{"id":62199,"date":"2021-09-04T12:48:00","date_gmt":"2021-09-04T07:18:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62199"},"modified":"2021-09-17T13:19:21","modified_gmt":"2021-09-17T07:49:21","slug":"rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 11 &#8211; Coordinate Geometry (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124461\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-11-Coordinate-Geometry.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 11\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-11-Coordinate-Geometry.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-11-Coordinate-Geometry-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 9 Maths Chapter 11:<\/strong> Are you confused about the relevance of purchasing <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 11 Coordinate Geometry<strong>? <\/strong>Yes<strong>. <\/strong>You need to calm down because in this blog we will be concentrating on the essential aspects of Class 9 Maths Coordinate Geometry Chapter 11 Solutions.\u00a0<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-6a038a8a78b1e\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-6a038a8a78b1e\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#download-rd-sharma-solutions-class-9-maths-chapter-11-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 11 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 11 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#exercise-wise-class-9-maths-rd-sharma-chapter-11\" title=\"Exercise-Wise Class 9 Maths RD Sharma Chapter 11\">Exercise-Wise Class 9 Maths RD Sharma Chapter 11<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#access-answers-of-rd-sharma-solutions-class-9-chapter-11\" title=\"Access answers of RD Sharma Solutions Class 9 Chapter 11\">Access answers of RD Sharma Solutions Class 9 Chapter 11<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#rd-sharma-class-9-solutions-chapter-11-coordinate-geometry-ex-111\" title=\"RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry Ex 11.1\">RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry Ex 11.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#rd-sharma-solutions-class-9-chapter-11-coordinate-geometry-ex-112\" title=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry Ex 11.2\">RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry Ex 11.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#rd-sharma-solutions-class-9-chapter-11-coordinate-geometry-vsaqs\" title=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry VSAQS\">RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#class-9-rd-sharma-solutions-chapter-11-coordinate-geometry-mcqs\" title=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry MCQS\">Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry MCQS<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#detailed-exercise-wise-explanation-with-listing-of-important-topics\" title=\"Detailed Exercise-wise Explanation with Listing of Important Topics\u00a0\">Detailed Exercise-wise Explanation with Listing of Important Topics\u00a0<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#explanation-for-coordinates-of-a\" title=\"Explanation for Coordinates of A:\">Explanation for Coordinates of A:<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 11 Coordinate Geometry\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 11 Coordinate Geometry<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-11\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 11?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 11?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-11\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 11?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 11?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-11-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 11\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 11\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-chapter-11-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 11 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/11-1.pdf\" target=\"_blank\" rel=\"noopener\">RS Sharma Class 9 Solutions Chapter 11<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/11-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-class-9-maths-rd-sharma-chapter-11\"><\/span><strong>Exercise-Wise Class 9 Maths RD Sharma Chapter 11<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-1\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 11 Exercise 11.1<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exhttps:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/ercise-11-1\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 11 Exercise 11.2<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 11 Exercise 11.3<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-9-chapter-11\"><\/span><strong>Access answers of RD Sharma Solutions Class 9 Chapter 11<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-11-coordinate-geometry-ex-111\"><\/span>RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry Ex 11.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>In a \u2206ABC, if \u2220A = 55\u00b0, \u2220B = 40\u00b0, find \u2220C.<br \/>Solution:<br \/>\u2235 Sum of three angles of a triangle is 180\u00b0<br \/>\u2234 In \u2206ABC, \u2220A = 55\u00b0, \u2220B = 40\u00b0<br \/>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1924\/30703040207_337b02451b_o.png\" alt=\"RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"225\" height=\"173\" \/><br \/>\u21d2 55\u00b0 + 40\u00b0 + \u2220C = 180\u00b0<br \/>\u21d2 95\u00b0 + \u2220C = 180\u00b0<br \/>\u2234 \u2220C= 180\u00b0 -95\u00b0 = 85\u00b0<\/p>\n<p>Question 2.<br \/>If the angles of a triangle are in the ratio 1:2:3, determine three angles.<br \/>Solution:<br \/>Ratio in three angles of a triangle =1:2:3<br \/>Let first angle = x<br \/>Then second angle = 2x<br \/>and third angle = 3x<br \/>\u2234 x + 2x + 3x = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d26x = 180\u00b0<br \/>\u21d2x =\u00a0<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"msubsup\"><span id=\"MathJax-Span-5\" class=\"texatom\"><span id=\"MathJax-Span-6\" class=\"mrow\"><span id=\"MathJax-Span-7\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-8\" class=\"texatom\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-11\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>\u00a0 = 30\u00b0<br \/>\u2234 First angle = x = 30\u00b0<br \/>Second angle = 2x = 2 x 30\u00b0 = 60\u00b0<br \/>and third angle = 3x = 3 x 30\u00b0 = 90\u00b0<br \/>\u2234 Angles are 30\u00b0, 60\u00b0, 90\u00b0<\/p>\n<p>Question 3.<br \/>The angles of a triangle are (x \u2013 40)\u00b0, (x \u2013 20)\u00b0 and (<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-12\" class=\"math\"><span id=\"MathJax-Span-13\" class=\"mrow\"><span id=\"MathJax-Span-14\" class=\"mfrac\"><span id=\"MathJax-Span-15\" class=\"mn\">1<\/span><span id=\"MathJax-Span-16\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x \u2013 10)\u00b0. Find the value of x.<br \/>Solution:<br \/>\u2235 Sum of three angles of a triangle = 180\u00b0<br \/>\u2234 (x \u2013 40)\u00b0 + (x \u2013 20)\u00b0 + (<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-17\" class=\"math\"><span id=\"MathJax-Span-18\" class=\"mrow\"><span id=\"MathJax-Span-19\" class=\"mfrac\"><span id=\"MathJax-Span-20\" class=\"mn\">1<\/span><span id=\"MathJax-Span-21\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x-10)0 = 180\u00b0<br \/>\u21d2 x \u2013 40\u00b0 + x \u2013 20\u00b0 +\u00a0<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-22\" class=\"math\"><span id=\"MathJax-Span-23\" class=\"mrow\"><span id=\"MathJax-Span-24\" class=\"mfrac\"><span id=\"MathJax-Span-25\" class=\"mn\">1<\/span><span id=\"MathJax-Span-26\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x \u2013 10\u00b0 = 180\u00b0<br \/>\u21d2 x + x+\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-27\" class=\"math\"><span id=\"MathJax-Span-28\" class=\"mrow\"><span id=\"MathJax-Span-29\" class=\"mfrac\"><span id=\"MathJax-Span-30\" class=\"mn\">1<\/span><span id=\"MathJax-Span-31\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x \u2013 70\u00b0 = 180\u00b0<br \/>\u21d2\u00a0<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-32\" class=\"math\"><span id=\"MathJax-Span-33\" class=\"mrow\"><span id=\"MathJax-Span-34\" class=\"mfrac\"><span id=\"MathJax-Span-35\" class=\"mn\">5<\/span><span id=\"MathJax-Span-36\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x = 180\u00b0 + 70\u00b0 = 250\u00b0<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-37\" class=\"math\"><span id=\"MathJax-Span-38\" class=\"mrow\"><span id=\"MathJax-Span-39\" class=\"mfrac\"><span id=\"MathJax-Span-40\" class=\"mrow\"><span id=\"MathJax-Span-41\" class=\"msubsup\"><span id=\"MathJax-Span-42\" class=\"texatom\"><span id=\"MathJax-Span-43\" class=\"mrow\"><span id=\"MathJax-Span-44\" class=\"mn\">250<\/span><\/span><\/span><span id=\"MathJax-Span-45\" class=\"texatom\"><span id=\"MathJax-Span-46\" class=\"mrow\"><span id=\"MathJax-Span-47\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-48\" class=\"mi\">x<\/span><span id=\"MathJax-Span-49\" class=\"mn\">2<\/span><\/span><span id=\"MathJax-Span-50\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>\u00a0 = 100\u00b0<br \/>\u2234 x = 100\u00b0<\/p>\n<p>Question 4.<br \/>Two angles of a triangle are equal and the third angle is greater than each of those angles by 30\u00b0. Determine all the angles of the triangle.<br \/>Solution:<br \/>Let each of the two equal angles = x<br \/>Then third angle = x + 30\u00b0<br \/>But sum of the three angles of a triangle is 180\u00b0<br \/>\u2234 x + x + x + 30\u00b0 = 180\u00b0<br \/>\u21d2 3x + 30\u00b0 = 180\u00b0<br \/>\u21d23x = 150\u00b0 \u21d2x =\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-51\" class=\"math\"><span id=\"MathJax-Span-52\" class=\"mrow\"><span id=\"MathJax-Span-53\" class=\"mfrac\"><span id=\"MathJax-Span-54\" class=\"msubsup\"><span id=\"MathJax-Span-55\" class=\"texatom\"><span id=\"MathJax-Span-56\" class=\"mrow\"><span id=\"MathJax-Span-57\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-58\" class=\"texatom\"><span id=\"MathJax-Span-59\" class=\"mrow\"><span id=\"MathJax-Span-60\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-61\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0= 50\u00b0<br \/>\u2234 Each equal angle = 50\u00b0<br \/>and third angle = 50\u00b0 + 30\u00b0 = 80\u00b0<br \/>\u2234 Angles are 50\u00b0, 50\u00b0 and 80\u00b0<\/p>\n<p>Question 5.<br \/>If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.<br \/>Solution:<br \/>In the triangle ABC,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1944\/30703040037_61825e9d82_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"222\" height=\"173\" \/><br \/>\u2220B = \u2220A + \u2220C<br \/>But \u2220A + \u2220B + \u2220C = 180\u00b0<br \/>\u21d2\u2220B + \u2220A + \u2220C = 180\u00b0<br \/>\u21d2\u2220B + \u2220B = 180\u00b0<br \/>\u21d22\u2220B = 180\u00b0<br \/>\u2234 \u2220B =\u00a0<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-62\" class=\"math\"><span id=\"MathJax-Span-63\" class=\"mrow\"><span id=\"MathJax-Span-64\" class=\"mfrac\"><span id=\"MathJax-Span-65\" class=\"msubsup\"><span id=\"MathJax-Span-66\" class=\"texatom\"><span id=\"MathJax-Span-67\" class=\"mrow\"><span id=\"MathJax-Span-68\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-69\" class=\"texatom\"><span id=\"MathJax-Span-70\" class=\"mrow\"><span id=\"MathJax-Span-71\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-72\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 90\u00b0<br \/>\u2235 One angle of the triangle is 90\u00b0<br \/>\u2234 \u2206ABC is a right triangle.<\/p>\n<p>Question 6.<br \/>Can a triangle have:<br \/>(i) Two right angles?<br \/>(ii) Two obtuse angles?<br \/>(iii) Two acute angles?<br \/>(iv) All angles more than 60\u00b0?<br \/>(v) All angles less than 60\u00b0?<br \/>(vi) All angles equal to 60\u00b0?<br \/>Justify your answer in each case.<br \/>Solution:<br \/>(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180\u00b0 and if there are two right-angles, then the third angle will be zero which is not possible.<br \/>(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180\u00b0 and if there are<br \/>two obtuse angle, then the third angle will be negative which is not possible.<br \/>(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180\u00b0.<br \/>(iv) All angles more than 60\u00b0, they are also not possible as the sum will be more than 180\u00b0.<br \/>(v) All angles less than 60\u00b0. They are also not possible as the sum will be less than 180\u00b0.<br \/>(vi) All angles equal to 60\u00b0. This is possible as the sum will be 60\u00b0 x 3 = 180\u00b0.<\/p>\n<p>Question 7.<br \/>The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10\u00b0, find the three angles.<br \/>Solution:<br \/>Let three angles of a triangle be x\u00b0, (x + 10)\u00b0, (x + 20)\u00b0<br \/>But sum of three angles of a triangle is 180\u00b0<br \/>\u2234 x + (x+ 10)\u00b0 + (x + 20) = 180\u00b0<br \/>\u21d2 x + x+10\u00b0+ x + 20 = 180\u00b0<br \/>\u21d2 3x + 30\u00b0 = 180\u00b0<br \/>\u21d2 3x = 180\u00b0 \u2013 30\u00b0 = 150\u00b0<br \/>\u2234 x =\u00a0<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-73\" class=\"math\"><span id=\"MathJax-Span-74\" class=\"mrow\"><span id=\"MathJax-Span-75\" class=\"mfrac\"><span id=\"MathJax-Span-76\" class=\"msubsup\"><span id=\"MathJax-Span-77\" class=\"texatom\"><span id=\"MathJax-Span-78\" class=\"mrow\"><span id=\"MathJax-Span-79\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-80\" class=\"texatom\"><span id=\"MathJax-Span-81\" class=\"mrow\"><span id=\"MathJax-Span-82\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-83\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 50\u00b0<br \/>\u2234 Angle are 50\u00b0, 50 + 10, 50 + 20<br \/>i.e. 50\u00b0, 60\u00b0, 70\u00b0<\/p>\n<p>Question 8.<br \/>ABC is a triangle is which \u2220A = 72\u00b0, the internal bisectors of angles B and C meet in O. Find the magnitude of \u2220BOC.<br \/>Solution:<br \/>In \u2206ABC, \u2220A = 12\u00b0 and bisectors of \u2220B and \u2220C meet at O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/30703039897_69de4ee965_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"174\" height=\"169\" \/><br \/>Now \u2220B + \u2220C = 180\u00b0 \u2013 12\u00b0 = 108\u00b0<br \/>\u2235 OB and OC are the bisectors of \u2220B and \u2220C respectively<br \/>\u2234 \u2220OBC + \u2220OCB =\u00a0<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-84\" class=\"math\"><span id=\"MathJax-Span-85\" class=\"mrow\"><span id=\"MathJax-Span-86\" class=\"mfrac\"><span id=\"MathJax-Span-87\" class=\"mn\">1<\/span><span id=\"MathJax-Span-88\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0(B + C)<br \/>=\u00a0<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-89\" class=\"math\"><span id=\"MathJax-Span-90\" class=\"mrow\"><span id=\"MathJax-Span-91\" class=\"mfrac\"><span id=\"MathJax-Span-92\" class=\"mn\">1<\/span><span id=\"MathJax-Span-93\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 108\u00b0 = 54\u00b0<br \/>But in \u2206OBC,<br \/>\u2234 \u2220OBC + \u2220OCB + \u2220BOC = 180\u00b0<br \/>\u21d2 54\u00b0 + \u2220BOC = 180\u00b0<br \/>\u2220BOC = 180\u00b0-54\u00b0= 126\u00b0<br \/>OR<br \/>According to corollary,<br \/>\u2220BOC = 90\u00b0+\u00a0<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-94\" class=\"math\"><span id=\"MathJax-Span-95\" class=\"mrow\"><span id=\"MathJax-Span-96\" class=\"mfrac\"><span id=\"MathJax-Span-97\" class=\"mn\">1<\/span><span id=\"MathJax-Span-98\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A<br \/>= 90+\u00a0<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-99\" class=\"math\"><span id=\"MathJax-Span-100\" class=\"mrow\"><span id=\"MathJax-Span-101\" class=\"mfrac\"><span id=\"MathJax-Span-102\" class=\"mn\">1<\/span><span id=\"MathJax-Span-103\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 72\u00b0 = 90\u00b0 + 36\u00b0 = 126\u00b0<\/p>\n<p>Question 9.<br \/>The bisectors of base angles of a triangle cannot enclose a right angle in any case.<br \/>Solution:<br \/>In right \u2206ABC, \u2220A is the vertex angle and OB and OC are the bisectors of \u2220B and \u2220C respectively<br \/>To prove : \u2220BOC cannot be a right angle<br \/>Proof: \u2235 OB and OC are the bisectors of \u2220B and \u2220C respectively<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/30703039697_361ceff1b1_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"261\" height=\"181\" \/><br \/>\u2234 \u2220BOC = 90\u00b0 x\u00a0<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-104\" class=\"math\"><span id=\"MathJax-Span-105\" class=\"mrow\"><span id=\"MathJax-Span-106\" class=\"mfrac\"><span id=\"MathJax-Span-107\" class=\"mn\">1<\/span><span id=\"MathJax-Span-108\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A<br \/>Let \u2220BOC = 90\u00b0, then<br \/><span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-109\" class=\"math\"><span id=\"MathJax-Span-110\" class=\"mrow\"><span id=\"MathJax-Span-111\" class=\"mfrac\"><span id=\"MathJax-Span-112\" class=\"mn\">1<\/span><span id=\"MathJax-Span-113\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A = O<br \/>\u21d2\u2220A = O<br \/>Which is not possible because the points A, B and C will be on the same line Hence, \u2220BOC cannot be a right angle.<\/p>\n<p>Question 10.<br \/>If the bisectors of the base angles of a triangle enclose an angle of 135\u00b0. Prove that the triangle is a right triangle.<br \/>Solution:<br \/>Given : In \u2206ABC, OB and OC are the bisectors of \u2220B and \u2220C and \u2220BOC = 135\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/30703039507_df9d86438a_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"217\" height=\"179\" \/><br \/>To prove : \u2206ABC is a right angled triangle<br \/>Proof: \u2235 Bisectors of base angles \u2220B and \u2220C of the \u2206ABC meet at O<br \/>\u2234 \u2220BOC = 90\u00b0+\u00a0<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-114\" class=\"math\"><span id=\"MathJax-Span-115\" class=\"mrow\"><span id=\"MathJax-Span-116\" class=\"mfrac\"><span id=\"MathJax-Span-117\" class=\"mn\">1<\/span><span id=\"MathJax-Span-118\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220A<br \/>But \u2220BOC =135\u00b0<br \/>\u2234 90\u00b0+\u00a0<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-119\" class=\"math\"><span id=\"MathJax-Span-120\" class=\"mrow\"><span id=\"MathJax-Span-121\" class=\"mfrac\"><span id=\"MathJax-Span-122\" class=\"mn\">1<\/span><span id=\"MathJax-Span-123\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A = 135\u00b0<br \/>\u21d2\u00a0<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-124\" class=\"math\"><span id=\"MathJax-Span-125\" class=\"mrow\"><span id=\"MathJax-Span-126\" class=\"mfrac\"><span id=\"MathJax-Span-127\" class=\"mn\">1<\/span><span id=\"MathJax-Span-128\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220A= 135\u00b0 -90\u00b0 = 45\u00b0<br \/>\u2234 \u2220A = 45\u00b0 x 2 = 90\u00b0<br \/>\u2234 \u2206ABC is a right angled triangle<\/p>\n<p>Question 11.<br \/>In a \u2206ABC, \u2220ABC = \u2220ACB and the bisectors of \u2220ABC and \u2220ACB intersect at O such that \u2220BOC = 120\u00b0. Show that \u2220A = \u2220B = \u2220C = 60\u00b0.<br \/>Solution:<br \/>Given : In \u2220ABC, BO and CO are the bisectors of \u2220B and \u2220C respectively and \u2220BOC = 120\u00b0 and \u2220ABC = \u2220ACB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/30703039197_da3a724242_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"183\" height=\"179\" \/><br \/>To prove : \u2220A = \u2220B = \u2220C = 60\u00b0<br \/>Proof : \u2235 BO and CO are the bisectors of \u2220B and \u2220C<br \/>\u2234 \u2220BOC = 90\u00b0 +\u00a0<span id=\"MathJax-Element-20-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-129\" class=\"math\"><span id=\"MathJax-Span-130\" class=\"mrow\"><span id=\"MathJax-Span-131\" class=\"mfrac\"><span id=\"MathJax-Span-132\" class=\"mn\">1<\/span><span id=\"MathJax-Span-133\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220A<br \/>But \u2220BOC = 120\u00b0<br \/>\u2234 90\u00b0+\u00a0<span id=\"MathJax-Element-21-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-134\" class=\"math\"><span id=\"MathJax-Span-135\" class=\"mrow\"><span id=\"MathJax-Span-136\" class=\"mfrac\"><span id=\"MathJax-Span-137\" class=\"mn\">1<\/span><span id=\"MathJax-Span-138\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A = 120\u00b0<br \/>\u2234\u00a0<span id=\"MathJax-Element-22-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-139\" class=\"math\"><span id=\"MathJax-Span-140\" class=\"mrow\"><span id=\"MathJax-Span-141\" class=\"mfrac\"><span id=\"MathJax-Span-142\" class=\"mn\">1<\/span><span id=\"MathJax-Span-143\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A = 120\u00b0 \u2013 90\u00b0 = 30\u00b0<br \/>\u2234 \u2220A = 60\u00b0<br \/>\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Angles of a triangle)<br \/>\u2220B + \u2220C = 180\u00b0 \u2013 60\u00b0 = 120\u00b0 and \u2220B = \u2220C<br \/>\u2235 \u2220B = \u2220C =\u00a0<span id=\"MathJax-Element-23-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-144\" class=\"math\"><span id=\"MathJax-Span-145\" class=\"mrow\"><span id=\"MathJax-Span-146\" class=\"mfrac\"><span id=\"MathJax-Span-147\" class=\"msubsup\"><span id=\"MathJax-Span-148\" class=\"texatom\"><span id=\"MathJax-Span-149\" class=\"mrow\"><span id=\"MathJax-Span-150\" class=\"mn\">120<\/span><\/span><\/span><span id=\"MathJax-Span-151\" class=\"texatom\"><span id=\"MathJax-Span-152\" class=\"mrow\"><span id=\"MathJax-Span-153\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-154\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 60\u00b0<br \/>Hence \u2220A = \u2220B = \u2220C = 60\u00b0<\/p>\n<p>Question 12.<br \/>If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.<br \/>Solution:<br \/>In a \u2206ABC,<br \/>Let \u2220A &lt; \u2220B + \u2220C<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1965\/45593358422_f37d88c7cf_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"214\" height=\"193\" \/><br \/>\u21d2\u2220A + \u2220A &lt; \u2220A + \u2220B + \u2220C<br \/>\u21d2 2\u2220A &lt; 180\u00b0<br \/>\u21d2 \u2220A &lt; 90\u00b0 (\u2235 Sum of angles of a triangle is 180\u00b0)<br \/>Similarly, we can prove that<br \/>\u2220B &lt; 90\u00b0 and \u2220C &lt; 90\u00b0<br \/>\u2234 Each angle of the triangle are acute angle.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-11-coordinate-geometry-ex-112\"><\/span>RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry Ex 11.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>The exterior angles obtained on producing the base of a triangle both ways are 104\u00b0 and 136\u00b0. Find all the angles of the triangle.<br \/>Solution:<br \/>In \u2206ABC, base BC is produced both ways to D and E respectivley forming \u2220ABE = 104\u00b0 and \u2220ACD = 136\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/30703039077_e49a07397c_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"353\" height=\"507\" \/><\/p>\n<p>Question 2.<br \/>In the figure, the sides BC, CA and AB of a \u2206ABC have been produced to D, E and F respectively. If \u2220ACD = 105\u00b0 and \u2220EAF = 45\u00b0, find all the angles of the \u2206ABC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/45593358102_a3d2c61d79_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"205\" height=\"202\" \/><br \/>Solution:<br \/>In \u2206ABC, sides BC, CA and BA are produced to D, E and F respectively.<br \/>\u2220ACD = 105\u00b0 and \u2220EAF = 45\u00b0<br \/>\u2220ACD + \u2220ACB = 180\u00b0 (Linear pair)<br \/>\u21d2 105\u00b0 + \u2220ACB = 180\u00b0<br \/>\u21d2 \u2220ACB = 180\u00b0- 105\u00b0 = 75\u00b0<br \/>\u2220BAC = \u2220EAF (Vertically opposite angles)<br \/>= 45\u00b0<br \/>But \u2220BAC + \u2220ABC + \u2220ACB = 180\u00b0<br \/>\u21d2 45\u00b0 + \u2220ABC + 75\u00b0 = 180\u00b0<br \/>\u21d2 120\u00b0 +\u2220ABC = 180\u00b0<br \/>\u21d2 \u2220ABC = 180\u00b0- 120\u00b0<br \/>\u2234 \u2220ABC = 60\u00b0<br \/>Hence \u2220ABC = 60\u00b0, \u2220BCA = 75\u00b0<br \/>and \u2220BAC = 45\u00b0<\/p>\n<p>Question 3.<br \/>Compute the value of x in each of the following figures:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/30703038737_b28e3e2e33_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"271\" height=\"514\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/45593357912_c8ecf52f5f_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"205\" height=\"256\" \/><br \/>Solution:<br \/>(i) In \u2206ABC, sides BC and CA are produced to D and E respectively<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/45643870021_5b1428fdbb_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"361\" height=\"531\" \/><br \/>(ii) In \u2206ABC, side BC is produced to either side to D and E respectively<br \/>\u2220ABE = 120\u00b0 and \u2220ACD =110\u00b0<br \/>\u2235 \u2220ABE + \u2220ABC = 180\u00b0 (Linear pair)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/45593357622_0e671134b5_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"340\" height=\"469\" \/><br \/>(iii) In the figure, BA || DC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1971\/45643869581_4277c744cb_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"342\" height=\"449\" \/><\/p>\n<p>Question 4.<br \/>In the figure, AC \u22a5 CE and \u2220A: \u2220B : \u2220C = 3:2:1, find the value of \u2220ECD.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/45643869171_a057c46844_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"268\" height=\"161\" \/><br \/>Solution:<br \/>In \u2206ABC, \u2220A : \u2220B : \u2220C = 3 : 2 : 1<br \/>BC is produced to D and CE \u22a5 AC<br \/>\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangles)<br \/>Let\u2220A = 3x, then \u2220B = 2x and \u2220C = x<br \/>\u2234 3x + 2x + x = 180\u00b0 \u21d2 6x = 180\u00b0<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-24-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-155\" class=\"math\"><span id=\"MathJax-Span-156\" class=\"mrow\"><span id=\"MathJax-Span-157\" class=\"mfrac\"><span id=\"MathJax-Span-158\" class=\"msubsup\"><span id=\"MathJax-Span-159\" class=\"texatom\"><span id=\"MathJax-Span-160\" class=\"mrow\"><span id=\"MathJax-Span-161\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-162\" class=\"texatom\"><span id=\"MathJax-Span-163\" class=\"mrow\"><span id=\"MathJax-Span-164\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-165\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>\u00a0 = 30\u00b0<br \/>\u2234 \u2220A = 3x = 3 x 30\u00b0 = 90\u00b0<br \/>\u2220B = 2x = 2 x 30\u00b0 = 60\u00b0<br \/>\u2220C = x = 30\u00b0<br \/>In \u2206ABC,<br \/>Ext. \u2220ACD = \u2220A + \u2220B<br \/>\u21d2 90\u00b0 + \u2220ECD = 90\u00b0 + 60\u00b0 = 150\u00b0<br \/>\u2234 \u2220ECD = 150\u00b0-90\u00b0 = 60\u00b0<\/p>\n<p>Question 5.<br \/>In the figure, AB || DE, find \u2220ACD.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/45593357032_3d6c2881aa_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry\" width=\"277\" height=\"221\" \/><br \/>Solution:<br \/>In the figure, AB || DE<br \/>AE and BD intersect each other at C \u2220BAC = 30\u00b0 and \u2220CDE = 40\u00b0<br \/>\u2235 AB || DE<br \/>\u2234 \u2220ABC = \u2220CDE (Alternate angles)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/45643869001_99ba566f78_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry\" width=\"270\" height=\"221\" \/><br \/>\u21d2 \u2220ABC = 40\u00b0<br \/>In \u2206ABC, BC is produced<br \/>Ext. \u2220ACD = Int. \u2220A + \u2220B<br \/>= 30\u00b0 + 40\u00b0 = 70\u00b0<\/p>\n<p>Question 6.<br \/>Which of the following statements are true (T) and which are false (F):<br \/>(i) Sum of the three angles of a triangle is 180\u00b0.<br \/>(ii) A triangle can have two right angles.<br \/>(iii) All the angles of a triangle can be less than 60\u00b0.<br \/>(iv) All the angles of a triangle can be greater than 60\u00b0.<br \/>(v) All the angles of a triangle can be equal to 60\u00b0.<br \/>(vi) A triangle can have two obtuse angles.<br \/>(vii) A triangle can have at most one obtuse angles.<br \/>(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.<br \/>(ix) An exterior angle of a triangle is less than either of its interior opposite angles.<br \/>(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.<br \/>(xi) An exterior angle of a triangle is greater than the opposite interior angles.<br \/>Solution:<br \/>(i) True.<br \/>(ii) False. A right triangle has only one right angle.<br \/>(iii) False. In this, the sum of three angles will be less than 180\u00b0 which is not true.<br \/>(iv) False. In this, the sum of three angles will be more than 180\u00b0 which is not true.<br \/>(v) True. As sum of three angles will be 180\u00b0 which is true.<br \/>(vi) False. A triangle has only one obtuse angle.<br \/>(vii) True.<br \/>(viii)True.<br \/>(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.<br \/>(x) True.<br \/>(xi) True.<\/p>\n<p>Question 7.<br \/>Fill in the blanks to make the following statements true:<br \/>(i) Sum of the angles of a triangle is \u2026\u2026\u2026<br \/>(ii) An exterior angle of a triangle is equal to the two \u2026\u2026.. opposite angles.<br \/>(iii) An exterior angle of a triangle is always \u2026\u2026.. than either of the interior opposite angles.<br \/>(iv) A triangle cannot have more than \u2026\u2026\u2026. right angles.<br \/>(v) A triangles cannot have more than \u2026\u2026\u2026 obtuse angles.<br \/>Solution:<br \/>(i) Sum of the angles of a triangle is 180\u00b0.<br \/>(ii) An exterior angle of a triangle is equal to the two interior opposite angles.<br \/>(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.<br \/>(iv) A triangle cannot have more than one right angles.<br \/>(v) A triangles cannot have more than one obtuse angles.<\/p>\n<p>Question 8.<br \/>In a \u2206ABC, the internal bisectors of \u2220B and \u2220C meet at P and the external bisectors of \u2220B and \u2220C meet at Q. Prove that \u2220BPC + \u2220BQC = 180\u00b0.<br \/>Solution:<br \/>Given : In \u2206ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior \u2220B and \u2220C meet at P and bisectors of exterior angles B and C meet at Q.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1918\/45593356782_846e6279f7_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"246\" height=\"256\" \/><br \/>To prove : \u2220BPC + \u2220BQC = 180\u00b0<br \/>Proof : \u2235 PB and PC are the internal bisectors of \u2220B and \u2220C<br \/>\u2220BPC = 90\u00b0+\u00a0<span id=\"MathJax-Element-25-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-166\" class=\"math\"><span id=\"MathJax-Span-167\" class=\"mrow\"><span id=\"MathJax-Span-168\" class=\"mfrac\"><span id=\"MathJax-Span-169\" class=\"mn\">1<\/span><span id=\"MathJax-Span-170\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A \u2026(i)<br \/>Similarly, QB and QC are the bisectors of exterior angles B and C<br \/>\u2234 \u2220BQC = 90\u00b0 +\u00a0<span id=\"MathJax-Element-26-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-171\" class=\"math\"><span id=\"MathJax-Span-172\" class=\"mrow\"><span id=\"MathJax-Span-173\" class=\"mfrac\"><span id=\"MathJax-Span-174\" class=\"mn\">1<\/span><span id=\"MathJax-Span-175\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A \u2026(ii)<br \/>Adding (i) and (ii),<br \/>\u2220BPC + \u2220BQC = 90\u00b0 +\u00a0<span id=\"MathJax-Element-27-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-176\" class=\"math\"><span id=\"MathJax-Span-177\" class=\"mrow\"><span id=\"MathJax-Span-178\" class=\"mfrac\"><span id=\"MathJax-Span-179\" class=\"mn\">1<\/span><span id=\"MathJax-Span-180\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A + 90\u00b0 \u2013\u00a0<span id=\"MathJax-Element-28-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-181\" class=\"math\"><span id=\"MathJax-Span-182\" class=\"mrow\"><span id=\"MathJax-Span-183\" class=\"mfrac\"><span id=\"MathJax-Span-184\" class=\"mn\">1<\/span><span id=\"MathJax-Span-185\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A<br \/>= 90\u00b0 + 90\u00b0 = 180\u00b0<br \/>Hence \u2220BPC + \u2220BQC = 180\u00b0<\/p>\n<p>Question 9.<br \/>In the figure, compute the value of x.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1948\/45593356622_a26904c9aa_o.png\" alt=\"RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"229\" height=\"229\" \/><br \/>Solution:<br \/>In the figure,<br \/>\u2220ABC = 45\u00b0, \u2220BAD = 35\u00b0 and \u2220BCD = 50\u00b0 Join BD and produce it E<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/45643868601_2fd5a4eac4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"340\" height=\"468\" \/><\/p>\n<p>Question 10.<br \/>In the figure, AB divides \u2220D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/45643868361_ff6f3b98d0_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"247\" height=\"173\" \/><br \/>Solution:<br \/>In the figure AB = DB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/45593356382_df4f219f84_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"259\" height=\"174\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1927\/30703037367_3fb8dd1e90_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"339\" height=\"423\" \/><\/p>\n<p>Question 11.<br \/>ABC is a triangle. The bisector of the exterior angle at B and the bisector of \u2220C intersect each other at D. Prove that \u2220D =\u00a0<span id=\"MathJax-Element-29-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-186\" class=\"math\"><span id=\"MathJax-Span-187\" class=\"mrow\"><span id=\"MathJax-Span-188\" class=\"mfrac\"><span id=\"MathJax-Span-189\" class=\"mn\">1<\/span><span id=\"MathJax-Span-190\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A.<br \/>Solution:<br \/>Given : In \u2220ABC, CB is produced to E bisectors of ext. \u2220ABE and into \u2220ACB meet at D.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/45593356172_40bff58f4c_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"280\" height=\"371\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45643868071_8f2185882d_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"357\" height=\"410\" \/><\/p>\n<p>Question 12.<br \/>In the figure, AM \u22a5 BC and AN is the bisector of \u2220A. If \u2220B = 65\u00b0 and \u2220C = 33\u00b0, find \u2220MAN.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45593355992_8103016355_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"232\" height=\"156\" \/><br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/30703037067_fb84003d03_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"349\" height=\"305\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1927\/45593355872_1d95c9f5bf_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"348\" height=\"352\" \/><\/p>\n<p>Question 13.<br \/>In a AABC, AD bisects \u2220A and \u2220C &gt; \u2220B. Prove that \u2220ADB &gt; \u2220ADC.<br \/>Solution:<br \/>Given : In \u2206ABC,<br \/>\u2220C &gt; \u2220B and AD is the bisector of \u2220A<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1962\/45593355462_3a93a0243d_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"231\" height=\"194\" \/><br \/>To prove : \u2220ADB &gt; \u2220ADC<br \/>Proof: In \u2206ABC, AD is the bisector of \u2220A<br \/>\u2234 \u22201 = \u22202<br \/>In \u2206ADC,<br \/>Ext. \u2220ADB = \u2220l+ \u2220C<br \/>\u21d2 \u2220C = \u2220ADB \u2013 \u22201 \u2026(i)<br \/>Similarly, in \u2206ABD,<br \/>Ext. \u2220ADC = \u22202 + \u2220B<br \/>\u21d2 \u2220B = \u2220ADC \u2013 \u22202 \u2026(ii)<br \/>From (i) and (ii)<br \/>\u2235 \u2220C &gt; \u2220B (Given)<br \/>\u2234 (\u2220ADB \u2013 \u22201) &gt; (\u2220ADC \u2013 \u22202)<br \/>But \u22201 = \u22202<br \/>\u2234 \u2220ADB &gt; \u2220ADC<\/p>\n<p>Question 14.<br \/>In \u2206ABC, BD \u22a5 AC and CE \u22a5 AB. If BD and CE intersect at O, prove that \u2220BOC = 180\u00b0-\u2220A.<br \/>Solution:<br \/>Given : In \u2206ABC, BD \u22a5 AC and CE\u22a5 AB BD and CE intersect each other at O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/45593355192_62f1936a30_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"280\" height=\"197\" \/><br \/>To prove : \u2220BOC = 180\u00b0 \u2013 \u2220A<br \/>Proof: In quadrilateral ADOE<br \/>\u2220A + \u2220D + \u2220DOE + \u2220E = 360\u00b0 (Sum of angles of quadrilateral)<br \/>\u21d2 \u2220A + 90\u00b0 + \u2220DOE + 90\u00b0 = 360\u00b0<br \/>\u2220A + \u2220DOE = 360\u00b0 \u2013 90\u00b0 \u2013 90\u00b0 = 180\u00b0<br \/>But \u2220BOC = \u2220DOE (Vertically opposite angles)<br \/>\u21d2 \u2220A + \u2220BOC = 180\u00b0<br \/>\u2234 \u2220BOC = 180\u00b0 \u2013 \u2220A<\/p>\n<p>Question 15.<br \/>In the figure, AE bisects \u2220CAD and \u2220B = \u2220C. Prove that AE || BC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/45593354882_d870a95ebe_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"207\" height=\"208\" \/><br \/>Solution:<br \/>Given : In AABC, BA is produced and AE is the bisector of \u2220CAD<br \/>\u2220B = \u2220C<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45593354722_ae04496774_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"202\" height=\"200\" \/><br \/>To prove : AE || BC<br \/>Proof: In \u2206ABC, BA is produced<br \/>\u2234 Ext. \u2220CAD = \u2220B + \u2220C<br \/>\u21d2 2\u2220EAC = \u2220C + \u2220C (\u2235 AE is the bisector of \u2220CAE) (\u2235 \u2220B = \u2220C)<br \/>\u21d2 2\u2220EAC = 2\u2220C<br \/>\u21d2 \u2220EAC = \u2220C<br \/>But there are alternate angles<br \/>\u2234 AE || BC<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-11-coordinate-geometry-vsaqs\"><\/span>RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Define a triangle.<br \/>Solution:<br \/>A figure bounded by three lines segments in a plane is called a triangle.<\/p>\n<p>Question 2.<br \/>Write the sum of the angles of an obtuse triangle.<br \/>Solution:<br \/>The sum of angles of an obtuse triangle is 180\u00b0.<\/p>\n<p>Question 3.<br \/>In \u2206ABC, if \u2220B = 60\u00b0, \u2220C = 80\u00b0 and the bisectors of angles \u2220ABC and \u2220ACB meet at a point O, then find the measure of \u2220BOC.<br \/>Solution:<br \/>In \u2206ABC, \u2220B = 60\u00b0, \u2220C = 80\u00b0<br \/>OB and OC are the bisectors of \u2220B and \u2220C<br \/>\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 \u2220A + 60\u00b0 + 80\u00b0 = 180\u00b0<br \/>\u21d2 \u2220A + 140\u00b0 = 180\u00b0<br \/>\u2234 \u2220A = 180\u00b0- 140\u00b0 = 40\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1936\/31771518608_97b01672cf_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"287\" height=\"316\" \/><br \/>= 90\u00b0 + \u2013 x 40\u00b0 = 90\u00b0 + 20\u00b0 = 110\u00b0<\/p>\n<p>Question 4.<br \/>If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.<br \/>Solution:<br \/>Sum of angles of a triangle = 180\u00b0<br \/>Ratio in the angles = 2 : 1 : 3<br \/>Let first angle = 2x<br \/>Second angle = x<br \/>and third angle = 3x<br \/>\u2234 2x + x + 3x = 180\u00b0 \u21d2 6x = 180\u00b0<br \/>\u2234 x =\u00a0<span id=\"MathJax-Element-30-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-191\" class=\"math\"><span id=\"MathJax-Span-192\" class=\"mrow\"><span id=\"MathJax-Span-193\" class=\"mfrac\"><span id=\"MathJax-Span-194\" class=\"msubsup\"><span id=\"MathJax-Span-195\" class=\"texatom\"><span id=\"MathJax-Span-196\" class=\"mrow\"><span id=\"MathJax-Span-197\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-198\" class=\"texatom\"><span id=\"MathJax-Span-199\" class=\"mrow\"><span id=\"MathJax-Span-200\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-201\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>\u00a0 = 30\u00b0<br \/>\u2234 First angle = 2x = 2 x 30\u00b0 = 60\u00b0<br \/>Second angle = x = 30\u00b0<br \/>and third angle = 3x = 3 x 30\u00b0 = 90\u00b0<br \/>Hence angles are 60\u00b0, 30\u00b0, 90\u00b0<\/p>\n<p>Question 5.<br \/>State exterior angle theorem.<br \/>Solution:<br \/>Given : In \u2206ABC, side BC is produced to D<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/45643873851_bd99b68b47_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"268\" height=\"207\" \/><br \/>To prove : \u2220ACD = \u2220A + \u2220B<br \/>Proof: In \u2206ABC,<br \/>\u2220A + \u2220B + \u2220ACB = 180\u00b0 \u2026(i) (Sum of angles of a triangle)<br \/>and \u2220ACD + \u2220ACB = 180\u00b0 \u2026(ii) (Linear pair)<br \/>From (i) and (ii)<br \/>\u2220ACD + \u2220ACB = \u2220A + \u2220B + \u2220ACB<br \/>\u2220ACD = \u2220A + \u2220B<br \/>Hence proved.<\/p>\n<p>Question 6.<br \/>The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.<br \/>Solution:<br \/>In \u2206ABC,<br \/>\u2220A + \u2220C = \u2220B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/45593360382_0f99b8b496_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"246\" height=\"189\" \/><br \/>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br \/>\u2234 \u2220B + \u2220A + \u2220C = 180\u00b0<br \/>\u21d2 \u2220B + \u2220B = 180\u00b0<br \/>\u21d2 2\u2220B = 180\u00b0<br \/>\u21d2 \u2220B =\u00a0<span id=\"MathJax-Element-31-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-202\" class=\"math\"><span id=\"MathJax-Span-203\" class=\"mrow\"><span id=\"MathJax-Span-204\" class=\"mfrac\"><span id=\"MathJax-Span-205\" class=\"msubsup\"><span id=\"MathJax-Span-206\" class=\"texatom\"><span id=\"MathJax-Span-207\" class=\"mrow\"><span id=\"MathJax-Span-208\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-209\" class=\"texatom\"><span id=\"MathJax-Span-210\" class=\"mrow\"><span id=\"MathJax-Span-211\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-212\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 90\u00b0<br \/>\u2234 Third angle = 90\u00b0<\/p>\n<p>Question 7.<br \/>In the figure, if AB || CD, EF || BC, \u2220BAC = 65\u00b0 and \u2220DHF = 35\u00b0, find \u2220AGH.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/31771518268_c234373109_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"314\" height=\"196\" \/><br \/>Solution:<br \/>Given : In figure, AB || CD, EF || BC \u2220BAC = 65\u00b0, \u2220DHF = 35\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45593360252_0f22b72aac_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"310\" height=\"203\" \/><br \/>\u2235 EF || BC<br \/>\u2234 \u2220A = \u2220ACH (Alternate angle)<br \/>\u2234 \u2220ACH = 65\u00b0<br \/>\u2235\u2220GHC = \u2220DHF<br \/>(Vertically opposite angles)<br \/>\u2234 \u2220GHC = 35\u00b0<br \/>Now in \u2206GCH,<br \/>Ext. \u2220AGH = \u2220GCH + \u2220GHC<br \/>= 65\u00b0 + 35\u00b0 = 100\u00b0<\/p>\n<p>Question 8.<br \/>In the figure, if AB || DE and BD || FG such that \u2220FGH = 125\u00b0 and \u2220B = 55\u00b0, find x and y.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45593360182_d55798124c_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"349\" height=\"262\" \/><br \/>Solution:<br \/>In the figure, AB || DF, BD || FG<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/30703040957_cd16679562_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"347\" height=\"257\" \/><br \/>\u2220FGH = 125\u00b0 and \u2220B = 55\u00b0<br \/>\u2220FGH + FGE = 180\u00b0 (Linear pair)<br \/>\u21d2 125\u00b0 + y \u2013 180\u00b0<br \/>\u21d2 y= 180\u00b0- 125\u00b0 = 55\u00b0<br \/>\u2235 BA || FD and BD || FG<br \/>\u2220B = \u2220F = 55\u00b0<br \/>Now in \u2206EFG,<br \/>\u2220F + \u2220FEG + \u2220FGE = 180\u00b0<br \/>(Angles of a triangle)<br \/>\u21d2 55\u00b0 + x + 55\u00b0 = 180\u00b0<br \/>\u21d2 x+ 110\u00b0= 180\u00b0<br \/>\u2234 x= 180\u00b0- 110\u00b0 = 70\u00b0<br \/>Hence x = 70, y = 55\u00b0<\/p>\n<p>Question 9.<br \/>If the angles A, B and C of \u2206ABC satisfy the relation B \u2013 A = C \u2013 B, then find the measure of \u2220B.<br \/>Solution:<br \/>In \u2206ABC,<br \/>\u2220A + \u2220B + \u2220C= 180\u00b0 \u2026(i)<br \/>and B \u2013 A = C \u2013 B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/45593360002_b52f470a73_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"225\" height=\"170\" \/><br \/>\u21d2 B + B = A + C \u21d2 2B = A + C<br \/>From (i),<br \/>B + 2B = 180\u00b0 \u21d2 3B = 180\u00b0<br \/>\u2220B =\u00a0<span id=\"MathJax-Element-32-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-213\" class=\"math\"><span id=\"MathJax-Span-214\" class=\"mrow\"><span id=\"MathJax-Span-215\" class=\"mfrac\"><span id=\"MathJax-Span-216\" class=\"msubsup\"><span id=\"MathJax-Span-217\" class=\"texatom\"><span id=\"MathJax-Span-218\" class=\"mrow\"><span id=\"MathJax-Span-219\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-220\" class=\"texatom\"><span id=\"MathJax-Span-221\" class=\"mrow\"><span id=\"MathJax-Span-222\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-223\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0= 60\u00b0<br \/>Hence \u2220B = 60\u00b0<\/p>\n<p>Question 10.<br \/>In \u2206ABC, if bisectors of \u2220ABC and \u2220ACB intersect at O at angle of 120\u00b0, then find the measure of \u2220A.<br \/>Solution:<br \/>In \u2206ABC, bisectors of \u2220B and \u2220C intersect at O and \u2220BOC = 120\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/30703040777_27f07730c5_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"282\" height=\"189\" \/><br \/>But \u2220BOC = 90\u00b0+\u00a0<span id=\"MathJax-Element-33-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-224\" class=\"math\"><span id=\"MathJax-Span-225\" class=\"mrow\"><span id=\"MathJax-Span-226\" class=\"mfrac\"><span id=\"MathJax-Span-227\" class=\"mn\">1<\/span><span id=\"MathJax-Span-228\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br \/>90\u00b0+\u00a0<span id=\"MathJax-Element-34-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-229\" class=\"math\"><span id=\"MathJax-Span-230\" class=\"mrow\"><span id=\"MathJax-Span-231\" class=\"mfrac\"><span id=\"MathJax-Span-232\" class=\"mn\">1<\/span><span id=\"MathJax-Span-233\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A= 120\u00b0<br \/>\u21d2\u00a0<span id=\"MathJax-Element-35-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-234\" class=\"math\"><span id=\"MathJax-Span-235\" class=\"mrow\"><span id=\"MathJax-Span-236\" class=\"mfrac\"><span id=\"MathJax-Span-237\" class=\"mn\">1<\/span><span id=\"MathJax-Span-238\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A= 120\u00b0-90\u00b0 = 30\u00b0<br \/>\u2234 \u2220A = 2 x 30\u00b0 = 60\u00b0<\/p>\n<p>Question 11.<br \/>If the side BC of \u2206ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and \u2220A.<br \/>Solution:<br \/>In \u2206ABC, side BC is produced on both sides forming exterior \u2220ABE and \u2220ACD<br \/>Ext. \u2220ABE = \u2220A + \u2220ACB<br \/>and Ext. \u2220ACD = \u2220ABC + \u2220A<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1925\/30703040647_79722b213b_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"272\" height=\"165\" \/><br \/>Adding we get,<br \/>\u2220ABE + \u2220ACD = \u2220A + \u2220ACB + \u2220A + \u2220ABC<br \/>\u21d2 \u2220ABE + \u2220ACD \u2013 \u2220A = \u2220A 4- \u2220ACB + \u2220A + \u2220ABC \u2013 \u2220A (Subtracting \u2220A from both sides)<br \/>= \u2220A + \u2220ABC + \u2220ACB = \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<\/p>\n<p>Question 12.<br \/>In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find \u2220ACD: \u2220ADC.<br \/>Solution:<br \/>In \u2206ABC, AB = AC<br \/>AB is produced to D such that BD = BC<br \/>DC are joined<br \/>In \u2206ABC, AB = AC<br \/>\u2234 \u2220ABC = \u2220ACB<br \/>In \u2206 BCD, BD = BC<br \/>\u2234 \u2220BDC = \u2220BCD<br \/>and Ext. \u2220ABC = \u2220BDC + \u2220BCD = 2\u2220BDC (\u2235 \u2220BDC = \u2220BCD)<br \/>\u21d2 \u2220ACB = 2\u2220BCD (\u2235 \u2220ABC = \u2220ACB)<br \/>Adding \u2220BDC to both sides<br \/>\u21d2 \u2220ACB + \u2220BDC = 2\u2220BDC + \u2220BDC<br \/>\u21d2 \u2220ACB + \u2220BCD = 3 \u2220BDC (\u2235 \u2220BDC = \u2220BCD)<br \/>\u21d2 \u2220ACB = 3\u2220BDC<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1916\/30703040537_22366857b0_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"335\" height=\"82\" \/><\/p>\n<p>Question 13.<br \/>In the figure, side BC of AABC is produced to point D such that bisectors of \u2220ABC and \u2220ACD meet at a point E. If \u2220BAC = 68\u00b0, find \u2220BEC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/45593359702_c97068d33b_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"318\" height=\"226\" \/><br \/>Solution:<br \/>In the figure,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/30703040377_07031df438_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry\" width=\"323\" height=\"238\" \/><br \/>side BC of \u2206ABC is produced to D such that bisectors of \u2220ABC and \u2220ACD meet at E<br \/>\u2220BAC = 68\u00b0<br \/>In \u2206ABC,<br \/>Ext. \u2220ACD = \u2220A + \u2220B<br \/>\u21d2\u00a0<span id=\"MathJax-Element-36-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-239\" class=\"math\"><span id=\"MathJax-Span-240\" class=\"mrow\"><span id=\"MathJax-Span-241\" class=\"mfrac\"><span id=\"MathJax-Span-242\" class=\"mn\">1<\/span><span id=\"MathJax-Span-243\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220ACD =\u00a0<span id=\"MathJax-Element-37-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-244\" class=\"math\"><span id=\"MathJax-Span-245\" class=\"mrow\"><span id=\"MathJax-Span-246\" class=\"mfrac\"><span id=\"MathJax-Span-247\" class=\"mn\">1<\/span><span id=\"MathJax-Span-248\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A +\u00a0<span id=\"MathJax-Element-38-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-249\" class=\"math\"><span id=\"MathJax-Span-250\" class=\"mrow\"><span id=\"MathJax-Span-251\" class=\"mfrac\"><span id=\"MathJax-Span-252\" class=\"mn\">1<\/span><span id=\"MathJax-Span-253\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220B<br \/>\u21d2 \u22202=\u00a0<span id=\"MathJax-Element-39-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-254\" class=\"math\"><span id=\"MathJax-Span-255\" class=\"mrow\"><span id=\"MathJax-Span-256\" class=\"mfrac\"><span id=\"MathJax-Span-257\" class=\"mn\">1<\/span><span id=\"MathJax-Span-258\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A + \u22201 \u2026(i)<br \/>But in \u2206BCE,<br \/>Ext. \u22202 = \u2220E + \u2220l<br \/>\u21d2 \u2220E + \u2220l = \u22202 =\u00a0<span id=\"MathJax-Element-40-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-259\" class=\"math\"><span id=\"MathJax-Span-260\" class=\"mrow\"><span id=\"MathJax-Span-261\" class=\"mfrac\"><span id=\"MathJax-Span-262\" class=\"mn\">1<\/span><span id=\"MathJax-Span-263\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A + \u2220l [From (i)]<br \/>\u21d2 \u2220E =\u00a0<span id=\"MathJax-Element-41-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-264\" class=\"math\"><span id=\"MathJax-Span-265\" class=\"mrow\"><span id=\"MathJax-Span-266\" class=\"mfrac\"><span id=\"MathJax-Span-267\" class=\"mn\">1<\/span><span id=\"MathJax-Span-268\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A =\u00a0<span id=\"MathJax-Element-42-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-269\" class=\"math\"><span id=\"MathJax-Span-270\" class=\"mrow\"><span id=\"MathJax-Span-271\" class=\"mfrac\"><span id=\"MathJax-Span-272\" class=\"msubsup\"><span id=\"MathJax-Span-273\" class=\"texatom\"><span id=\"MathJax-Span-274\" class=\"mrow\"><span id=\"MathJax-Span-275\" class=\"mn\">68<\/span><\/span><\/span><span id=\"MathJax-Span-276\" class=\"texatom\"><span id=\"MathJax-Span-277\" class=\"mrow\"><span id=\"MathJax-Span-278\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-279\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 =34\u00b0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-rd-sharma-solutions-chapter-11-coordinate-geometry-mcqs\"><\/span>Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Mark the correct alternative in each of the following:<br \/>Question 1.<br \/>If all the three angles of a triangle are equal, then each one of them is equal to<br \/>(a) 90\u00b0<br \/>(b) 45\u00b0<br \/>(c) 60\u00b0<br \/>(d) 30\u00b0<br \/>Solution:<br \/>\u2235 Sum of three angles of a triangle = 180\u00b0<br \/>\u2234 Each angle =\u00a0<span id=\"MathJax-Element-43-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-280\" class=\"math\"><span id=\"MathJax-Span-281\" class=\"mrow\"><span id=\"MathJax-Span-282\" class=\"mfrac\"><span id=\"MathJax-Span-283\" class=\"msubsup\"><span id=\"MathJax-Span-284\" class=\"texatom\"><span id=\"MathJax-Span-285\" class=\"mrow\"><span id=\"MathJax-Span-286\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-287\" class=\"texatom\"><span id=\"MathJax-Span-288\" class=\"mrow\"><span id=\"MathJax-Span-289\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-290\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0 = 60\u00b0 (c)<\/p>\n<p>Question 2.<br \/>If two acute angles of a right triangle are equal, then each acute is equal to<br \/>(a) 30\u00b0<br \/>(b) 45\u00b0<br \/>(c) 60\u00b0<br \/>(d) 90\u00b0<br \/>Solution:<br \/>In a right triangle, one angle = 90\u00b0<br \/>\u2234 Sum of other two acute angles = 180\u00b0 \u2013 90\u00b0 = 90\u00b0<br \/>\u2235 Both angles are equal<br \/>\u2234 Each angle will be =\u00a0<span id=\"MathJax-Element-44-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-291\" class=\"math\"><span id=\"MathJax-Span-292\" class=\"mrow\"><span id=\"MathJax-Span-293\" class=\"mfrac\"><span id=\"MathJax-Span-294\" class=\"msubsup\"><span id=\"MathJax-Span-295\" class=\"texatom\"><span id=\"MathJax-Span-296\" class=\"mrow\"><span id=\"MathJax-Span-297\" class=\"mn\">90<\/span><\/span><\/span><span id=\"MathJax-Span-298\" class=\"texatom\"><span id=\"MathJax-Span-299\" class=\"mrow\"><span id=\"MathJax-Span-300\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-301\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 45\u00b0 (b)<\/p>\n<p>Question 3.<br \/>An exterior angle of a triangle is equal to 100\u00b0 and two interior opposite angles are equal. Each of these angles is equal to<br \/>(a) 75\u00b0<br \/>(b) 80\u00b0<br \/>(c) 40\u00b0<br \/>(d) 50\u00b0<br \/>Solution:<br \/>In a triangle, exterior angles is equal to the sum of its interior opposite angles<br \/>\u2234 Sum of interior opposite angles = 100\u00b0<br \/>\u2235 Both angles are equal<br \/>\u2234 Each angle will be =\u00a0<span id=\"MathJax-Element-45-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-302\" class=\"math\"><span id=\"MathJax-Span-303\" class=\"mrow\"><span id=\"MathJax-Span-304\" class=\"mfrac\"><span id=\"MathJax-Span-305\" class=\"msubsup\"><span id=\"MathJax-Span-306\" class=\"texatom\"><span id=\"MathJax-Span-307\" class=\"mrow\"><span id=\"MathJax-Span-308\" class=\"mn\">100<\/span><\/span><\/span><span id=\"MathJax-Span-309\" class=\"texatom\"><span id=\"MathJax-Span-310\" class=\"mrow\"><span id=\"MathJax-Span-311\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-312\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 50\u00b0 (d)<\/p>\n<p>Question 4.<br \/>If one angle of a triangle is equal to the sum of the other two angles, then the triangle is<br \/>(a) an isosceles triangle<br \/>(b) an obtuse triangle<br \/>(c) an equilateral triangle<br \/>(d) a right triangle<br \/>Solution:<br \/>Let \u2220A, \u2220B, \u2220C be the angles of a \u2206ABC and let \u2220A = \u2220B + \u2220C<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/31771526488_2d8ddd2a38_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"263\" height=\"183\" \/><br \/>But \u2220A + \u2220B + \u2220C = 180\u00b0<br \/>( Sum of angles of a triangle)<br \/>\u2234 \u2220A + \u2220A = 180\u00b0 \u21d2 2\u2220A = 180\u00b0<br \/>\u21d2 \u2220A =\u00a0<span id=\"MathJax-Element-46-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-313\" class=\"math\"><span id=\"MathJax-Span-314\" class=\"mrow\"><span id=\"MathJax-Span-315\" class=\"mfrac\"><span id=\"MathJax-Span-316\" class=\"msubsup\"><span id=\"MathJax-Span-317\" class=\"texatom\"><span id=\"MathJax-Span-318\" class=\"mrow\"><span id=\"MathJax-Span-319\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-320\" class=\"texatom\"><span id=\"MathJax-Span-321\" class=\"mrow\"><span id=\"MathJax-Span-322\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-323\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 90\u00b0<br \/>\u2234 \u2206 is a right triangle (d)<\/p>\n<p>Question 5.<br \/>Side BC of a triangle ABC has been produced to a point D such that \u2220ACD = 120\u00b0. If \u2220B =\u00a0<span id=\"MathJax-Element-47-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-324\" class=\"math\"><span id=\"MathJax-Span-325\" class=\"mrow\"><span id=\"MathJax-Span-326\" class=\"mfrac\"><span id=\"MathJax-Span-327\" class=\"mn\">1<\/span><span id=\"MathJax-Span-328\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220A, then \u2220A is equal to<br \/>(a) 80\u00b0<br \/>(b) 75\u00b0<br \/>(c) 60\u00b0<br \/>(d) 90\u00b0<br \/>Solution:<br \/>Side BC of \u2206ABC is produced to D, then<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/31771526298_f7872ee4f0_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"298\" height=\"187\" \/><br \/>Ext. \u2220ACB = \u2220A + \u2220B<br \/>(Exterior angle of a triangle is equal to the sum of its interior opposite angles)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/31771526078_a4c500df3b_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"340\" height=\"206\" \/><\/p>\n<p>Question 6.<br \/>In \u2206ABC, \u2220B = \u2220C and ray AX bisects the exterior angle \u2220DAC. If \u2220DAX = 70\u00b0, then \u2220ACB =<br \/>(a) 35\u00b0<br \/>(b) 90\u00b0<br \/>(c) 70\u00b0<br \/>(d) 55\u00b0<br \/>Solution:<br \/>In \u2206ABC, \u2220B = \u2220C<br \/>AX is the bisector of ext. \u2220CAD<br \/>\u2220DAX = 70\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/31771525848_a55c843599_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"223\" height=\"237\" \/><br \/>\u2234 \u2220DAC = 70\u00b0 x 2 = 140\u00b0<br \/>But Ext. \u2220DAC = \u2220B + \u2220C<br \/>= \u2220C + \u2220C (\u2235 \u2220B = \u2220C)<br \/>= 2\u2220C<br \/>\u2234 2\u2220C = 140\u00b0 \u21d2 \u2220C =\u00a0<span id=\"MathJax-Element-48-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-329\" class=\"math\"><span id=\"MathJax-Span-330\" class=\"mrow\"><span id=\"MathJax-Span-331\" class=\"mfrac\"><span id=\"MathJax-Span-332\" class=\"msubsup\"><span id=\"MathJax-Span-333\" class=\"texatom\"><span id=\"MathJax-Span-334\" class=\"mrow\"><span id=\"MathJax-Span-335\" class=\"mn\">140<\/span><\/span><\/span><span id=\"MathJax-Span-336\" class=\"texatom\"><span id=\"MathJax-Span-337\" class=\"mrow\"><span id=\"MathJax-Span-338\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-339\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 70\u00b0<br \/>\u2234 \u2220ACB = 70\u00b0 (c)<\/p>\n<p>Question 7.<br \/>In a triangle, an exterior angle at a vertex is 95\u00b0 and its one of the interior opposite angle is 55\u00b0, then the measure of the other interior angle is<br \/>(a) 55\u00b0<br \/>(b) 85\u00b0<br \/>(c) 40\u00b0<br \/>(d) 9.0\u00b0<br \/>Solution:<br \/>In \u2206ABC, BA is produced to D such that \u2220CAD = 95\u00b0<br \/>and let \u2220C = 55\u00b0 and \u2220B = x\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/31771525548_0d22741b02_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"245\" height=\"225\" \/><br \/>\u2235 Exterior angle of a triangle is equal to the sum of its opposite interior angle<br \/>\u2234 \u2220CAD = \u2220B + \u2220C \u21d2 95\u00b0 = x + 55\u00b0<br \/>\u21d2 x = 95\u00b0 \u2013 55\u00b0 = 40\u00b0<br \/>\u2234 Other interior angle = 40\u00b0 (c)<\/p>\n<p>Question 8.<br \/>If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is<br \/>(a) 90\u00b0<br \/>(b) 180\u00b0<br \/>(c) 270\u00b0<br \/>(d) 360\u00b0<br \/>Solution:<br \/>In \u2206ABC, sides AB, BC and CA are produced in order, then exterior \u2220FAB, \u2220DBC and \u2220ACE are formed<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1918\/31771525318_3697f42151_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"223\" height=\"253\" \/><br \/>We know an exterior angles of a triangle is equal to the sum of its interior opposite angles<br \/>\u2234 \u2220FAB = \u2220B + \u2220C<br \/>\u2220DBC = \u2220C + \u2220A and<br \/>\u2220ACE = \u2220A + \u2220B Adding we get,<br \/>\u2220FAB + \u2220DBC + \u2220ACE = \u2220B + \u2220C + \u2220C + \u2220A + \u2220A + \u2220B<br \/>= 2(\u2220A + \u2220B + \u2220C)<br \/>= 2 x 180\u00b0 (Sum of angles of a triangle)<br \/>= 360\u00b0 (d)<\/p>\n<p>Question 9.<br \/>In \u2206ABC, if \u2220A = 100\u00b0, AD bisects \u2220A and AD\u22a5 BC. Then, \u2220B =<br \/>(a) 50\u00b0<br \/>(b) 90\u00b0<br \/>(c) 40\u00b0<br \/>(d) 100\u00b0<br \/>Solution:<br \/>In \u2206ABC, \u2220A = 100\u00b0<br \/>AD is bisector of \u2220A and AD \u22a5 BC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/31771525078_187e994d2b_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"262\" height=\"222\" \/><br \/>Now, \u2220BAD =\u00a0<span id=\"MathJax-Element-49-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-340\" class=\"math\"><span id=\"MathJax-Span-341\" class=\"mrow\"><span id=\"MathJax-Span-342\" class=\"mfrac\"><span id=\"MathJax-Span-343\" class=\"msubsup\"><span id=\"MathJax-Span-344\" class=\"texatom\"><span id=\"MathJax-Span-345\" class=\"mrow\"><span id=\"MathJax-Span-346\" class=\"mn\">100<\/span><\/span><\/span><span id=\"MathJax-Span-347\" class=\"texatom\"><span id=\"MathJax-Span-348\" class=\"mrow\"><span id=\"MathJax-Span-349\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-350\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 50\u00b0<br \/>In \u2206ABD,<br \/>\u2220BAD + \u2220B + \u2220D= 180\u00b0<br \/>(Sum of angles of a triangle)<br \/>\u21d2 \u222050\u00b0 + \u2220B + 90\u00b0 = 180\u00b0<br \/>\u2220B + 140\u00b0 = 180\u00b0<br \/>\u21d2 \u2220B = 180\u00b0 \u2013 140\u00b0 \u2220B = 40\u00b0 (c)<\/p>\n<p>Question 10.<br \/>An exterior angle of a triangle is 108\u00b0 and its interior opposite angles are in the ratio 4:5. The angles of the triangle are<br \/>(a) 48\u00b0, 60\u00b0, 72\u00b0<br \/>(b) 50\u00b0, 60\u00b0, 70\u00b0<br \/>(c) 52\u00b0, 56\u00b0, 72\u00b0<br \/>(d) 42\u00b0, 60\u00b0, 76\u00b0<br \/>Solution:<br \/>In \u2206ABC, BC is produced to D and \u2220ACD = 108\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/31771524628_b32f57a76b_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"254\" height=\"217\" \/><br \/>Ratio in \u2220A : \u2220B = 4:5<br \/>\u2235 Exterior angle of a triangle is equal to the sum of its opposite interior angles<br \/>\u2234 \u2220ACD = \u2220A + \u2220B = 108\u00b0<br \/>Ratio in \u2220A : \u2220B = 4:5<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1947\/31771524898_06d6d23cb6_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"343\" height=\"184\" \/><\/p>\n<p>Question 11.<br \/>In a \u2206ABC, if \u2220A = 60\u00b0, \u2220B = 80\u00b0 and the bisectors of \u2220B and \u2220C meet at O, then \u2220BOC =<br \/>(a) 60\u00b0<br \/>(b) 120\u00b0<br \/>(c) 150\u00b0<br \/>(d) 30\u00b0<br \/>Solution:<br \/>In \u2206ABC, \u2220A = 60\u00b0, \u2220B = 80\u00b0<br \/>\u2234 \u2220C = 180\u00b0 \u2013 (\u2220A + \u2220B)<br \/>= 180\u00b0 \u2013 (60\u00b0 + 80\u00b0)<br \/>= 180\u00b0 \u2013 140\u00b0 = 40\u00b0<br \/>Bisectors of \u2220B and \u2220C meet at O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/31771524508_00675cf207_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"239\" height=\"165\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/31771524338_04cca3e841_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"343\" height=\"106\" \/><\/p>\n<p>Question 12.<br \/>Line segments AB and CD intersect at O such that AC || DB. If \u2220CAB = 45\u00b0 and \u2220CDB = 55\u00b0, then \u2220BOD =<br \/>(a) 100\u00b0<br \/>(b) 80\u00b0<br \/>(c) 90\u00b0<br \/>(d) 135\u00b0<br \/>Solution:<br \/>In the figure,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1952\/31771524098_fb2d48cb8d_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry\" width=\"172\" height=\"227\" \/><br \/>AB and CD intersect at O<br \/>and AC || DB, \u2220CAB = 45\u00b0<br \/>and \u2220CDB = 55\u00b0<br \/>\u2235 AC || DB<br \/>\u2234 \u2220CAB = \u2220ABD (Alternate angles)<br \/>In \u2206OBD,<br \/>\u2220BOD = 180\u00b0 \u2013 (\u2220CDB + \u2220ABD)<br \/>= 180\u00b0 \u2013 (55\u00b0 + 45\u00b0)<br \/>= 180\u00b0 \u2013 100\u00b0 = 80\u00b0 (b)<\/p>\n<p>Question 13.<br \/>In the figure, if EC || AB, \u2220ECD = 70\u00b0 and \u2220BDO = 20\u00b0, then \u2220OBD is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1951\/31771523918_bb7d8369e3_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry\" width=\"253\" height=\"233\" \/><br \/>(a) 20\u00b0<br \/>(b) 50\u00b0<br \/>(c) 60\u00b0<br \/>(d) 70\u00b0<br \/>Solution:<br \/>In the figure, EC || AB<br \/>\u2220ECD = 70\u00b0, \u2220BDO = 20\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/31771523768_fcea2cc549_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"256\" height=\"230\" \/><br \/>\u2235 EC || AB<br \/>\u2220AOD = \u2220ECD (Corresponding angles)<br \/>\u21d2 \u2220AOD = 70\u00b0<br \/>In \u2206OBD,<br \/>Ext. \u2220AOD = \u2220OBD + \u2220BDO<br \/>70\u00b0 = \u2220OBD + 20\u00b0<br \/>\u21d2 \u2220OBD = 70\u00b0 \u2013 20\u00b0 = 50\u00b0 (b)<\/p>\n<p>Question 14.<br \/>In the figure, x + y =<br \/>(a) 270<br \/>(b) 230<br \/>(c) 210<br \/>(d) 190\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/31771523578_0f93a80d33_o.png\" alt=\"RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"360\" height=\"183\" \/><br \/>Solution:<br \/>In the figure<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/31771523278_1759b99d41_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"362\" height=\"196\" \/><br \/>Ext. \u2220OAE = \u2220AOC + \u2220ACO<br \/>\u21d2 x = 40\u00b0 + 80\u00b0 = 120\u00b0<br \/>Similarly,<br \/>Ext. \u2220DBF = \u2220ODB + \u2220DOB<br \/>y = 70\u00b0 + \u2220DOB<br \/>[(\u2235 \u2220AOC = \u2220DOB) (vertically opp. angles)]<br \/>= 70\u00b0 + 40\u00b0 = 110\u00b0<br \/>\u2234 x+y= 120\u00b0+ 110\u00b0 = 230\u00b0 (b)<\/p>\n<p>Question 15.<br \/>If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?<br \/>(a) 25\u00b0<br \/>(b) 30\u00b0<br \/>(c) 45\u00b0<br \/>(d) 60\u00b0<br \/>Solution:<br \/>Ratio in the measures of the triangle =3:4:5<br \/>Sum of angles of a triangle = 180\u00b0<br \/>Let angles be 3x, 4x, 5x<br \/>Sum of angles = 3x + 4x + 5x = 12x<br \/>\u2234 Smallest angle =\u00a0<span id=\"MathJax-Element-50-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-351\" class=\"math\"><span id=\"MathJax-Span-352\" class=\"mrow\"><span id=\"MathJax-Span-353\" class=\"mfrac\"><span id=\"MathJax-Span-354\" class=\"mrow\"><span id=\"MathJax-Span-355\" class=\"mn\">180<\/span><span id=\"MathJax-Span-356\" class=\"mi\">x<\/span><span id=\"MathJax-Span-357\" class=\"mn\">3<\/span><span id=\"MathJax-Span-358\" class=\"mi\">x<\/span><\/span><span id=\"MathJax-Span-359\" class=\"mrow\"><span id=\"MathJax-Span-360\" class=\"mn\">12<\/span><span id=\"MathJax-Span-361\" class=\"mi\">x<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= 45\u00b0 (c)<\/p>\n<p>Question 16.<br \/>In the figure, if AB \u22a5 BC, then x =<br \/>(a) 18<br \/>(b) 22<br \/>(c) 25<br \/>(d) 32<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/31771523138_c86807b0a9_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"255\" height=\"236\" \/><br \/>Solution:<br \/>In the figure, AB \u22a5 BC<br \/>\u2220AGF = 32\u00b0<br \/>\u2234 \u2220CGB = \u2220AGF (Vertically opposite angles)<br \/>= 32\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/31771522888_b62f5c3e0b_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"296\" height=\"234\" \/><br \/>In \u2206GCB, \u2220B = 90\u00b0<br \/>\u2234 \u2220CGB + \u2220GCB = 90\u00b0<br \/>\u21d2 32\u00b0 + \u2220GCB = 90\u00b0<br \/>\u21d2 \u2220GCB = 90\u00b0 \u2013 32\u00b0 = 58\u00b0<br \/>Now in \u2206GDC,<br \/>Ext. \u2220GCB = \u2220CDG + \u2220DGC<br \/>\u21d2 58\u00b0 = x + 14\u00b0 + x<br \/>\u21d2 2x + 14\u00b0 = 58\u00b0<br \/>\u21d2 2x = 58 \u2013 14\u00b0 = 44<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-51-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-362\" class=\"math\"><span id=\"MathJax-Span-363\" class=\"mrow\"><span id=\"MathJax-Span-364\" class=\"mfrac\"><span id=\"MathJax-Span-365\" class=\"mn\">44<\/span><span id=\"MathJax-Span-366\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 22\u00b0<br \/>\u2234 x = 22\u00b0 (b)<\/p>\n<p>Question 17.<br \/>In the figure, what is \u2220 in terms of x and y?<br \/>(a) x + y + 180<br \/>(b) x + y \u2013 180<br \/>(c) 180\u00b0 -(x+y)<br \/>(d) x+y + 360\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1931\/31771522658_fc4cf24a86_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"272\" height=\"200\" \/><br \/>Solution:<br \/>In the figure, BC is produced both sides CA and BA are also produced<br \/>In \u2206ABC,<br \/>\u2220B = 180\u00b0 -y<br \/>and \u2220C 180\u00b0 \u2013 x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1965\/31771522378_1181db9809_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"277\" height=\"196\" \/><br \/>\u2234 z = \u2220A = 180\u00b0 \u2013 (B + C)<br \/>= 180\u00b0 \u2013 (180 \u2013 y + 180 -x)<br \/>= 180\u00b0 \u2013 (360\u00b0 \u2013 x \u2013 y)<br \/>= 180\u00b0 \u2013 360\u00b0 + x + y = x + y \u2013 180\u00b0 (b)<\/p>\n<p>Question 18.<br \/>In the figure, for which value of x is l<sub>1<\/sub>\u00a0|| l<sub>2<\/sub>?<br \/>(a) 37<br \/>(b) 43<br \/>(c) 45<br \/>(d) 47<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1965\/31771522158_03d94f2e14_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"314\" height=\"172\" \/><br \/>Solution:<br \/>In the figure, l<sub>1<\/sub>\u00a0|| l<sub>2<\/sub><br \/>\u2234 \u2220EBA = \u2220BAH (Alternate angles)<br \/>\u2234 \u2220BAH = 78\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/31771521998_5b82383db2_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"309\" height=\"181\" \/><br \/>\u21d2 \u2220BAC + \u2220CAH = 78\u00b0<br \/>\u21d2 \u2220BAC + 35\u00b0 = 78\u00b0<br \/>\u21d2 \u2220BAC = 78\u00b0 \u2013 35\u00b0 = 43\u00b0<br \/>In \u2206ABC, \u2220C = 90\u00b0<br \/>\u2234 \u2220ABC + \u2220BAC = 90\u00b0<br \/>\u21d2 x + 43\u00b0 = 90\u00b0 \u21d2 x = 90\u00b0 \u2013 43\u00b0<br \/>\u2234 x = 47\u00b0 (d)<\/p>\n<p>Question 19.<br \/>In the figure, what is y in terms of x?<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/31771521978_bfcc734741_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"319\" height=\"322\" \/><br \/>Solution:<br \/>In \u2206ABC,<br \/>\u2220ACB = 180\u00b0 \u2013 (x + 2x)<br \/>= 180\u00b0 \u2013 3x \u2026(i)<br \/>and in \u2206BDG,<br \/>\u2220BED = 180\u00b0 \u2013 (2x + y) \u2026(ii)<br \/>\u2220EGC = \u2220AGD (Vertically opposite angles)<br \/>= 3y<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/31771521488_40f9e97c51_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"249\" height=\"201\" \/><br \/>In quad. BCGE,<br \/>\u2220B + \u2220ACB + \u2220CGE + \u2220BED = 360\u00b0 (Sum of angles of a quadrilateral)<br \/>\u21d2 2x+ 180\u00b0 \u2013 3x + 3y + 180\u00b0- 2x-y = 360\u00b0<br \/>\u21d2 -3x + 2y = 0<br \/>\u21d2 3x = 2y \u21d2 y =\u00a0<span id=\"MathJax-Element-52-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-367\" class=\"math\"><span id=\"MathJax-Span-368\" class=\"mrow\"><span id=\"MathJax-Span-369\" class=\"mfrac\"><span id=\"MathJax-Span-370\" class=\"mn\">3<\/span><span id=\"MathJax-Span-371\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x (a)<\/p>\n<p>Question 20.<br \/>In the figure, what is the value of x?<br \/>(a) 35<br \/>(b) 45<br \/>(c) 50<br \/>(d) 60<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/31771521268_5cd454812e_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"351\" height=\"229\" \/><br \/>Solution:<br \/>In the figure, side AB is produced to D<br \/>\u2234 \u2220CBA + \u2220CBD = 180\u00b0 (Linear pair)<br \/>\u21d2 7y + 5y = 180\u00b0<br \/>\u21d2 12y = 180\u00b0<br \/>\u21d2 y =\u00a0<span id=\"MathJax-Element-53-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-372\" class=\"math\"><span id=\"MathJax-Span-373\" class=\"mrow\"><span id=\"MathJax-Span-374\" class=\"mfrac\"><span id=\"MathJax-Span-375\" class=\"mn\">180<\/span><span id=\"MathJax-Span-376\" class=\"mn\">12<\/span><\/span><\/span><\/span><\/span>\u00a0= 15<br \/>and Ext. \u2220CBD = \u2220A + \u2220C<br \/>\u21d2 7y = 3y + x<br \/>\u21d2 7y -3y = x<br \/>\u21d2 4y = x<br \/>\u2234 x = 4 x 15 = 60 (d)<\/p>\n<p>Question 21.<br \/>In the figure, the value of x is<br \/>(a) 65\u00b0<br \/>(b) 80\u00b0<br \/>(c) 95\u00b0<br \/>(d) 120\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/31771521088_9e33e185b6_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"267\" height=\"219\" \/><br \/>Solution:<br \/>In the figure, \u2220A = 55\u00b0, \u2220D = 25\u00b0 and \u2220C = 40\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1977\/31771520978_ddeab49558_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"274\" height=\"222\" \/><br \/>Now in \u2206ABD,<br \/>Ext. \u2220DBC = \u2220A + \u2220D<br \/>= 55\u00b0 + 25\u00b0 = 80\u00b0<br \/>Similarly, in \u2206BCE,<br \/>Ext. \u2220DEC = \u2220EBC + \u2220ECB<br \/>= 80\u00b0 + 40\u00b0 = 120\u00b0 (d)<\/p>\n<p>Question 22.<br \/>In the figure, if BP || CQ and AC = BC, then the measure of x is<br \/>(a) 20\u00b0<br \/>(b) 25\u00b0<br \/>(c) 30\u00b0<br \/>(d) 35\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/31771520848_d8afc5e31c_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"277\" height=\"202\" \/><br \/>Solution:<br \/>In the figure, AC = BC, BP || CQ<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/31771520668_50044bd710_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry\" width=\"266\" height=\"202\" \/><br \/>\u2235 BP || CQ<br \/>\u2234 \u2220PBC \u2013 \u2220QCD<br \/>\u21d2 20\u00b0 + \u2220ABC = 70\u00b0<br \/>\u21d2 \u2220ABC = 70\u00b0 \u2013 20\u00b0 = 50\u00b0<br \/>\u2235 BC = AC<br \/>\u2234 \u2220ACB = \u2220ABC (Angles opposite to equal sides)<br \/>= 50\u00b0<br \/>Now in \u2206ABC,<br \/>Ext. \u2220ACD = \u2220B + \u2220A<br \/>\u21d2 x + 70\u00b0 = 50\u00b0 + 50\u00b0<br \/>\u21d2 x + 70\u00b0 = 100\u00b0<br \/>\u2234 x = 100\u00b0 \u2013 70\u00b0 = 30\u00b0 (c)<\/p>\n<p>Question 23.<br \/>In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If \u2220APR = 25\u00b0, \u2220RQC = 30\u00b0 and \u2220CQF = 65\u00b0, then<br \/>(a) x = 55\u00b0, y = 40\u00b0<br \/>(b) x = 50\u00b0, y = 45\u00b0<br \/>(c) x = 60\u00b0, y = 35\u00b0<br \/>(d) x = 35\u00b0, y = 60\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/31771520458_6a960b38ba_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry\" width=\"320\" height=\"279\" \/><br \/>Solution:<br \/>In the figure,<br \/>\u2235 AB || CD, EF intersects them at P and Q respectively,<br \/>\u2220APR = 25\u00b0, \u2220RQC = 30\u00b0, \u2220CQF = 65\u00b0<br \/>\u2235 AB || CD<br \/>\u2234 \u2220APQ = \u2220CQF (Corresponding anlges)<br \/>\u21d2 y + 25\u00b0 = 65\u00b0<br \/>\u21d2 y = 65\u00b0 \u2013 25\u00b0 = 40\u00b0<br \/>and APQ + PQC = 180\u00b0 (Co-interior angles)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/31771520308_2cb60e2410_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"324\" height=\"278\" \/><br \/>y + 25\u00b0 + \u22201 +30\u00b0= 180\u00b0<br \/>40\u00b0 + 25\u00b0 + \u22201 + 30\u00b0 = 180\u00b0<br \/>\u21d2 \u22201 + 95\u00b0 = 180\u00b0<br \/>\u2234 \u22201 = 180\u00b0 \u2013 95\u00b0 = 85\u00b0<br \/>Now, \u2206PQR,<br \/>\u2220RPQ + \u2220PQR + \u2220PRQ = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 40\u00b0 + x + 85\u00b0 = 180\u00b0<br \/>\u21d2 125\u00b0 + x = 180\u00b0<br \/>\u21d2 x = 180\u00b0 \u2013 125\u00b0 = 55\u00b0<br \/>\u2234 x = 55\u00b0, y = 40\u00b0 (a)<\/p>\n<p>Question 24.<br \/>The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94\u00b0 and 126\u00b0. Then, \u2220BAC = ?<br \/>(a) 94\u00b0<br \/>(b) 54\u00b0<br \/>(c) 40\u00b0<br \/>(d) 44\u00b0<br \/>Solution:<br \/>In \u2206ABC, base BC is produced both ways and \u2220ACD = 94\u00b0, \u2220ABE = 126\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1961\/31771520108_b305ef5a4e_o.png\" alt=\"RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"273\" height=\"192\" \/><br \/>Ext. \u2220ACD = \u2220BAC + \u2220ABC<br \/>\u21d2 94\u00b0 = \u2220BAC + \u2220ABC<br \/>Similarly, \u2220ABE = \u2220BAC + \u2220ACB<br \/>\u21d2 126\u00b0 = \u2220BAC + \u2220ACB<br \/>Adding,<br \/>94\u00b0 + 126\u00b0 = \u2220BAC + \u2220ABC + \u2220ACB + \u2220BAC<br \/>220\u00b0 = 180\u00b0 + \u2220BAC<br \/>\u2234 \u2220BAC = 220\u00b0 -180\u00b0 = 40\u00b0 (c)<\/p>\n<p>Question 25.<br \/>If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is<br \/>(a) 45\u00b0<br \/>(b) 95\u00b0<br \/>(c) 135\u00b0<br \/>(d) 90\u00b0<br \/>Solution:<br \/>In right \u2206ABC, \u2220A = 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1944\/31771519998_96b95d39cd_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"215\" height=\"184\" \/><br \/>Bisectors of \u2220B and \u2220C meet at O, then 1<br \/>\u2220BOC = 90\u00b0 +\u00a0<span id=\"MathJax-Element-54-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-377\" class=\"math\"><span id=\"MathJax-Span-378\" class=\"mrow\"><span id=\"MathJax-Span-379\" class=\"mfrac\"><span id=\"MathJax-Span-380\" class=\"mn\">1<\/span><span id=\"MathJax-Span-381\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A<br \/>= 90\u00b0+\u00a0<span id=\"MathJax-Element-55-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-382\" class=\"math\"><span id=\"MathJax-Span-383\" class=\"mrow\"><span id=\"MathJax-Span-384\" class=\"mfrac\"><span id=\"MathJax-Span-385\" class=\"mn\">1<\/span><span id=\"MathJax-Span-386\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 90\u00b0 = 90\u00b0 + 45\u00b0= 135\u00b0 (c)<\/p>\n<p>Question 26.<br \/>The bisects of exterior angles at B and C of \u2206ABC, meet at O. If \u2220A = .x\u00b0, then \u2220BOC=<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1954\/31771519768_b3b179a4e3_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"136\" height=\"181\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/31771519658_6754b797ed_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"136\" height=\"52\" \/><br \/>Solution:<br \/>In \u2206ABC, \u2220A = x\u00b0<br \/>and bisectors of \u2220B and \u2220C meet at O.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1952\/31771519608_7e05c6f4be_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"337\" height=\"316\" \/><\/p>\n<p>Question 27.<br \/>In a \u2206ABC, \u2220A = 50\u00b0 and BC is produced to a point D. If the bisectors of \u2220ABC and \u2220ACD meet at E, then \u2220E =<br \/>(a) 25\u00b0<br \/>(b) 50\u00b0<br \/>(c) 100\u00b0<br \/>(d) 75\u00b0<br \/>Solution:<br \/>In \u2206ABC, \u2220A = 50\u00b0<br \/>BC is produced<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1961\/31771519508_09ba910c51_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"335\" height=\"195\" \/><br \/>Bisectors of \u2220ABC and \u2220ACD meet at \u2220E<br \/>\u2234 \u2220E =\u00a0<span id=\"MathJax-Element-56-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-387\" class=\"math\"><span id=\"MathJax-Span-388\" class=\"mrow\"><span id=\"MathJax-Span-389\" class=\"mfrac\"><span id=\"MathJax-Span-390\" class=\"mn\">1<\/span><span id=\"MathJax-Span-391\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A =\u00a0<span id=\"MathJax-Element-57-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-392\" class=\"math\"><span id=\"MathJax-Span-393\" class=\"mrow\"><span id=\"MathJax-Span-394\" class=\"mfrac\"><span id=\"MathJax-Span-395\" class=\"mn\">1<\/span><span id=\"MathJax-Span-396\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 50\u00b0 = 25\u00b0 (a)<\/p>\n<p>Question 28.<br \/>The side BC of AABC is produced to a point D. The bisector of \u2220A meets side BC in L. If \u2220ABC = 30\u00b0 and \u2220ACD =115\u00b0,then \u2220ALC =<br \/>(a) 85\u00b0<br \/>(b) 72<span id=\"MathJax-Element-58-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-397\" class=\"math\"><span id=\"MathJax-Span-398\" class=\"mrow\"><span id=\"MathJax-Span-399\" class=\"mfrac\"><span id=\"MathJax-Span-400\" class=\"mn\">1<\/span><span id=\"MathJax-Span-401\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u00b0<br \/>(c) 145\u00b0<br \/>(d) none of these<br \/>Solution:<br \/>In \u2206ABC, BC is produced to D<br \/>\u2220B = 30\u00b0, \u2220ACD = 115\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/31771519448_09e57b4e11_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"360\" height=\"669\" \/><\/p>\n<p>Question 29.<br \/>In the figure , if l1 || l2, the value of x is<br \/>(a) 22\u00a0<span id=\"MathJax-Element-59-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-402\" class=\"math\"><span id=\"MathJax-Span-403\" class=\"mrow\"><span id=\"MathJax-Span-404\" class=\"mfrac\"><span id=\"MathJax-Span-405\" class=\"mn\">1<\/span><span id=\"MathJax-Span-406\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br \/>(b) 30<br \/>(c) 45<br \/>(d) 60<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/31771519138_ef64ed8fca_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"318\" height=\"239\" \/><br \/>Solution:<br \/>In the figure, l<sub>1<\/sub>\u00a0|| l<sub>2<\/sub><br \/>EC, EB are the bisectors of \u2220DCB and \u2220CBA respectively EF is the bisector of \u2220GEB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1955\/31771519038_2bbc4442a0_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"328\" height=\"258\" \/><br \/>\u2235 EC and EB are the bisectors of \u2220DCB and \u2220CBA respectively<br \/>\u2234 \u2220CEB = 90\u00b0<br \/>\u2234 a + b = 90\u00b0 ,<br \/>and \u2220GEB = 90\u00b0 (\u2235 \u2220CEB = 90\u00b0)<br \/>2x = 90\u00b0 \u21d2 x =\u00a0<span id=\"MathJax-Element-60-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-407\" class=\"math\"><span id=\"MathJax-Span-408\" class=\"mrow\"><span id=\"MathJax-Span-409\" class=\"mfrac\"><span id=\"MathJax-Span-410\" class=\"mn\">90<\/span><span id=\"MathJax-Span-411\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 45 (c)<\/p>\n<p>Question 30.<br \/>In \u2206RST (in the figure), what is the value of x?<br \/>(a) 40\u00b0<br \/>(b) 90\u00b0<br \/>(c) 80\u00b0<br \/>(d) 100\u00b0<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1952\/31771518918_c1a81d0182_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"281\" height=\"202\" \/><br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1936\/31771518758_b9e21cbae4_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"705\" height=\"259\" \/><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7391\/16349894020_4459ddc3b5_o.png\" alt=\"RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-1\" width=\"446\" height=\"467\" \/><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7293\/16535659701_b62e03a471_o.png\" alt=\"RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-2\" width=\"464\" height=\"615\" \/><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7312\/16351088669_cf49c5c405_o.png\" alt=\"RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-3\" width=\"429\" height=\"441\" \/><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7286\/16351461337_6b7f00d4ac_o.png\" alt=\"RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-4\" width=\"473\" height=\"615\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-listing-of-important-topics\"><\/span><strong>Detailed Exercise-wise Explanation with Listing of Important Topics\u00a0<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>A student must remember to check all the significant details of Class 9 Chapter 11 RD Sharma Solutions while getting ready for the final Class 9 Maths exam. In Chapter 11 Solutions RD Sharma the students will cover from all kinds of tricky questions.<\/p>\n<p>A basic idea of the parts falling under Coordinate Geometry can play an important role in securing a good position. You will get all of that in the RD Sharma Class 9 Solutions Chapter 11. Practicing on a regular basis holds the key and here we have tried to throw light on the exercise wise explanations. Some solutions from the various exercises have been presented here take a look-<\/p>\n<p><strong>Question 1: Plot the following points on the graph paper:<\/strong><\/p>\n<p><strong>(i) (2,5) (ii) (4,-3) (iii) (-5,-7) (iv) (7,-4) (v) (-3,2)<\/strong><\/p>\n<p><strong>(vi) (7,0) (vii) (-4,0) (viii) (0,7) (ix) (0,-4) (x) (0,0)<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p><strong>Question 2: Write the coordinates of each of the following points marked in the graph paper.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<table style=\"width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>Point<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>Distance from the y-axis (units)<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>Distance from x-axis (units)<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>Coordinates of Point<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>A<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>3<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>1<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(3,1)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>B<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>6<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>0<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(6, 0)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>C<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>0<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>6<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(0, 6)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>D<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>-3<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>0<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(-3,0)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>E<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>-4<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>3<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(-4,3)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>F<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>-2<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>-4<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(-2,-4)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>G<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>0<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>-5<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(0,-5)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>H<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>3<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>-6<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(3,-6)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>P<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>7<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>-3<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(7,-3)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 17.6398%;\">\n<p>Q<\/p>\n<\/td>\n<td style=\"width: 26.087%;\">\n<p>7<\/p>\n<\/td>\n<td style=\"width: 31.677%;\">\n<p>6<\/p>\n<\/td>\n<td style=\"width: 24.472%;\">\n<p>(7,6)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3><span class=\"ez-toc-section\" id=\"explanation-for-coordinates-of-a\"><\/span>Explanation for Coordinates of A:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>The distance of point A from y-axis is 3 units and from x-axis is 1 unit.<\/p>\n<p>A lies in the first quadrant, so both the coordinates are positive.<\/p>\n<p>This implies coordinates of A are (3,1).<\/p>\n<p>Similarly, other points are,<\/p>\n<p>A (3,1), B (6,0), C (0,6), D (-3,0), E (-4,3), F (-2,-4), G (0,-5), H (3,-6), P (7,-3), Q (7,6)<\/p>\n<p>Important concepts<\/p>\n<ul>\n<li>Rectangular or cartesian coordinates of a point<\/li>\n<li>Cartesian co-ordinates axes<\/li>\n<li>Quadrants<\/li>\n<li>Cartesian coordinates of a point<\/li>\n<li>Plotting of points<\/li>\n<\/ul>\n<p>This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 11. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 11 Coordinate Geometry<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630741134799\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-11\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 11?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630741179415\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-11\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 11?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630741199786\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-11-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 11\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 9 Maths Chapter 11: Are you confused about the relevance of purchasing RD Sharma Solutions Class 9 Maths Chapter 11 Coordinate Geometry? Yes. You need to calm down because in this blog we will be concentrating on the essential aspects of Class 9 Maths Coordinate Geometry Chapter 11 Solutions.\u00a0 Download RD &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 11 &#8211; Coordinate Geometry (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 11 &#8211; Coordinate Geometry (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124461,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3081,3048,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62199"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=62199"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62199\/revisions"}],"predecessor-version":[{"id":128728,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62199\/revisions\/128728"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124461"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62199"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62199"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}