{"id":25849,"date":"2021-06-24T12:56:02","date_gmt":"2021-06-24T07:26:02","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25849"},"modified":"2022-03-02T10:36:05","modified_gmt":"2022-03-02T05:06:05","slug":"ncert-solutions-for-class-9-maths-chapter-10-ex-10-6","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-10-ex-10-6\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 10 Circles Ex 10.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6<\/h2>\n

Question 1.
\nProve that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
\nSolution:
\nGiven: Two circles of centre O and O’ intersect each other at A and B.
\nTo prove that: \u2220OAO’ = \u2220OBO’
\nConstruction: Join OO’.
\nProof: In \u2206AOO’ and \u2206BOO’
\nAO = BO (Radii of the same circle)
\nAO’ = BO’ (Radii of the same circle)
\nOO’ = OO’ (Common)
\nSo, by S-S-S congruency condition
\n\u2206AOO’ \u2245 \u2206BOO’
\n\u2234 \u2220OAO’ = \u2220OBO’ (By CPCT)<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nTwo 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 centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
\nSolution:
\nLet O be the centre of the given circle and let its radius be r cm.
\nDraw OM \u22a5 AB and ON \u22a5 CD.
\nSince OM \u22a5 AB, ON \u22a5 CD, and AB || CD.
\nTherefore point M, O and N are collinear. So MN = 6 cm.
\nLet OM = x cm, Then ON = 6 – x cm
\nJoin OA and OC, then OA = OC = r.
\nSince the perpendicular from the centre to a chord of the circle bisect the chord.
\n\"NCERT
\n\u2234 AM = MB = \\(\\frac {5}{2}\\) cm
\nand CN = ND = \\(\\frac {11}{2}\\) cm.
\nIn right triangle OAM,
\nOA2<\/sup> = OM2<\/sup> + AM2<\/sup>
\nr2<\/sup> = x2<\/sup> + \\(\\left(\\frac{5}{2}\\right)^{2}\\) ……(i)
\nNow, in right triangle OCN
\nOC2<\/sup> = ON2<\/sup> + CN2<\/sup>
\nr2<\/sup> = (6 – x)2<\/sup> + \\(\\left(\\frac{11}{2}\\right)^{2}\\) …….(ii)
\nFrom equation (i) and (ii)
\n\\(x^{2}+\\left(\\frac{5}{2}\\right)^{2}=(6-x)^{2}+\\left(\\frac{11}{2}\\right)^{2}\\)
\n\"NCERT
\nHence, the radius of the circle is \\(\\frac{5 \\sqrt{5}}{2}\\) cm.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nThe length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance of 4 cm from the centre, what is the distance of the other chord from the centre.
\nSolution:
\nLet AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.
\nLet the radius of the circle be r cm.
\nDraw OP \u22a5 AB and OQ \u22a5 CD.
\nThe length of OP is 4 cm.
\n\"NCERT
\nIn right triangle OAP,
\nr2<\/sup> = OP2<\/sup> + AP2<\/sup>
\n\u21d2 r2<\/sup> = (4)2<\/sup> + (3)2<\/sup> [\u2235 AP = \\(\\frac {1}{2}\\) AB = 3 cm]
\n\u21d2 r2<\/sup> = 16 + 9 = 25
\n\u21d2 r = 5 cm ……(i)
\nNow, in right triangle OQC,
\nr2<\/sup> = (OQ)2<\/sup> + (CQ)2<\/sup>
\n\u21d2 (5)2 <\/sup>= (OQ)2<\/sup> + (4)2<\/sup>
\n[\u2235 r = 5 cm prove above and CQ = \\(\\frac {1}{2}\\) CD = 4 cm]
\n\u21d2 25 = (OQ)2<\/sup> + 16
\n\u21d2 (OQ)2<\/sup> = 25 – 16 = 9
\n\u21d2 OQ = 3 cm
\nTherefore the distance of the chord CD from the centre is 3 cm.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nLet the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with a circle. Prove that \u2220ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
\nSolution:
\nGiven: Vertex of an angle ABC be located outside a circle, and AD = CE.
\nLet \u2220AOC = \u2220x, \u2220AOD = \u2220z and \u2220DOE = \u2220y.
\n\"NCERT
\nTo prove that: \u2220ABC = \\(\\frac {1}{2}\\) (x – y)
\nProof: AD = CE (given)
\n\u2234 \u2220AOD = \u2220COE (Angle made by equal chord at the centre are equal)
\nTherefore,
\n\u2220x + \u2220y + \u2220z + \u2220z = 360\u00b0
\n\u21d2 \u2220x + \u2220y + 2\u2220z = 360\u00b0
\n\u21d2 2\u2220z = 360\u00b0 – \u2220x – \u2220y
\n\u21d2 \u2220z = 180 – \\(\\frac {1}{2}\\) \u2220x – \\(\\frac {1}{2}\\) \u2220y ……(A)
\nNow, \u2220ODB = \u2220OAD + \u2220DOA
\n(Exterior angle is equal to sum of opposite interior angles)
\n\u2220ODB = \u2220OAD + \u2220z ……(i)
\nAgain, \u2220OAD + \u2220z + \u2220ODA = 180\u00b0
\n(Sum of all three angles of triangles)
\nor, \u2220OAD + \u2220z + \u2220OAD = 180\u00b0
\n(Angle opposite to equal sides are equal and OA = OD radii of circle)
\nor, 2\u2220OAD = 180\u00b0 – \u2220z
\nor, \u2220AOD = \\(\\frac {1}{2}\\) (180\u00b0 – \u2220z) = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220z
\nPutting the value of \u2220AOD in equation (i)
\n\u2234 \u2220ODB = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220z + \u2220z
\n(\u2235 \u2220OAD = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220z)
\nor, \u2220ODB = 90\u00b0 + \\(\\frac {1}{2}\\) \u2220z ……..(ii)
\nSimilarly, \u2220OEB = 90\u00b0 + \\(\\frac {1}{2}\\) \u2220z ……(iii)
\nNow, in ODBE
\n\u2220ODB + \u2220B + \u2220OEB + \u2220y = 360\u00b0
\n(Sum of all angle of \u2206 is 360\u00b0)
\n90 + \\(\\frac {1}{2}\\) \u2220z + \u2220B + 90 + \\(\\frac {1}{2}\\) \u2220z + \u2220y = 360\u00b0
\n(Use equation (ii) and (iii))
\nor, 180 + \u2220z + \u2220B + \u2220y = 360
\nor, \u2220B = 180 – \u2220y – \u2220z …..(iv)
\n\u21d2 \u2220ABC = \\(\\frac {1}{2}\\) [(Angle subtended by the chorde DE at the centre) – (Angle subtended by the chord AC at the centre)]
\n\u21d2 \u2220ABC = \\(\\frac {1}{2}\\) [(Difference of the angles subtended by the chorde DE and AC at the centre)]<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nProve that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
\nSolution:
\nWe have a rhombus ABCD such that its diagonals AC and BD intersect at O.
\nP, Q, R and S are the mid-points of DC, AB, AD and BC respectively
\n\u2234 \\(\\frac {1}{2}\\) AD = \\(\\frac {1}{2}\\) BC
\n\u21d2 RA = SB
\n\u21d2 RA = OQ ……(ii)
\n[\u2234 PQ is drawn parallel to AD and AD = BC]
\n\u21d2 \\(\\frac {1}{2}\\) AB = \\(\\frac {1}{2}\\) AD
\n\u21d2 AQ = AR ……(iii)
\nFrom (i), (ii) and (iii), we have
\nAQ = QB = OQ
\ni.e., A circle drawn with Q as centre, will pass through A, B and O.
\nThus, the circle passes through the intersection ‘O’ of the diagonals rhombus ABCD.<\/p>\n

Question 6.
\nABCD is a parallelogram. The circle through A, B, and C intersect CD (produced if necessary) at E. Prove that AE = AD.
\nSolution:
\nWe have a circle passing through A, B, and C is drawn such that it intersects CD at E.
\n\u2234 \u2220AEC + \u2220B = 180\u00b0 ……(i)
\n[Opposite angles of a cyclic quadrilateral are supplementary]
\nBut ABCD is a prallelogram (Given)
\n\u2234 \u2220D = \u2220B ……(ii)
\n[Opposite angles of a parallelogram are equal]
\nFrom (i) and (ii), we have
\n\u2220AEC + \u2220D = 180\u00b0 …….(iii)
\nBut \u2220AEC + \u2220AED = 180\u00b0 (Linear pair) ……(iv)
\n\"NCERT
\nFrom (iii) and (iv),
\nWe have \u2220D = \u2220AED
\ni.e., The base angles of \u2206ADE are euqal.
\n\u2234 AE = AD<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nAC and BD are chords of a circle which bisects each other. Prove that
\n(i) AC and BD are diameters
\n(ii) ABCD is a rectangle.
\nSolution:
\nGiven: A circle in which two chords AC and BD are such that they bisect each other.
\nLet their point of intersection be O.
\nTo Prove: (i) AC and BD are diameters.
\n(ii) ABCD is a rectangle.
\nConstruction: Join AB, BC, CD, and DA.
\nProof: (i) In \u2206AOB and \u2206COD, we have
\n\"NCERT
\n(\u2235 \u2220ADC = \u2220ABC, opposite angles of || gm ABCD)
\nFrom equation (i) and (ii)
\n\u2220AED + \u2220ABC = \u2220ADE + \u2220ABC
\nor, \u2220AED = \u2220ADE
\nThus, in \u2206AED, we have
\n\u2220AED = \u2220ADE
\n\u2234 AD = AE. (Side opposite to equal angles are equal)<\/p>\n

Question 8.
\nBisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angle of the triangle DEF are 90\u00b0 – \\(\\frac {1}{2}\\) \u2220A, 90\u00b0 – \\(\\frac {1}{2}\\) \u2220B and 90\u00b0 – \\(\\frac {1}{2}\\) \u2220C.
\nSolution:
\nGiven: In \u2206ABC, AD, BE and CF is the angle bisector of \u2220A, \u2220B, and \u2220C respectively.
\nWhere D, E, and F lie on the circumcircle of \u2206ABC.
\n\"NCERT
\nTo prove that:
\n(i) \u2220FDE = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220A
\n(ii) \u2220DEF = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220B
\n(iii) \u2220EFD = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220C
\nProof:
\n(i) \u2220ADF = \u2220ACF ……(i) (Angles in the same segments of a circle are equal)
\nAgain, \u2220ADE = \u2220ABE ……(ii) (Angles in the same segment of atcircle are equal)
\nAdding (i) and (ii)
\n\u2220APF + \u2220ACF = \u2220ACF + \u2220ABE
\nor, \u2220FDE = \\(\\frac {1}{2}\\) \u2220C + \\(\\frac {1}{2}\\) \u2220B ……(iii)
\n(\u2235 CF and BE are the bisector of \u2220B and \u2220C respectively)
\nNow, In \u2206ABC
\n\u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of all angles of a \u2206 is 180\u00b0)
\nor, \\(\\frac {1}{2}\\) (\u2220A + \u2220B + \u2220C) = 90\u00b0
\nor, \\(\\frac {1}{2}\\) \u2220B + \\(\\frac {1}{2}\\) \u2220C = 90 – \\(\\frac {1}{2}\\) \u2220A ……(iv)
\nFrom equation (iii) and (iv)
\n\u2220FDE = 90\u00b0 – \\(\\frac {1}{2}\\) \u2220A
\nSimilarly we can prove
\n(ii) \u2220DEF = 90 – \\(\\frac {1}{2}\\) \u2220B
\n(iii) \u2220EFD = 90 – \\(\\frac {1}{2}\\) \u2220C.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nTwo congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q, lie on the two circles. Prove that BP = BQ.
\nSolution:
\nLet C(0, r) and C(O’, r) be two congruent circles.
\n\"NCERT
\nSince AB is a common chord of two congruent circles. Therefore,
\narc ACB = arc ADB
\n\u2234 \u2220BPA = \u2220BQA
\nThus, in \u2206BPQ, we have \u2220BPA = \u2220BQA
\n\u2234 BP = BC (Sides opposite to equal angles of a triangle are equal)<\/p>\n

Question 10.
\nIn any triangle ABC, if the angle bisector of \u2220A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
\nSolution:
\nGiven: ABC is a triangle in which the bisector of \u2220A intersects the circumcircle of \u2206ABC at D.
\nTo prove that: Perpendicular bisector of side BC intersects the angle bisector of \u2220A at D.
\n\"NCERT
\nConstruction: Join BD and CD.
\nProof: \u2220BCD = \u2220BAD ……(i)
\n(Angles in the same segment of a circle are equal)
\nAgain, \u2220DBC = \u2220DAC …….(ii)
\n(Angles in the same segment of a circle are equal)
\nBut, \u2220BAD = \u2220DAC …..(iii)
\n(\u2235 AD is the bisector of \u2220A)
\nFrom equation (i), (ii) and (iii)
\n\u2220BCD = \u2220DBC
\nor, BD = CD (Side opposite to equal angles of a \u2206 are equal)
\nSo, D must be lying on the perpendicular bisector of BC.
\nTherefore, the perpendicular bisector of side BC intersects the angle bisector of \u2220A).<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6 Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection. …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6<\/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 10.6 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-10-ex-10-6\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6 Question 1. 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