{"id":126696,"date":"2026-04-04T00:33:38","date_gmt":"2026-04-03T19:03:38","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126696"},"modified":"2026-04-04T06:03:47","modified_gmt":"2026-04-04T00:33:47","slug":"rd-sharma-class-9-solutions-chapter-11-mcqs","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/","title":{"rendered":"RD Sharma Class 9 Solutions: Complete Guide [2026]"},"content":{"rendered":"<p><script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"Article\",\n  \"headline\": \"RD Sharma Class 9 Solutions: Complete Guide [2026]\",\n  \"description\": \"Comprehensive guide to RD Sharma Class 9 Solutions for Maths. Includes chapter-wise solutions and MCQs to help students excel in their exams.\",\n  \"author\": {\n    \"@type\": \"Organization\",\n    \"name\": \"KopyKitab\"\n  },\n  \"publisher\": {\n    \"@type\": \"Organization\",\n    \"name\": \"KopyKitab\",\n    \"url\": \"https:\/\/www.kopykitab.com\"\n  },\n  \"datePublished\": \"2026-04-04\",\n  \"dateModified\": \"2026-04-04\"\n}\n<\/script><\/p>\n<div class=\"freshness-block\" style=\"background:#e8f5e9;padding:15px;border-left:4px solid #4caf50;margin:20px 0;border-radius:4px;\">\n<strong>Last Updated:<\/strong> April 04, 2026 | This article has been updated with the latest information for 2026.\n<\/div>\n<p><img alt=\"RD Sharma Class 9 Solutions Chapter 11 MCQs\" class=\"alignnone size-full wp-image-126698\" height=\"675\" loading=\"eager\" sizes=\"(max-width: 1200px) 100vw, 1200px\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-VSAQS-1.jpg\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-VSAQS-1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-VSAQS-1-768x432.jpg 768w\" width=\"1200\"\/><\/p>\n<p><strong>Read more:<\/strong> <a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\">RD Sharma Class 9 \u2014 Complete Guide<\/a><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 11 MCQs:<\/strong> MCQ type questions ko solve karna kuch students ke liye tricky ho sakta hai. But tension mat lo kyunki humne <a href=\"\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/\" rel=\"noopener\" target=\"_blank\">RD Sharma Class 9 Solutions Chapter 11<\/a> MCQs aapke liye tayyar kiye hain. Saare MCQs easy aur understandable manner mein solve kiye gaye hain. Aap dusre exercises ke solutions <a href=\"\/blog\/rd-sharma-class-9-solutions-for-maths\/\" rel=\"noopener\" target=\"_blank\">RD Sharma Class 9 Solutions Maths<\/a> mein dekh sakte hain. RD Sharma Class 9 Solutions Chapter 11 MCQs ke baare mein aur jaanne ke liye, pura blog padhiye.<\/p>\n<div class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\" id=\"ez-toc-container\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<p><span class=\"ez-toc-title-toggle\"><a aria-label=\"ez-toc-toggle-icon-1\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" href=\"#\"><label aria-label=\"Table of Content\" for=\"item-69d05b980631a\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg class=\"list-377408\" fill=\"none\" height=\"20px\" style=\"fill: #000000;color:#000000\" viewbox=\"0 0 24 24\" width=\"20px\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg baseprofile=\"tiny\" class=\"arrow-unsorted-368013\" height=\"10px\" style=\"fill: #000000;color:#000000\" version=\"1.2\" viewbox=\"0 0 24 24\" width=\"10px\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><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\"><\/path><\/svg><\/span><\/label><input id=\"item-69d05b980631a\" type=\"checkbox\"\/><\/a><\/span><\/div>\n<nav>\n<ul class=\"ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default\">\n<li class=\"ez-toc-page-1 ez-toc-heading-level-2\"><a class=\"ez-toc-link ez-toc-heading-1\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#access-answers-of-rd-sharma-class-9-solutions-chapter-11-mcqs\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 11 MCQs \">Access answers of RD Sharma Class 9 Solutions Chapter 11 MCQs <\/a><\/li>\n<li class=\"ez-toc-page-1 ez-toc-heading-level-2\"><a class=\"ez-toc-link ez-toc-heading-2\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#chapter-11-coordinate-geometry-ka-importance\" title=\"Chapter 11 Coordinate Geometry ka Importance\">Chapter 11 Coordinate Geometry ka Importance<\/a><\/li>\n<li class=\"ez-toc-page-1 ez-toc-heading-level-2\"><a class=\"ez-toc-link ez-toc-heading-3\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#mcq-solving-tips-aur-strategies\" title=\"MCQ Solving Tips aur Strategies\">MCQ Solving Tips aur Strategies<\/a><\/li>\n<li class=\"ez-toc-page-1 ez-toc-heading-level-2\"><a class=\"ez-toc-link ez-toc-heading-4\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#faqs-on-rd-sharma-class-9-solutions-chapter-11-mcqs\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 11 MCQs\">FAQs on RD Sharma Class 9 Solutions Chapter 11 MCQs<\/a>\n<ul class=\"ez-toc-list-level-3\">\n<li class=\"ez-toc-heading-level-3\"><a class=\"ez-toc-link ez-toc-heading-5\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#how-many-questions-are-there-in-rd-sharma-class-9-solutions-chapter-11-mcqs\" title=\"How many questions are there in RD Sharma Class 9 Solutions Chapter 11 MCQs?\">How many questions are there in RD Sharma Class 9 Solutions Chapter 11 MCQs?<\/a><\/li>\n<li class=\"ez-toc-page-1 ez-toc-heading-level-3\"><a class=\"ez-toc-link ez-toc-heading-6\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#is-it-even-beneficial-to-study-rd-sharma-for-class-9-solutions-chapter-11-mcqs\" title=\"Is it even beneficial to study RD Sharma for Class 9 Solutions Chapter 11 MCQs?\">Is it even beneficial to study RD Sharma for Class 9 Solutions Chapter 11 MCQs?<\/a><\/li>\n<li class=\"ez-toc-page-1 ez-toc-heading-level-3\"><a class=\"ez-toc-link ez-toc-heading-7\" href=\"\/blog\/rd-sharma-class-9-solutions-chapter-11-mcqs\/#are-the-solutions-rd-sharma-class-9-solutions-chapter-11-mcqs-relevant\" title=\"Are the solutions RD Sharma Class 9 Solutions Chapter 11 MCQs relevant?\">Are the solutions RD Sharma Class 9 Solutions Chapter 11 MCQs relevant?<\/a><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"related-articles\" style=\"background:#fff3e0;padding:20px;border-left:4px solid #ff9800;margin:24px 0;border-radius:4px;\">\n<h3 style=\"margin:0 0 12px;color:#e65100;\"><span class=\"ez-toc-section\" id=\"you-may-also-like\"><\/span>You May Also Like<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul style=\"list-style:none;margin:0;padding:0;\">\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 RD Sharma Solutions<\/a><\/li>\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 Maths Solutions<\/a><\/li>\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/cbse\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 CBSE Preparation<\/a><\/li>\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/exam-preparation\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 Exam Preparation Guide<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"cta-box\" style=\"background:linear-gradient(135deg,#e8f5e9,#c8e6c9);padding:24px;border-radius:8px;margin:24px 0;text-align:center;border:1px solid #a5d6a7;\">\n<h3 style=\"margin:0 0 8px;color:#2e7d32;\"><span class=\"ez-toc-section\" id=\"explore-study-materials\"><\/span>Explore Study Materials<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"margin:0 0 16px;color:#333;\">Get comprehensive study materials, practice tests, and expert guides for your exam preparation.<\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/competitive-exam-books\" style=\"display:inline-block;background:#2e7d32;color:#fff;padding:12px 32px;border-radius:6px;text-decoration:none;font-weight:600;\">Browse Study Materials \u2192<\/a>\n<\/div>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-11-mcqs\"><\/span><span class=\"ez-toc-section\" id=\"access-answers-of-RD-sharma-class-9-solutions-chapter-11-mcqs\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 11 MCQs<\/strong> <span class=\"ez-toc-section-end\"><\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Yahan par hum aapko Chapter 11 ke saare MCQ questions ke detailed <a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\">solutions<\/a> provide kar rahe hain. Ye questions coordinate geometry ke basic concepts par based hain jo ki Class 9 maths ka ek important part hai.<\/p>\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 = <span class=\"MathJax\" id=\"MathJax-Element-1-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-1\"><span class=\"mrow\" id=\"MathJax-Span-2\"><span class=\"mfrac\" id=\"MathJax-Span-3\"><span class=\"msubsup\" id=\"MathJax-Span-4\"><span class=\"texatom\" id=\"MathJax-Span-5\"><span class=\"mrow\" id=\"MathJax-Span-6\"><span class=\"mn\" id=\"MathJax-Span-7\">180<\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-8\"><span class=\"mrow\" id=\"MathJax-Span-9\"><span class=\"mo\" id=\"MathJax-Span-10\">\u2218<\/span><\/span><\/span><\/span><span class=\"mn\" id=\"MathJax-Span-11\">3<\/span><\/span><\/span><\/span><\/span>  = 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 = <span class=\"MathJax\" id=\"MathJax-Element-2-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-12\"><span class=\"mrow\" id=\"MathJax-Span-13\"><span class=\"mfrac\" id=\"MathJax-Span-14\"><span class=\"msubsup\" id=\"MathJax-Span-15\"><span class=\"texatom\" id=\"MathJax-Span-16\"><span class=\"mrow\" id=\"MathJax-Span-17\"><span class=\"mn\" id=\"MathJax-Span-18\">90<\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-19\"><span class=\"mrow\" id=\"MathJax-Span-20\"><span class=\"mo\" id=\"MathJax-Span-21\">\u2218<\/span><\/span><\/span><\/span><span class=\"mn\" id=\"MathJax-Span-22\">2<\/span><\/span><\/span><\/span><\/span>  = 45\u00b0 (b)<\/p>\n<div class=\"code-block code-block-2\"> <\/div>\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 = <span class=\"MathJax\" id=\"MathJax-Element-3-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-23\"><span class=\"mrow\" id=\"MathJax-Span-24\"><span class=\"mfrac\" id=\"MathJax-Span-25\"><span class=\"msubsup\" id=\"MathJax-Span-26\"><span class=\"texatom\" id=\"MathJax-Span-27\"><span class=\"mrow\" id=\"MathJax-Span-28\"><span class=\"mn\" id=\"MathJax-Span-29\">100<\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-30\"><span class=\"mrow\" id=\"MathJax-Span-31\"><span class=\"mo\" id=\"MathJax-Span-32\">\u2218<\/span><\/span><\/span><\/span><span class=\"mn\" id=\"MathJax-Span-33\">2<\/span><\/span><\/span><\/span><\/span>  = 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 &quot;=\"\" blog=\"\" category=\"\" class=\"alignnone size-full wp-image-66559\" height=\"450\" https:=\"\" loading=\"lazy\" rd-sharma-solutions=\"\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/&lt;a href=\" width=\"800\" www.kopykitab.com=\"\" alt=\"\">RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q4.1.png&#8221; alt=&#8221;RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q4.1&#8243; width=&#8221;254&#8243; height=&#8221;176&#8243; \/&gt;<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 = <span class=\"MathJax\" id=\"MathJax-Element-4-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-34\"><span class=\"mrow\" id=\"MathJax-Span-35\"><span class=\"mfrac\" id=\"MathJax-Span-36\"><span class=\"msubsup\" id=\"MathJax-Span-37\"><span class=\"texatom\" id=\"MathJax-Span-38\"><span class=\"mrow\" id=\"MathJax-Span-39\"><span class=\"mn\" id=\"MathJax-Span-40\">180<\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-41\"><span class=\"mrow\" id=\"MathJax-Span-42\"><span class=\"mo\" id=\"MathJax-Span-43\">\u2218<\/span><\/span><\/span><\/span><span class=\"mn\" id=\"MathJax-Span-44\">2<\/span><\/span><\/span><\/span><\/span>  = 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 = <span class=\"MathJax\" id=\"MathJax-Element-5-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-45\"><span class=\"mrow\" id=\"MathJax-Span-46\"><span class=\"mfrac\" id=\"MathJax-Span-47\"><span class=\"mn\" id=\"MathJax-Span-48\">1<\/span><span class=\"mn\" id=\"MathJax-Span-49\">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 &quot;=\"\" blog=\"\" category=\"\" class=\"alignnone size-full wp-image-66560\" height=\"450\" https:=\"\" loading=\"lazy\" rd-sharma-solutions=\"\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/&lt;a href=\" width=\"800\" www.kopykitab.com=\"\" alt=\"\">RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q5.1.png&#8221; sizes=&#8221;(max-width: 305px) 100vw, 305px&#8221; srcset=&#8221;https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q5.1.png 305w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q5.1-300&#215;188.png 300w&#8221; alt=&#8221;RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q5.1&#8243; width=&#8221;305&#8243; height=&#8221;191&#8243; \/&gt;<br \/>Ext. \u2220ACB = \u2220A + \u2220B<br \/>(Exterior angle of a triangle is equal to the sum of its interior opposite angles)<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q5.2\" class=\"alignnone size-full wp-image-66561\" height=\"212\" loading=\"lazy\" sizes=\"(max-width: 345px) 100vw, 345px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q5.2.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q5.2.png 345w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q5.2-300x184.png 300w\" width=\"345\"\/><\/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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q6.1\" class=\"alignnone size-full wp-image-66562\" height=\"241\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q6.1.png\" width=\"196\"\/><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 = <span class=\"MathJax\" id=\"MathJax-Element-6-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-50\"><span class=\"mrow\" id=\"MathJax-Span-51\"><span class=\"mfrac\" id=\"MathJax-Span-52\"><span class=\"msubsup\" id=\"MathJax-Span-53\"><span class=\"texatom\" id=\"MathJax-Span-54\"><span class=\"mrow\" id=\"MathJax-Span-55\"><span class=\"mn\" id=\"MathJax-Span-56\">140<\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-57\"><span class=\"mrow\" id=\"MathJax-Span-58\"><span class=\"mo\" id=\"MathJax-Span-59\">\u2218<\/span><\/span><\/span><\/span><span class=\"mn\" id=\"MathJax-Span-60\">2<\/span><\/span><\/span><\/span><\/span> = 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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q7.1\" class=\"alignnone size-full wp-image-66563\" height=\"222\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q7.1.png\" width=\"218\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q8.1\" class=\"alignnone size-full wp-image-66564\" height=\"252\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q8.1.png\" width=\"216\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q9.1\" class=\"alignnone size-full wp-image-66565\" height=\"222\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q9.1.png\" width=\"256\"\/><br \/>Now, \u2220BAD = <span class=\"MathJax\" id=\"MathJax-Element-7-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-61\"><span class=\"mrow\" id=\"MathJax-Span-62\"><span class=\"mfrac\" id=\"MathJax-Span-63\"><span class=\"msubsup\" id=\"MathJax-Span-64\"><span class=\"texatom\" id=\"MathJax-Span-65\"><span class=\"mrow\" id=\"MathJax-Span-66\"><span class=\"mn\" id=\"MathJax-Span-67\">100<\/span><\/span><\/span><span class=\"texatom\" id=\"MathJax-Span-68\"><span class=\"mrow\" id=\"MathJax-Span-69\"><span class=\"mo\" id=\"MathJax-Span-70\">\u2218<\/span><\/span><\/span><\/span><span class=\"mn\" id=\"MathJax-Span-71\">2<\/span><\/span><\/span><\/span><\/span> = 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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q10.1\" class=\"alignnone size-full wp-image-66566\" height=\"210\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q10.1.png\" width=\"249\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q10.2\" class=\"alignnone size-full wp-image-66567\" height=\"194\" loading=\"lazy\" sizes=\"(max-width: 353px) 100vw, 353px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q10.2.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q10.2.png 353w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q10.2-300x165.png 300w\" width=\"353\"\/><\/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<\/p>\n<p><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q11.1\" class=\"alignnone size-full wp-image-66568\" height=\"286\" loading=\"lazy\" sizes=\"(max-width: 357px) 100vw, 357px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q11.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q11.1.png 357w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q11.1-300x240.png 300w\" width=\"357\"\/><\/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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q12.1\" class=\"alignnone size-full wp-image-66569\" height=\"229\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q12.1.png\" width=\"198\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q13.1\" class=\"alignnone size-full wp-image-66570\" height=\"235\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q13.1.png\" width=\"276\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q13.2\" class=\"alignnone size-full wp-image-66571\" height=\"234\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q13.2.png\" width=\"268\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q14.1\" class=\"alignnone size-full wp-image-66572\" height=\"190\" loading=\"lazy\" sizes=\"(max-width: 361px) 100vw, 361px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q14.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q14.1.png 361w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q14.1-300x158.png 300w\" width=\"361\"\/><br \/>Solution:<br \/>In the figure<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q14.2\" class=\"alignnone size-full wp-image-66573\" height=\"195\" loading=\"lazy\" sizes=\"(max-width: 373px) 100vw, 373px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q14.2.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q14.2.png 373w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q14.2-300x157.png 300w\" width=\"373\"\/><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 = <span class=\"MathJax\" id=\"MathJax-Element-8-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-72\"><span class=\"mrow\" id=\"MathJax-Span-73\"><span class=\"mfrac\" id=\"MathJax-Span-74\"><span class=\"mrow\" id=\"MathJax-Span-75\"><span class=\"mn\" id=\"MathJax-Span-76\">180<\/span><span class=\"mi\" id=\"MathJax-Span-77\">x<\/span><span class=\"mn\" id=\"MathJax-Span-78\">3<\/span><span class=\"mi\" id=\"MathJax-Span-79\">x<\/span><\/span><span class=\"mrow\" id=\"MathJax-Span-80\"><span class=\"mn\" id=\"MathJax-Span-81\">12<\/span><span class=\"mi\" id=\"MathJax-Span-82\">x<\/span><\/span><\/span><\/span><\/span><\/span> = 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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q16.1\" class=\"alignnone size-full wp-image-66574\" height=\"240\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q16.1.png\" width=\"270\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q16.2\" class=\"alignnone size-full wp-image-66575\" height=\"240\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q16.2.png\" width=\"297\"\/><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 = <span class=\"MathJax\" id=\"MathJax-Element-9-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-83\"><span class=\"mrow\" id=\"MathJax-Span-84\"><span class=\"mfrac\" id=\"MathJax-Span-85\"><span class=\"mn\" id=\"MathJax-Span-86\">44<\/span><span class=\"mn\" id=\"MathJax-Span-87\">2<\/span><\/span><\/span><\/span><\/span> = 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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q17.1\" class=\"alignnone size-full wp-image-66576\" height=\"203\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q17.1.png\" width=\"274\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q17.2\" class=\"alignnone size-full wp-image-66577\" height=\"196\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q17.2.png\" width=\"280\"\/><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> || l<sub>2<\/sub>?<br \/>(a) 37<br \/>(b) 43<br \/>(c) 45<br \/>(d) 47<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q18.1\" class=\"alignnone size-full wp-image-66578\" height=\"174\" loading=\"lazy\" sizes=\"(max-width: 326px) 100vw, 326px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q18.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q18.1.png 326w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q18.1-300x160.png 300w\" width=\"326\"\/><br \/>Solution:<br \/>In the figure, l<sub>1<\/sub> || l<sub>2<\/sub><br \/>\u2234 \u2220EBA = \u2220BAH (Alternate angles)<br \/>\u2234 \u2220BAH = 78\u00b0<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q18.2\" class=\"alignnone size-full wp-image-66579\" height=\"181\" loading=\"lazy\" sizes=\"(max-width: 337px) 100vw, 337px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q18.2.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q18.2.png 337w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q18.2-300x161.png 300w\" width=\"337\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q19.1\" class=\"alignnone size-full wp-image-66580\" height=\"325\" loading=\"lazy\" sizes=\"(max-width: 309px) 100vw, 309px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q19.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q19.1.png 309w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q19.1-285x300.png 285w\" width=\"309\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q19.2\" class=\"alignnone size-full wp-image-66581\" height=\"205\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q19.2.png\" width=\"252\"\/><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 = <span class=\"MathJax\" id=\"MathJax-Element-10-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-88\"><span class=\"mrow\" id=\"MathJax-Span-89\"><span class=\"mfrac\" id=\"MathJax-Span-90\"><span class=\"mn\" id=\"MathJax-Span-91\">3<\/span><span class=\"mn\" id=\"MathJax-Span-92\">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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q20.1\" class=\"alignnone size-full wp-image-66582\" height=\"237\" loading=\"lazy\" sizes=\"(max-width: 363px) 100vw, 363px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q20.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q20.1.png 363w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q20.1-300x196.png 300w\" width=\"363\"\/><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 = <span class=\"MathJax\" id=\"MathJax-Element-11-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-93\"><span class=\"mrow\" id=\"MathJax-Span-94\"><span class=\"mfrac\" id=\"MathJax-Span-95\"><span class=\"mn\" id=\"MathJax-Span-96\">180<\/span><span class=\"mn\" id=\"MathJax-Span-97\">12<\/span><\/span><\/span><\/span><\/span> = 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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q21.1\" class=\"alignnone size-full wp-image-66584\" height=\"224\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q21.1.png\" width=\"270\"\/><br \/>Solution:<br \/>In the figure, \u2220A = 55\u00b0, \u2220D = 25\u00b0 and \u2220C = 40\u00b0<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q21.2\" class=\"alignnone size-full wp-image-66585\" height=\"225\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q21.2.png\" width=\"276\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q22.1\" class=\"alignnone size-full wp-image-66586\" height=\"205\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q22.1.png\" width=\"268\"\/><br \/>Solution:<br \/>In the figure, AC = BC, BP || CQ<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q22.2\" class=\"alignnone size-full wp-image-66587\" height=\"206\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q22.2.png\" width=\"277\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q23.1\" class=\"alignnone size-full wp-image-66588\" height=\"283\" loading=\"lazy\" sizes=\"(max-width: 328px) 100vw, 328px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q23.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q23.1.png 328w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q23.1-300x259.png 300w\" width=\"328\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q23.2\" class=\"alignnone size-full wp-image-66589\" height=\"280\" loading=\"lazy\" sizes=\"(max-width: 326px) 100vw, 326px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q23.2.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q23.2.png 326w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q23.2-300x258.png 300w\" width=\"326\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q24.1\" class=\"alignnone size-full wp-image-66590\" height=\"194\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q24.1.png\" width=\"275\"\/><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 alt=\"RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q25.1\" class=\"alignnone size-full wp-image-66591\" height=\"184\" loading=\"lazy\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-11-Co-ordinate-Geometry-MCQS-Q25.1.png\" width=\"216\"\/><br \/>Bisectors of \u2220B and \u2220C meet at O, then 1<br \/>\u2220BOC = 90\u00b0 + <span class=\"MathJax\" id=\"MathJax-Element-12-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-98\"><\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Last Updated: April 04, 2026 | This article has been updated with the latest information for 2026. Read more: RD Sharma Class 9 \u2014 Complete Guide RD Sharma Class 9 Solutions Chapter 11 MCQs: MCQ type questions ko solve karna kuch students ke liye tricky ho sakta hai. 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