup> – \\(\\frac{x^{2}}{4}\\) = 900 \n\u21d2 \\(\\frac{4 x^{2}-x^{2}}{4}\\) = 900 \n\u21d2 \\(\\frac{3 x^{2}}{4}\\) = 900 \n\u21d2 x2<\/sup> = \\(\\frac{900 \\times 4}{3}\\) \n\u21d2 x2<\/sup> = 1200 \n\u21d2 x = \u221a1200 \n\u21d2 x = 20\u221a3 meteres \nTherefore, the length of the string of each phone is 20\u221a3 meters.<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.4 Question 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n