{"id":4319,"date":"2023-02-08T05:41:20","date_gmt":"2023-02-08T05:41:20","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4319"},"modified":"2023-02-08T05:42:26","modified_gmt":"2023-02-08T05:42:26","slug":"ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 6 Lines And Angles Ex 6.2"},"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 6.2 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-6-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 6.1<\/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; 6.2\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Fig. 6.28, find the values of x and y and then show that AB || CD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4334\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"232\" height=\"239\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>We know that a linear pair is equal to 180\u00b0.<br \/>\nSo, x + 50\u00b0 = 180\u00b0<br \/>\n\u2234 x = 180\u00b0 &#8211; 50\u00b0<br \/>\nx = 130\u00b0<br \/>\nWe also know that vertically opposite angles are equal.<br \/>\nSo, y = 130\u00b0<br \/>\nIn two parallel lines, the alternate interior angles are equal. In this,<br \/>\nx = y = 130\u00b0<br \/>\nThis proves that alternate interior angles are equal and so, AB || CD.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4335\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"177\" height=\"170\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>It is known that AB || CD and CD||EF<br \/>\nAs the angles on the same side of a transversal line sum up to 180\u00b0,<br \/>\nx + y = 180\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8211;(i)<br \/>\nAlso,<br \/>\n\u2220O = z (Since they are corresponding angles)<br \/>\nand, y +\u2220O = 180\u00b0 (Since they are a linear pair)<br \/>\nSo, y+z = 180\u00b0<br \/>\nNow, let y = 3w and hence, z = 7w (As y : z = 3 : 7)<br \/>\n\u2234 3w + 7w = 180\u00b0<br \/>\n\u21d2 10 w = 180\u00b0<br \/>\n\u21d2 w = 18\u00b0<br \/>\nNow, y = 3 \u00d7 18\u00b0 = 54\u00b0<br \/>\nand, z = 7 \u00d7 18\u00b0 = 126\u00b0<br \/>\nNow, angle x can be calculated from equation (i)<br \/>\nx + y = 180\u00b0<br \/>\n\u21d2 x + 54\u00b0 = 180\u00b0<br \/>\n\u21d2 x = 126\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Fig. 6.30, if AB || CD, EF \u22a5 CD and \u2220GED = 126\u00b0, find \u2220AGE, \u2220GEF and \u2220FGE.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4336\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"212\" height=\"164\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Since AB || CD, GE is a transversal.<br \/>\nIt is given that \u2220GED = 126\u00b0<br \/>\nSo, \u2220GED = \u2220AGE = 126\u00b0 (as they are alternate interior angles)<br \/>\nAlso,<br \/>\n\u2220GED = \u2220GEF +\u2220FED<br \/>\nAs EF\u22a5 CD, \u2220FED = 90\u00b0<br \/>\n\u2234 \u2220GED = \u2220GEF+90\u00b0<br \/>\n\u21d2 \u2220GEF = 126\u00b0 \u2013 90\u00b0 = 36\u00b0<br \/>\n\u2220FGE + \u2220GED = 180\u00b0 (transversal)<br \/>\n\u2220FGE + 126\u00b0 = 180\u00b0<br \/>\n\u2220FGE = 180\u00b0 &#8211; 126\u00b0<br \/>\n\u2220FGE = 54\u00b0<br \/>\nSo,<br \/>\n\u2220AGE = 126\u00b0,<br \/>\n\u2220GEF = 36\u00b0 and<br \/>\n\u2220FGE = 54\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 6.31, if PQ || ST, \u2220PQR = 110\u00b0 and \u2220RST = 130\u00b0, find \u2220QRS.<br \/>\n<\/strong><strong>[Hint : Draw a line parallel to ST through point R.]<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4337\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"220\" height=\"137\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>First, construct a line XY parallel to PQ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4338\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Ans-4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"214\" height=\"107\" \/><br \/>\nWe know that the angles on the same side of a transversal is equal to 180\u00b0.<br \/>\nSo, \u2220PQR + \u2220QRX = 180\u00b0<br \/>\n\u2220QRX = 180\u00b0-110\u00b0<br \/>\n\u2234 \u2220QRX = 70\u00b0<br \/>\nSimilarly,<br \/>\n\u2220RST +\u2220SRY = 180\u00b0<br \/>\n\u2220SRY = 180\u00b0- 130\u00b0<br \/>\n\u2234 \u2220SRY = 50\u00b0<br \/>\nNow, for the linear pairs on the line XY,<br \/>\n\u2220QRX + \u2220QRS + \u2220SRY = 180\u00b0<br \/>\nPutting their respective values, we get,<br \/>\n\u2220QRS = 180\u00b0 \u2013 70\u00b0 \u2013 50\u00b0<br \/>\nHence, \u2220QRS = 60\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Fig. 6.32, if AB || CD, \u2220APQ = 50\u00b0 and \u2220PRD = 127\u00b0, find x and y.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4339\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"226\" height=\"154\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>From the diagram,<br \/>\n\u2220APQ = \u2220PQR (alternate interior angles)<br \/>\nNow, putting the value of \u2220APQ = 50\u00b0 and \u2220PQR = x we get,<br \/>\nx = 50\u00b0<br \/>\nAlso,<br \/>\n\u2220APR = \u2220PRD (alternate interior angles)<br \/>\n\u2220APR = 127\u00b0 (as it is given that \u2220PRD = 127\u00b0)<br \/>\nWe know that<br \/>\n\u2220APR = \u2220APQ + \u2220QPR<br \/>\nNow, putting values of \u2220QPR = y and \u2220APR = 127\u00b0 we get,<br \/>\n127\u00b0 = 50\u00b0+ y<br \/>\ny = 77\u00b0<br \/>\nThus, the values of x and y are calculated as:<br \/>\nx = 50\u00b0 and y = 77\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. In 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.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4340\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"297\" height=\"244\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>First, draw two lines BE and CF such that BE \u22a5 PQ and CF \u22a5 RS.<br \/>\nNow, since PQ || RS,<br \/>\nSo, BE || CF<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4342\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.2-Answer6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"283\" height=\"217\" \/><br \/>\nWe know that,<br \/>\nAngle of incidence = Angle of reflection (By the law of reflection)<br \/>\nSo,<br \/>\n\u22201 = \u22202 and<br \/>\n\u22203 = \u22204<br \/>\nWe also know that alternate interior angles are equal. Here, BE \u22a5 CF and the transversal line BC cuts them at B and C<br \/>\nSo, \u22202 = \u22203 (As they are alternate interior angles)<br \/>\nNow, \u22201 +\u22202 = \u22203 +\u22204<br \/>\nOr, \u2220ABC = \u2220DCB<br \/>\nSo, AB || CD (alternate interior angles are equal)<\/span><\/p>\n<p>&nbsp;<\/p>\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 6.2 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-4319","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 v27.4) - 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