NCERT Solutions for Class 9 Maths<\/a> Chapter 6 Lines and Angles Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1<\/h2>\n Question 1. \nIn Fig. 6.13 lines AB and CD intersect at O. If \u2220AOC + \u2220BOE = 70\u00b0 and \u2220BOD = 40\u00b0. Find \u2220BOE and reflex \u2220COE. \n \nSolution: \nIn this fig. we have given that \n\u2220AOC + \u2220BOE = 70\u00b0 and \u2220BOD = 40\u00b0 \nBut, \u2220BOD = \u2220AOC (Vertical opposite angles) \n\u2220AOC = 40 ……(i) (\u2235 \u2220BOD = 40\u00b0) \nAgain, \u2220AOC + \u2220BOE = 70\u00b0 (Given) \n40 + \u2220ZBOE = 70\u00b0 (from equation (i)) \n\u2220BOE = 70 – 40 = 30\u00b0 \nAgain, \u2220AOC + \u2220COE + \u2220EOB = 180\u00b0 (Because AB is a st. line) \n40\u00b0 + \u2220COE + 30\u00b0 = 180\u00b0 (\u2235 \u2220AOC = 40\u00b0 and \u2220EOB = 30\u00b0) \nTherefore, \u2220BOE = 30\u00b0, and Reflex \u2220COE = 360\u00b0 – 110\u00b0 = 250\u00b0 \nTherefore, \u2220BOE = 30\u00b0, and Reflex \u2220COE = 250\u00b0.<\/p>\n
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Question 2. \nIn Fig. 6.14 lines XY and MN intersect at O. If \u2220POY = 90\u00b0 and a : b = 2 : 3, find c. \n \nSolution: \nWe have given \n\u2220POY = 90\u00b0, and a : b = 2 : 3 \nor, \\(\\frac{a}{b}=\\frac{2}{3}\\) \n\u2234 a = \\(\\frac {2}{3}\\) b ……(i) \nAgain, \u2220POY + a + b = 180\u00b0 (Because OXY is a straight line) \n90 + \\(\\frac {2}{3}\\) b + b = 180\u00b0 \nor, \\(\\frac {2}{3}\\) b + b = 90 \nb = \\(\\frac{90 \\times 3}{5}\\) = 54\u00b0 ……(ii) \nAgain, c + b = 180\u00b0 (because MN is a straight line) \nc + 54\u00b0 = 180\u00b0 (from equation (ii)) \nTherefore, c = 126\u00b0<\/p>\n
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Question 3. \nIn Fig. 6.15, \u2220PQR = \u2220PRQ then prove that \u2220PQS = \u2220PRT. \n \nSolution: \nWe have given \n\u2220PQR = \u2220PRQ \nAgain, \n\u2220PQR + \u2220PQS = 180\u00b0 ……(i) (Linear pair) \n\u2220PRQ = \u2220PRT = 180\u00b0 ……(ii) (Linear pair) \nFrom equation (i) and (ii) \n\u2220PQR + \u2220PQS = \u2220PRQ + \u2220PRT \nBut, \u2220PQR = \u2220PRQ \nTherefore, \u2220PQS = \u2220PRT<\/p>\n
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Question 4. \nIn Fig. 6.16, if x + y = w + z, then prove that AOB is a line. \n \nSolution: \nWe have given x + y = w + z \nAgain, x + y + w + z = 360 \nor, x + y + x + y = 360 (\u2235 x + y = w + z) \nor, 2(x + y) = 360\u00b0 \nor, x + y = 180\u00b0 \n\u2220AOB = 180\u00b0 \nwhich is straight line angle. \n\u2234 AOB is a straight line.<\/p>\n
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Question 5. \nIn Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that \u2220ROS = \\(\\frac {1}{2}\\) (\u2220QOS – \u2220POS) \n \nSolution: \nWe have given, \nRay OR is perpendicular to line PQ \nTherefore, \u2220QOR = \u2220POR = 90\u00b0 ……(i) \nNow, \u2220QOS = \u2220QOR + \u2220ROS \nor, \u2220QOS = 90\u00b0 + \u2220ROS ……(ii) (from equation(i)) \nAgain, \u2220POS = \u2220POR – \u2220ROS \nor, \u2220PSO = 90 – \u2220ROS ……(iii) (from equation (i)) \nSubtract equation (iii) from equation (ii), \n\u2220QOS – \u2220POS \n\u2220QOS – \u2220POS = 90\u00b0 + \u2220ROS – (90\u00b0 – \u2220ROS) \n\u2220QOS – \u2220POS = 90\u00b0 + \u2220ROS – 90\u00b0 + \u2220ROS \n\u2220QOS – \u2220POS = 2\u2220ROS \n\u2220ROS = \\(\\frac {1}{2}\\) (\u2220QOS – \u2220POS)<\/p>\n
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Question 6. \nIt is given that \u2220XYZ = 64\u00b0 and xy is produced to point P. Draw a figure from the given information. If ray YQ bisects \u2220ZYP, find \u2220XYQ and reflex \u2220QYP. \n \nSolution: \nWe have given that \u2220XYZ = 64\u00b0 \nand \u2220ZYQ = \u2220QYP (\u2235 YQ is bisector of \u2220ZYP) \nNow, \u2220XYZ + \u2220ZYQ + \u2220QYP = 180\u00b0 (because XYP is a straight line) \nor, 64\u00b0 + \u2220ZYQ + \u2220ZYQ = 180\u00b0 (\u2235 \u2220ZYQ + \u2220QYP) \nor, 2\u2220ZYQ = 180\u00b0 – 64\u00b0 = 116\u00b0 \nor, \u2220ZYQ = 58\u00b0 \nNow, \u2220XYQ = \u2220XYZ + \u2220ZYQ = 64\u00b0 + 58\u00b0 \n\u2220XYQ = 122\u00b0 \nAgain, \u2220QYP = \u2220ZYQ (from equation (i)) \n\u2234 \u2220QYP = 58\u00b0 (\u2235 \u2220ZYQ = 58\u00b0) \n\u2234 Reflex \u2220QYP = 360\u00b0 – 58\u00b0 = 302\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1 Question 1. In Fig. 6.13 lines AB and CD intersect at O. If \u2220AOC + \u2220BOE = …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1<\/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 6.1 - 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