{"id":4318,"date":"2023-02-08T05:41:17","date_gmt":"2023-02-08T05:41:17","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4318"},"modified":"2023-02-08T05:42:16","modified_gmt":"2023-02-08T05:42:16","slug":"ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-1\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 6 Lines And Angles Ex 6.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 6 (Lines And Angles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 6 Lines And Angles <\/strong>Exercise 6.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 6: Lines and Angles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 6.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 6.3<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.1\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Fig. 6.13, lines AB and CD intersect at O. If \u2220AOC +\u2220BOE = 70\u00b0 and \u2220BOD = 40\u00b0, find \u2220BOE and reflex \u2220COE.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4327\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"222\" height=\"203\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>From the diagram, we have<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOC + \u2220BOE +\u2220COE and \u2220COE +\u2220BOD + \u2220BOE form a straight line.<\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220AOC + \u2220BOE +\u2220COE = \u2220COE +\u2220BOD + \u2220BOE = 180<sup>o<\/sup><\/span><br \/>\n<span style=\"color: #000000;\">Now, by putting the values of \u2220AOC + \u2220BOE = 70<sup>o<\/sup>\u00a0and \u2220BOD = 40<sup>o<\/sup>\u00a0we get,<\/span><br \/>\n<span style=\"color: #000000;\">70<sup>o<\/sup>\u00a0+\u2220COE = 180<sup>o<\/sup><\/span><br \/>\n<span style=\"color: #000000;\">\u2220COE = 110\u00b0 and \u2220BOE = 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">So, reflex \u2220COE = 360<sup>o<\/sup>\u00a0\u2013 110<sup>o<\/sup>\u00a0= 250<sup>o<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In Fig. 6.14, lines XY and MN intersect at O. If \u2220POY = 90\u00b0 and a : b = 2 : 3, find c.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4328\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"329\" height=\"257\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q2.png 329w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q2-300x234.png 300w\" sizes=\"auto, (max-width: 329px) 100vw, 329px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Line OP is perpendicular to line XY. Hence \u2220POY = \u2220POX = 90\u00b0<br \/>\n\u2220POX = \u2220POM + \u2220MOX<\/span><br \/>\n<span style=\"color: #000000;\">90\u00b0 = a + b\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;(1)<\/span><br \/>\n<span style=\"color: #000000;\">Since a and b are in the ratio 2 : 3 that is,<\/span><br \/>\n<span style=\"color: #000000;\">a = 2x and b = 3x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;(2)<\/span><br \/>\n<span style=\"color: #000000;\">Substituting (2) in (1),<\/span><br \/>\n<span style=\"color: #000000;\">a + b = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">2x + 3x = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">5x = 90\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">x = 90\u00b0\/5 = 18\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">a = 2x = 2 \u00d7 18\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">a = 36\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">b = 3x = 3 \u00d7 18\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">b = 54\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Also , \u2220MOY= \u2220MOP + \u2220POY <\/span><br \/>\n<span style=\"color: #000000;\">= a + 90\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">= 36\u00b0 + 90\u00b0 = 126\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Lines MN and XY intersect at point O and the\u00a0vertically opposite angles\u00a0formed are equal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XON = \u2220MOY <\/span><br \/>\n<span style=\"color: #000000;\">c = 126\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Fig. 6.15, \u2220PQR = \u2220PRQ, then prove that \u2220PQS = \u2220PRT.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4329\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"239\" height=\"174\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Since ST is a straight line so,<\/span><br \/>\n<span style=\"color: #000000;\"><b>\u2220<\/b>PQS + <b>\u2220<\/b>PQR = 180\u00b0 (linear pair) and<\/span><br \/>\n<span style=\"color: #000000;\"><b>\u2220<\/b>PRT + <b>\u2220<\/b>PRQ = 180\u00b0 (linear pair)<\/span><br \/>\n<span style=\"color: #000000;\">Now,\u00a0<b>\u2220<\/b>PQS +\u00a0<b>\u2220<\/b>PQR =\u00a0<b>\u2220<\/b>PRT+<b>\u2220<\/b>PRQ = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Since\u00a0<b>\u2220<\/b>PQR =<b>\u2220<\/b>PRQ (as given in the question) <\/span><br \/>\n<span style=\"color: #000000;\"><b>\u2220<\/b>PQS =\u00a0<b>\u2220<\/b>PRT. (Hence proved).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4330\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"202\" height=\"229\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>For proving AOB is a straight line, we will have to prove x+y is a linear pair<\/span><br \/>\n<span style=\"color: #000000;\">i.e. x + y = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">We know that the angles around a point are 360\u00b0 so, <\/span><br \/>\n<span style=\"color: #000000;\">x + y + w + z = 360\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">In the question, it is given that, <\/span><br \/>\n<span style=\"color: #000000;\">x + y = w + z<\/span><br \/>\n<span style=\"color: #000000;\">So, (x+y) + (x+y) = 360\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">2 (x+y) = 360\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 (x+y) = 180\u00b0 (Hence proved).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In 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 = \u00bd (\u2220QOS \u2013 \u2220POS).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4331\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"213\" height=\"195\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>In the question, it is given that (OR \u22a5 PQ) and \u2220POQ = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">We can write it as \u2220ROP = \u2220ROQ = 90<sup>0<br \/>\n<\/sup>We know that <\/span><br \/>\n<span style=\"color: #000000;\">\u2220ROP = \u2220ROQ <\/span><br \/>\n<span style=\"color: #000000;\">It can be written as <\/span><br \/>\n<span style=\"color: #000000;\">\u2220POS + \u2220ROS = \u2220ROQ<\/span><br \/>\n<span style=\"color: #000000;\">\u2220POS + \u2220ROS = \u2220QOS \u2013 \u2220ROS<\/span><br \/>\n<span style=\"color: #000000;\">\u2220SOR + \u2220ROS = \u2220QOS \u2013 \u2220POS<\/span><br \/>\n<span style=\"color: #000000;\">So we get <\/span><br \/>\n<span style=\"color: #000000;\">2\u2220ROS = \u2220QOS \u2013 \u2220POS <\/span><br \/>\n<span style=\"color: #000000;\">Or, \u2220ROS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (\u2220QOS \u2013 \u2220POS) (Hence proved).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. It 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.<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Here, XP is a straight line<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4332\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.1-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"294\" height=\"197\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220XYZ + \u2220ZYP = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Putting the value of \u2220XYZ = 64\u00b0 we get, <\/span><br \/>\n<span style=\"color: #000000;\">64\u00b0 +\u2220ZYP = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ZYP = 116\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">From the diagram, we also know that \u2220ZYP = \u2220ZYQ + \u2220QYP<\/span><br \/>\n<span style=\"color: #000000;\">Now, as YQ bisects \u2220ZYP, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220ZYQ = \u2220QYP <\/span><br \/>\n<span style=\"color: #000000;\">Or, \u2220ZYP = 2\u2220ZYQ<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ZYQ = \u2220QYP = 58\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">Again, \u2220XYQ = \u2220XYZ + \u2220ZYQ <\/span><br \/>\n<span style=\"color: #000000;\">By putting the value of \u2220XYZ = 64\u00b0 and \u2220ZYQ = 58\u00b0 we get.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XYQ = 64\u00b0+58\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">Or, \u2220XYQ = 122\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Now, reflex \u2220QYP = 180\u00b0+XYQ <\/span><br \/>\n<span style=\"color: #000000;\">We computed that the value of \u2220XYQ = 122\u00b0. <\/span><br \/>\n<span style=\"color: #000000;\">So, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220QYP = 180\u00b0+122\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220QYP = 302\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 6 (Lines And Angles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 6 Lines And Angles Exercise 6.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 6: Lines and Angles NCERT Solution Class 9 Maths Ex &#8211; [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[711,702,186,712,701,5],"class_list":["post-4318","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-6-lines-and-angles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-6-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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