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 <\/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 \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
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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 \nFrom (iii) and (iv), \nWe have \u2220D = \u2220AED \ni.e., The base angles of \u2206ADE are euqal. \n\u2234 AE = AD<\/p>\n
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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 \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 \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
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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 \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 \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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n