{"id":25124,"date":"2021-06-22T10:09:11","date_gmt":"2021-06-22T04:39:11","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25124"},"modified":"2021-06-22T10:09:11","modified_gmt":"2021-06-22T04:39:11","slug":"ncert-solutions-for-class-9-maths-chapter-6-ex-6-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-6-ex-6-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1"},"content":{"rendered":"

These 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>\n

NCERT 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\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\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

\"NCERT<\/p>\n

Question 3.
\nIn Fig. 6.15, \u2220PQR = \u2220PRQ then prove that \u2220PQS = \u2220PRT.
\n\"NCERT
\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

\"NCERT<\/p>\n

Question 4.
\nIn Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
\n\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\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":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-6-ex-6-1\/\" \/>\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 6.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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