{"id":25135,"date":"2021-06-22T10:55:50","date_gmt":"2021-06-22T05:25:50","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25135"},"modified":"2022-03-02T10:36:34","modified_gmt":"2022-03-02T05:06:34","slug":"ncert-solutions-for-class-9-maths-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-6-ex-6-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 6 Lines and Angles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.2<\/h2>\n

Question 1.
\nIn Fig. 6.28, find the values of x and y and then show that AB || CD.
\n\"NCERT
\nSolution:
\nAccording to fig.
\n50\u00b0 + x = 180\u00b0 (Linear pair)
\n\u21d2 x = 180\u00b0 – 50\u00b0
\n\u21d2 x = 130\u00b0 …..(i)
\nAgain, x = 130\u00b0 …..(ii)
\n(Vertically opposite angles are equal)
\nFrom equation (i) and (ii)
\nx = y (each 130\u00b0)
\nwhich is the pair of alternate interior angles. And we know that if the pair of alternate interior angles are equal, then the given two lines are parallel.
\nAB || CD<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIn Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
\n\"NCERT
\nSolution:
\nWe have given AB || CD || EF and y : z = 3 : 7
\nor, \\(\\frac{y}{z}=\\frac{3}{7}\\)
\nor, y = \\(\\frac {3}{7}\\) z …….(i)
\nAgain AB || EF
\n\u2234 x = z ……(ii)
\n(Pair of alternate interior angle)
\nNow, AB || CD
\n\u2234 x + y = 180\u00b0
\n(Sum of interior angle of the same side of transversal)
\nz + \\(\\frac {3}{7}\\) z = 180 (\u2234 x = z and y = \\(\\frac {3}{7}\\) z)
\n\u21d2 z = \\(\\frac{180 \\times 7}{10}\\)
\n\u21d2 z = 126
\nFrom, equation (ii)
\nx = z
\nor, x = 126<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn Fig. 6.30, if AB || CD, EF || CD and \u2220GED = 126\u00b0, Find \u2220AGE, \u2220CEF and \u2220FGE.
\n\"NCERT
\nSolution:
\nWe have given AB || CD, EF || CD and \u2220GED = 126\u00b0
\nAgain \u2220GED = \u2220AGE (Alternate interior angle)
\n\u2220AGE = 126\u00b0 (i) (\u2234 Given \u2220GED = 126\u00b0)
\nNow, \u2220GED = \u2220DEF + \u2220GEF
\n\u21d2 126 = 90 + \u2220GEF (\u2234 \u2220GED = 126 & EF \u22a5 CD)
\n\u21d2 \u2220GEF = 126\u00b0 – 90\u00b0 = 36\u00b0
\nAgain, \u2220AGE + \u2220FGE = 180\u00b0 (Linear pair)
\n126\u00b0 + \u2220FGE = 180\u00b0 (From equ. (i) \u2220AGE = 126\u00b0)
\n\u2220FGE = 180\u00b0 – 126\u00b0 = 54\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nIn Fig. 6.31, if PQ || ST, \u2220PQR = 110\u00b0, and \u2220RST = 130\u00b0, find \u2220QRS.
\n\"NCERT
\nSolution:
\nWe have given that PQ || ST,
\n\u2220PQR = 110\u00b0 and \u2220RST = 130\u00b0
\nConstruction:
\nThrough R draw a line MN | ST
\n\"NCERT
\nNow, ST || RN (by construction)
\nFigure
\nTherefore,
\n\u2220RST + \u2220SRN = 180\u00b0 (Sum of interior angle of the same side of transversal)
\nor, 130\u00b0 + \u2220SRN = 180\u00b0
\n\u2220SRN = 180\u00b0 – 130\u00b0 = 50\u00b0 ……(i)
\nGiven \u2220PQR = 110\u00b0
\n\u2220QRN = 110\u00b0
\nAgain, \u2220QRN = \u2220QRS + \u2220SRN
\n110 = \u2220QRS + 50 (\u2235 \u2220QRN = 110\u00b0 and \u2220SRN = 50\u00b0 from equ. (i))
\n\u2234 \u2220QRS = 110\u00b0 – 50\u00b0 = 60\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nIn Fig. 6.32, if AB || CD, \u2220APQ = 50\u00b0 and \u2220PRD = 127\u00b0, Find x and y.
\n\"NCERT
\nSolution:
\nWe have given AB || CD
\nand \u2220APQ = 50\u00b0 & \u2220PRD = 127\u00b0
\nNow, AB || CD (given)
\n\u2234 \u2220APQ = x (Pair of alternate interior angle)
\n\u2234 x = 50\u00b0 (\u2234 \u2220APQ = 50\u00b0)
\nAgain, \u2220APR = \u2220PRD (Pair of alternate interior angles)
\n\u2234 \u2220APQ = 127\u00b0
\nBut, \u2220APR = 50\u00b0 + y
\n\u21d2 127\u00b0 = 50\u00b0 + y
\n\u21d2 y = 127\u00b0 – 50\u00b0
\n\u21d2 y = 77\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nIn Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
\n\"NCERT
\nSolution:
\nTwo plane mirrors PQ and RS such that PQ || RS. An incident ray AB after reflections takes the path BC and CD. BM and CN are the normals to the plane mirror PQ and RS respectively.
\n\"NCERT
\nTo prove: AB || CD
\nProof: Since BM \u22a5 PQ, CN \u22a5 RS, and PQ || RS
\nTherefore, CN \u22a5 PQ \u21d2 BM || CN
\nThus, BM and CN are two parallel lines and a transversal BC cuts them at B and C respectively.
\n\u2234 \u22202 = \u22203 (Alternate interior angles)
\nBut, \u22201 = \u22202 and \u22203 = \u22204 (By law of reflection)
\n\u21d2 \u22201 + \u22202 = \u22202 + \u22202 and \u22203 + \u22204 = \u22203 + \u22203
\n\u21d2 \u22201 + \u22202 = 2(\u22202) and \u22203 + \u22204 = 2(\u22203)
\n\u21d2 \u22201 + \u22202 = \u22203 + \u22204 [\u2235 \u22202 + \u22203 \u21d2 2(\u22202) = (\u22203)]
\n\u2234 \u2220ABC = \u2220BCD
\nThus, lines AB and CD are intersected by transversal BC such that \u2220ABC = \u2220BCD i.e. alternate interior angles are equal.
\nTherefore, AB || CD
\nHence, AB || CD.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.2 Question 1. In Fig. 6.28, find the values of x and y and then show that AB …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2<\/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 6 Lines and Angles Ex 6.2 - 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-6-ex-6-2\/\" \/>\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 6 Lines and Angles Ex 6.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. 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