{"id":61717,"date":"2023-09-03T17:30:00","date_gmt":"2023-09-03T12:00:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=61717"},"modified":"2023-12-12T11:31:24","modified_gmt":"2023-12-12T06:01:24","slug":"rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 10 &#8211; Congruent Triangles (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124434\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/0202\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-10-Congruent-Triangles.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 10\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/0202\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-10-Congruent-Triangles.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/0202\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-10-Congruent-Triangles-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles:<\/strong> Start practisng for your Class 9 Maths exam with the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 9. All the solutions are designed by subject matter experts and that too as per the latest CBSE syllabus.&nbsp;<\/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-6a01c0f0b405a\" 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-6a01c0f0b405a\"><\/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-10-congruent-triangles\/#download-rd-sharma-solutions-class-9-maths-%e2%80%93-chapter-10-congruent-triangles-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths &#8211; Chapter 10 Congruent Triangles PDF\">Download RD Sharma Solutions Class 9 Maths &#8211; Chapter 10 Congruent Triangles 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-10-congruent-triangles\/#rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles-exercise-wise\" title=\"RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles: Exercise-Wise&nbsp;\">RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles: Exercise-Wise&nbsp;<\/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-10-congruent-triangles\/#detailed-exercise-wise-explanation-with-important-topics-in-the-exercise\" title=\"Detailed Exercise-wise Explanation with Important Topics in the Exercise\">Detailed Exercise-wise Explanation with Important Topics in the Exercise<\/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-10-congruent-triangles\/#rd-sharma-class-9-solutions-chapter-10-ex-10a\" title=\"RD Sharma Class 9 Solutions Chapter 10 Ex 10A\">RD Sharma Class 9 Solutions Chapter 10 Ex 10A<\/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-10-congruent-triangles\/#rd-sharma-class-9-chapter-10-exercise-10b\" title=\"RD Sharma Class 9 Chapter 10 Exercise 10B\">RD Sharma Class 9 Chapter 10 Exercise 10B<\/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-10-congruent-triangles\/#rd-sharma-solutions-class-9-chapter-10c\" title=\"RD Sharma Solutions Class 9 Chapter 10C\">RD Sharma Solutions Class 9 Chapter 10C<\/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-10-congruent-triangles\/#rd-sharma-class-9-solutions-chapter-10-ex10d\" title=\"RD Sharma Class 9 Solutions Chapter 10 Ex10D\">RD Sharma Class 9 Solutions Chapter 10 Ex10D<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#rd-sharma-class-9-chapter-10-exercise-10e\" title=\"RD Sharma Class 9 Chapter 10 Exercise 10E\">RD Sharma Class 9 Chapter 10 Exercise 10E<\/a><\/li><li class='ez-toc-page-1 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-10-congruent-triangles\/#rd-sharma-class-9-solutions-chapter-10-ex-10f\" title=\"RD Sharma Class 9 Solutions Chapter 10 Ex 10F\">RD Sharma Class 9 Solutions Chapter 10 Ex 10F<\/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-10-congruent-triangles\/#acess-answers-of-rd-sharma-solutions-class-9-chapter-10\" title=\"Acess answers of RD Sharma Solutions Class 9 Chapter 10\">Acess answers of RD Sharma Solutions Class 9 Chapter 10<\/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-10-congruent-triangles\/#rd-sharma-class-9-solutions-chapter-10-congruent-triangles-ex-101\" title=\"RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles&nbsp;Ex 10.1\">RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles&nbsp;Ex 10.1<\/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-10-congruent-triangles\/#class-9-rd-sharma-solutions-chapter-10-congruent-triangles-ex-102\" title=\"Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles Ex 10.2\">Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles Ex 10.2<\/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-10-congruent-triangles\/#class-9-maths-chapter-10-congruent-triangles-rd-sharma-solutions-ex-103\" title=\"Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions Ex 10.3\">Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions Ex 10.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#rd-sharma-class-9-maths-book-questions-chapter-10-congruent-triangles-ex-104\" title=\"RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles Ex 10.4\">RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles Ex 10.4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#rd-sharma-class-9-solution-chapter-10-congruent-triangles-vsaqs\" title=\"RD Sharma Class 9 Solution Chapter 10 Congruent Triangles VSAQS\">RD Sharma Class 9 Solution Chapter 10 Congruent Triangles VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#rd-sharma-solutions-class-9-chapter-10-congruent-triangles-mcqs\" title=\"RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles MCQS\">RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles MCQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#important-concepts-from-rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\" title=\"Important concepts from RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles\">Important concepts from RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangle\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangle\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangle<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-10\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 10?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 10?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-10\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 10?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 10?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-21\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-10-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 10\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 10\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-%e2%80%93-chapter-10-congruent-triangles-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths &#8211; Chapter 10 Congruent Triangles PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/0202\/09\/rd-10-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 10<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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\/0202\/09\/rd-10-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles-exercise-wise\"><\/span><strong>RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles: Exercise-Wise&nbsp;<\/strong><u><\/u><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-chapter-10-class-9-maths-exercise-10-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 10 Exercise 10.1<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-10-class-9-maths-exercise-10-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 10 Exercise 10.2<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-10-class-9-maths-exercise-10-3-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 10 Exercise 10.3<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-10-class-9-maths-exercise-10-4-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 10 Exercise 10.4<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-10-class-9-maths-exercise-10-5-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 10 Exercise 10.5<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-10-class-9-maths-exercise-10-6-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 10 Exercise 10.6<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-important-topics-in-the-exercise\"><\/span>Detailed Exercise-wise Explanation with Important Topics in the Exercise<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-10-ex-10a\"><\/span>RD Sharma Class 9 Solutions Chapter 10 Ex 10A<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the RD Sharma Class 9 Solutions Chapter 10 Ex 10A the students will have to deal with the questions from the Congruent Triangles part. The experts have given extra bit of attention to Congruence of Line segments, Congruence of Angles, Congruence of Triangles, Congruence Relation, and Congruence Criteria-<\/p>\n<p>-SAS (Side-Angle-Side)<\/p>\n<p>-SSS (Side-Side-Side)<\/p>\n<p>-ASA (Angle-Side-Angle)<\/p>\n<p>\u2013 RHS (Right angle- Hypotenuse-Side)<\/p>\n<p>as well as Congruence Criterion for fulfilling the needs of the students. Apart from these demands, you will also get to see things like in-depth explanations. Each part of RD Sharma Class 9 Chapter 10A has been covered to meet the requirements of the students. All the concepts have been explained in form of analysis.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-10-exercise-10b\"><\/span>RD Sharma Class 9 Chapter 10 Exercise 10B<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In Chapter 10 Ex-10B RD Sharma Class 9 the students will be able to examine their performances by attempting objective type questions which are associated with Congruent Triangles. After going through the solutions of RD Sharma Class 9 Chapter 10B the students will be able to take risks fearlessly.<\/p>\n<p>The preparation style of the students will undergo a lot of changes which will be beneficial for their performances in the final Maths Exam. Hence, you must buy a copy for yourself right now. The main aspects are based on the topic- Criterion of Congruence \u2013 ASA ( Angle-Side-Angle).<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-10c\"><\/span>RD Sharma Solutions Class 9 Chapter 10C<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the RD Sharma Solutions Class 9 Maths Chapter 10 Exercise 10C the students will find step by step explanations which have been done by different experts. &nbsp;The main areas which the experts have tried to focus on are problems based on the Criterion of Congruence \u2013 SAS (Side-Angle-Side).<\/p>\n<p>Once you examine the contents you will be able to figure out the parts which need your immediate attention.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-10-ex10d\"><\/span>RD Sharma Class 9 Solutions Chapter 10 Ex10D<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the RD Sharma Class 9 Chapter 10 Exercise 10D Solutions you will be exposed to all kinds of tricky problems from Congruent Triangles. Try to answer all the questions from RD Sharma Class 9 Chapter 10D without seeing the answers from Chapter 10 solutions. Assess your performance and make a note of it. After some time you check all the answers from RD Sharma Solutions Class 9 Chapter 10D.<\/p>\n<p>If you make a habit you will get the maximum benefit in your final Maths exam. While going through the concept of Congruence Criterion \u2013 SSS (Side-Side-Side).from RD Sharma Class 9 Solutions Chapter 10 Exercise 10D look for the ways the concepts have been presented. You will be in a commendable position if you look for the little things at the time of practising questions from Class 9 RD Sharma Chapter 10D.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-10-exercise-10e\"><\/span>RD Sharma Class 9 Chapter 10 Exercise 10E<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the Chapter 10E Solution of RD Sharma Class 9, the students will get all kind of explanations which will clear the basic concepts of Congruent Triangles chapter. The main area which has been dealt with here are connected with the Congruence Criterion \u2013 Right angle- Hypotenuse-Side.<\/p>\n<p>The step-wise explanations have been provided by the experts to satisfy the needs of the students. The important concepts will become crystal clear and the students will be confident enough to solve their own problems. The students are required to solve problems on a daily basis for maximum benefit.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-10-ex-10f\"><\/span>RD Sharma Class 9 Solutions Chapter 10 Ex 10F<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the exercise 10F of RD Sharma Solutions, students will get to know about the techniques which can be applied in case of Congruent Triangles. Parts like Inequality Relations in a Triangle have been addressed by the experts with the extra bit of care and all you need to do is sit with the relevant parts after understanding the concepts. You can also include the parts in your study materials while examining the essential aspects of Congruent Triangles.<\/p>\n<p>In the last exercise, you will have to focus on the various kinds of questions from Congruent Triangles. Students often face problems while answering objective-based questions and the RD Sharma Class 9 solutions will bring you back on track. The in-depth explanations and smart approach will help the students to plan ahead and score good marks in this part.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"acess-answers-of-rd-sharma-solutions-class-9-chapter-10\"><\/span>Acess answers of RD Sharma Solutions Class 9 Chapter 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-10-congruent-triangles-ex-101\"><\/span>RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles&nbsp;Ex 10.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Write the complement of each of the following angles:<br>(i) 20\u00b0<br>(ii) 35\u00b0<br>(iii) 90\u00b0<br>(iv) 77\u00b0<br>(v) 30\u00b0<br>Solution:<br>We know that two angles are complement to each other if their sum is 90\u00b0. Therefore,<br>(i) Complement of 20\u00b0 is (90\u00b0 \u2013 20\u00b0) = 70\u00b0<br>(ii) Complement of 35\u00b0 is (90\u00b0 \u2013 35\u00b0) = 55\u00b0<br>(iii) Complement of 90\u00b0 is (90\u00b0 \u2013 90\u00b0) = 0\u00b0<br>(iv) Complement of 77\u00b0 is (90\u00b0 \u2013 77\u00b0) = 13\u00b0<br>(v) Complement of 30\u00b0 is (90\u00b0 \u2013 30\u00b0) = 60\u00b0<\/p>\n<p>Question 2.<br>Write the supplement of each of the following angles:<br>(i) 54\u00b0<br>(ii) 132\u00b0<br>(iii) 138\u00b0<br>Solution:<br>We know that two angles are supplement to each other if their sum if 180\u00b0. Therefore,<br>(i) Supplement of 54\u00b0 is (180\u00b0 \u2013 54\u00b0) = 126\u00b0<br>(ii) Supplement of 132\u00b0 is (180\u00b0 \u2013 132\u00b0) = 48\u00b0<br>(iii) Supplement of 138\u00b0 is (180\u00b0 \u2013 138\u00b0) = 42\u00b0<\/p>\n<p>Question 3.<br>If an angle is 28\u00b0 less than its complement, find its measure.<br>Solution:<br>Let required angle = x, then<br>Its complement = x + 28\u00b0<br>\u2234&nbsp; x + x + 28\u00b0 = 90\u00b0 \u21d2&nbsp; 2x = 90\u00b0 \u2013 28\u00b0 = 62\u00b0<br>\u2234 x =&nbsp;<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\">62<\/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\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 31\u00b0<br>\u2234 Required angle = 31\u00b0<\/p>\n<p>Question 4.<br>If an angle is 30\u00b0 more than one half of its complement, find the measure of the angle.<br>Solution:<br>Let the measure of the required angle = x<br>\u2234&nbsp; Its complement =&nbsp; 90\u00b0 \u2013 x<br>\u2234&nbsp; x =&nbsp;<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>&nbsp;(90\u00b0 \u2013 x) + 30\u00b0<br>2x = 90\u00b0 \u2013 x + 60\u00b0<br>\u21d2 2x + x = 90\u00b0 + 60\u00b0<br>\u21d2&nbsp; 3x = 150\u00b0<br>\u21d2 x = &nbsp;<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=\"msubsup\"><span id=\"MathJax-Span-21\" class=\"texatom\"><span id=\"MathJax-Span-22\" class=\"mrow\"><span id=\"MathJax-Span-23\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-24\" class=\"texatom\"><span id=\"MathJax-Span-25\" class=\"mrow\"><span id=\"MathJax-Span-26\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-27\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp; = 50\u00b0<br>\u2234 Required angle = 50\u00b0<\/p>\n<p>Question 5.<br>Two supplementary angles are in the ratio 4 : 5. Find the angles.<br>Solution:<br>Ratio in two supplementary angles = 4 : 5<br>Let first angle = 4x<br>Then second angle = 5x<br>\u2234&nbsp; 4x + 5x = 180<br>\u21d2&nbsp; 9x = 180\u00b0<br>\u2234 x&nbsp; =&nbsp;<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-28\" class=\"math\"><span id=\"MathJax-Span-29\" class=\"mrow\"><span id=\"MathJax-Span-30\" class=\"mfrac\"><span id=\"MathJax-Span-31\" class=\"msubsup\"><span id=\"MathJax-Span-32\" class=\"texatom\"><span id=\"MathJax-Span-33\" class=\"mrow\"><span id=\"MathJax-Span-34\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-35\" class=\"texatom\"><span id=\"MathJax-Span-36\" class=\"mrow\"><span id=\"MathJax-Span-37\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-38\" class=\"mn\">9<\/span><\/span><\/span><\/span><\/span>&nbsp;= 20\u00b0<br>\u2234&nbsp; First angle = 4x = 4 x 20\u00b0 = 80\u00b0<br>and second angle = 5x<br>= 5 x 20\u00b0 = 100\u00b0<\/p>\n<p>Question 6.<br>Two supplementary angles differ by 48\u00b0. Find the angles.<br>Solution:<br>Let first angle = x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u201d<br>Then second angle = x + 48\u00b0<br>\u2234&nbsp; x + x + 48\u00b0 = 180\u00b0\u21d2&nbsp; 2x + 48\u00b0 = 180\u00b0<br>\u21d2&nbsp; 2x = 180\u00b0 \u2013 48\u00b0 = 132\u00b0<br>x=&nbsp;<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-39\" class=\"math\"><span id=\"MathJax-Span-40\" class=\"mrow\"><span id=\"MathJax-Span-41\" class=\"mfrac\"><span id=\"MathJax-Span-42\" class=\"msubsup\"><span id=\"MathJax-Span-43\" class=\"texatom\"><span id=\"MathJax-Span-44\" class=\"mrow\"><span id=\"MathJax-Span-45\" class=\"mn\">132<\/span><\/span><\/span><span id=\"MathJax-Span-46\" class=\"texatom\"><span id=\"MathJax-Span-47\" class=\"mrow\"><span id=\"MathJax-Span-48\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-49\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;=66\u00b0<br>\u2234&nbsp; First angle = 66\u00b0<br>and second angle = x + 48\u00b0 = 66\u00b0 + 48\u00b0 = 114\u00b0<br>\u2234 Angles are 66\u00b0, 114\u00b0<\/p>\n<p>Question 7.<br>An angle is equal to 8 times its complement. Determine its measure.<br>Solution:<br>Let the required angle = x<br>Then its complement angle = 90\u00b0 \u2013 x<br>\u2234 x = 8(90\u00b0 \u2013 x)<br>\u21d2 x = 720\u00b0 \u2013 8x \u21d2&nbsp; x + 8x = 720\u00b0<br>\u21d2 9x = 720\u00b0&nbsp;\u21d2 x =&nbsp;&nbsp;<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-50\" class=\"math\"><span id=\"MathJax-Span-51\" class=\"mrow\"><span id=\"MathJax-Span-52\" class=\"mfrac\"><span id=\"MathJax-Span-53\" class=\"msubsup\"><span id=\"MathJax-Span-54\" class=\"texatom\"><span id=\"MathJax-Span-55\" class=\"mrow\"><span id=\"MathJax-Span-56\" class=\"mn\">720<\/span><\/span><\/span><span id=\"MathJax-Span-57\" class=\"texatom\"><span id=\"MathJax-Span-58\" class=\"mrow\"><span id=\"MathJax-Span-59\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-60\" class=\"mn\">9<\/span><\/span><\/span><\/span><\/span>&nbsp;= 80\u00b0<br>\u2234&nbsp; Required angle = 80\u00b0<\/p>\n<p>Question 8.<br>If the angles (2x \u2013 10)\u00b0 and (x \u2013 5)\u00b0 are complementary angles, find x.<br>Solution:<br>First complementary angle = (2x \u2013 10\u00b0) and second = (x \u2013 5)\u00b0<br>\u2234 2x \u2013 10\u00b0 + x \u2013 5\u00b0 = 90\u00b0<br>\u21d2 3x \u2013 15\u00b0 = 90\u00b0 \u21d2&nbsp; 3x = 90\u00b0 + 15\u00b0 = 105\u00b0<br>\u2234 x =&nbsp;<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-61\" class=\"math\"><span id=\"MathJax-Span-62\" class=\"mrow\"><span id=\"MathJax-Span-63\" class=\"mfrac\"><span id=\"MathJax-Span-64\" class=\"msubsup\"><span id=\"MathJax-Span-65\" class=\"texatom\"><span id=\"MathJax-Span-66\" class=\"mrow\"><span id=\"MathJax-Span-67\" class=\"mn\">105<\/span><\/span><\/span><span id=\"MathJax-Span-68\" class=\"texatom\"><span id=\"MathJax-Span-69\" class=\"mrow\"><span id=\"MathJax-Span-70\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-71\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 35\u00b0<br>\u2234&nbsp; First angle = 2x \u2013 10\u00b0 = 2 x 35\u00b0 \u2013 10\u00b0<br>= 70\u00b0 \u2013 10\u00b0 = 60\u00b0<br>and second angle = x \u2013 5 = 35\u00b0 \u2013 5 = 30\u00b0<\/p>\n<p>Question 9.<br>If an angle differ from its complement by 10\u00b0, find the angle.<br>Solution:<br>Let required angle = x\u00b0<br>Then its complement angle = 90\u00b0 \u2013 x\u00b0<br>\u2234 x \u2013 (90\u00b0 \u2013 x) = 10<br>\u21d2&nbsp; x \u2013 90\u00b0 + x = 10\u00b0\u21d2&nbsp; 2x = 10\u00b0 + 90\u00b0 = 100\u00b0 100\u00b0<br>\u21d2 x =&nbsp;&nbsp;<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-72\" class=\"math\"><span id=\"MathJax-Span-73\" class=\"mrow\"><span id=\"MathJax-Span-74\" class=\"mfrac\"><span id=\"MathJax-Span-75\" class=\"msubsup\"><span id=\"MathJax-Span-76\" class=\"texatom\"><span id=\"MathJax-Span-77\" class=\"mrow\"><span id=\"MathJax-Span-78\" class=\"mn\">100<\/span><\/span><\/span><span id=\"MathJax-Span-79\" class=\"texatom\"><span id=\"MathJax-Span-80\" class=\"mrow\"><span id=\"MathJax-Span-81\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-82\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 50\u00b0<br>\u2234 Required angle = 50\u00b0<\/p>\n<p>Question 10.<br>If the supplement of an angle is two-third of itself Determine the angle and its supplement.<br>Solution:<br>Let required angle = x<br>Then its supplement angle = 180\u00b0 \u2013 x<br>\u2234&nbsp; (180\u00b0-x)=&nbsp;<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-83\" class=\"math\"><span id=\"MathJax-Span-84\" class=\"mrow\"><span id=\"MathJax-Span-85\" class=\"mfrac\"><span id=\"MathJax-Span-86\" class=\"mn\">2<\/span><span id=\"MathJax-Span-87\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>x<br>540\u00b0 \u2013 3x = 2x \u21d2 2x + 3x = 540\u00b0<br>\u21d2 5x = 540\u00b0\u21d2&nbsp; x =&nbsp;<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-88\" class=\"math\"><span id=\"MathJax-Span-89\" class=\"mrow\"><span id=\"MathJax-Span-90\" class=\"mfrac\"><span id=\"MathJax-Span-91\" class=\"msubsup\"><span id=\"MathJax-Span-92\" class=\"texatom\"><span id=\"MathJax-Span-93\" class=\"mrow\"><span id=\"MathJax-Span-94\" class=\"mn\">540<\/span><\/span><\/span><span id=\"MathJax-Span-95\" class=\"texatom\"><span id=\"MathJax-Span-96\" class=\"mrow\"><span id=\"MathJax-Span-97\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-98\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp;= 108\u00b0<br>-. Supplement angle = 180\u00b0 \u2013 108\u00b0 = 72\u00b0<\/p>\n<p>Question 11.<br>An angle is 14\u00b0 more than its complementary angle. What is its measure?<br>Solution:<br>Let required angle = x<br>Then its complementary angle = 90\u00b0 \u2013 x<br>\u2234&nbsp; x + 14\u00b0 = 90\u00b0 \u2013 x<br>x + x = 90\u00b0 \u2013 14\u00b0 \u21d2&nbsp; 2x = 76\u00b0<br>\u21d2 x =&nbsp;&nbsp;<span id=\"MathJax-Element-11-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=\"msubsup\"><span id=\"MathJax-Span-103\" class=\"texatom\"><span id=\"MathJax-Span-104\" class=\"mrow\"><span id=\"MathJax-Span-105\" class=\"mn\">76<\/span><\/span><\/span><span id=\"MathJax-Span-106\" class=\"texatom\"><span id=\"MathJax-Span-107\" class=\"mrow\"><span id=\"MathJax-Span-108\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-109\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 38\u00b0<br>\u2234&nbsp; Required angle = 38\u00b0<\/p>\n<p>Question 12.<br>The measure of an angle is twice the measure of its supplementary angle. Find its measure.<br>Solution:<br>Let the required angle = x<br>\u2234&nbsp; Its supplementary angle = 180\u00b0 \u2013 x<br>\u2234&nbsp; x = 2(180\u00b0-x) = 360\u00b0-2x<br>\u21d2&nbsp; x +&nbsp;2x = 360\u00b0<br>\u21d2 3x = 360\u00b0<br>\u21d2&nbsp; x =&nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-110\" class=\"math\"><span id=\"MathJax-Span-111\" class=\"mrow\"><span id=\"MathJax-Span-112\" class=\"mfrac\"><span id=\"MathJax-Span-113\" class=\"msubsup\"><span id=\"MathJax-Span-114\" class=\"texatom\"><span id=\"MathJax-Span-115\" class=\"mrow\"><span id=\"MathJax-Span-116\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-117\" class=\"texatom\"><span id=\"MathJax-Span-118\" class=\"mrow\"><span id=\"MathJax-Span-119\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-120\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 120\u00b0<br>\u2234&nbsp; Required angle = 120\u00b0<\/p>\n<p>Question 13.<br>If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.<br>Solution:<br>Let required angle = x<br>Then its complement angle = 90\u00b0 \u2013 x<br>and supplement angle = 180\u00b0 \u2013 x<br>\u2234&nbsp; 3(90\u00b0 \u2013 x) = 180\u00b0 \u2013 x<br>\u21d2 270\u00b0 \u2013 3x = 180\u00b0 \u2013 x<br>\u21d2270\u00b0 \u2013 180\u00b0 = -x + 3x =&gt; 2x = 90\u00b0<br>\u21d2 x = 45\u00b0<br>\u2234&nbsp; Required angle = 45\u00b0<\/p>\n<p>Question 14.<br>If the supplement of an angle is three times its complement, find the angle.<br>Solution:<br>Let required angle = x<br>Then its complement = 90\u00b0-&nbsp; x<br>and supplement = 180\u00b0 \u2013 x<br>\u2234&nbsp; 180\u00b0-x = 3(90\u00b0-x)<br>\u21d2&nbsp; 180\u00b0 \u2013 x = 270\u00b0 \u2013 3x<br>\u21d2&nbsp; -x + 3x = 270\u00b0 \u2013 180\u00b0<br>\u21d2 2x = 90\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-121\" class=\"math\"><span id=\"MathJax-Span-122\" class=\"mrow\"><span id=\"MathJax-Span-123\" class=\"mfrac\"><span id=\"MathJax-Span-124\" class=\"msubsup\"><span id=\"MathJax-Span-125\" class=\"texatom\"><span id=\"MathJax-Span-126\" class=\"mrow\"><span id=\"MathJax-Span-127\" class=\"mn\">90<\/span><\/span><\/span><span id=\"MathJax-Span-128\" class=\"texatom\"><span id=\"MathJax-Span-129\" class=\"mrow\"><span id=\"MathJax-Span-130\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-131\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;=45\u00b0<br>\u2234 Required angle = 45\u00b0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-rd-sharma-solutions-chapter-10-congruent-triangles-ex-102\"><\/span>Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles Ex 10.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In the figure, OA and OB are opposite rays:<br>(i) If x = 25\u00b0, what is the value of y?<br>(ii) If y = 35\u00b0, what is the value of x?<\/p>\n<p>Solution:<br>(i) If x = 25\u00b0<br>\u2234 3x = 3 x 25\u00b0 = 75\u00b0<br>But \u2220AOC + \u2220BOC = 180\u00b0 (Linear pair)<br>\u21d2 \u2220AOC + 75\u00b0 = 180\u00b0<br>\u21d2 \u2220AOC = 180\u00b0 \u2013 75\u00b0<br>\u21d2 \u2220AOC = 105\u00b0<br>\u2234 2y + 5 = 105\u00b0 \u21d2 2y= 105\u00b0 \u2013 5\u00b0 = 100\u00b0<br>\u21d2 y =&nbsp;<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-132\" class=\"math\"><span id=\"MathJax-Span-133\" class=\"mrow\"><span id=\"MathJax-Span-134\" class=\"mfrac\"><span id=\"MathJax-Span-135\" class=\"msubsup\"><span id=\"MathJax-Span-136\" class=\"texatom\"><span id=\"MathJax-Span-137\" class=\"mrow\"><span id=\"MathJax-Span-138\" class=\"mn\">100<\/span><\/span><\/span><span id=\"MathJax-Span-139\" class=\"texatom\"><span id=\"MathJax-Span-140\" class=\"mrow\"><span id=\"MathJax-Span-141\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-142\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 50\u00b0<br>\u2234 If x = 25\u00b0 then y = 50\u00b0<br>(ii) If y = 35\u00b0, then \u2220AOC = 2y + 5<br>\u2234 2y + 5 = 2 x 35\u00b0 + 5 = 70\u00b0 + 5 = 75\u00b0<br>But \u2220AOC + \u2220BOC = 180\u00b0 (Linear pair)<br>\u21d2 75\u00b0 + \u2220BOC = 180\u00b0<br>\u21d2 \u2220BOC = 180\u00b0-75\u00b0= 105\u00b0<br>\u2234 3x = 105\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-143\" class=\"math\"><span id=\"MathJax-Span-144\" class=\"mrow\"><span id=\"MathJax-Span-145\" class=\"mfrac\"><span id=\"MathJax-Span-146\" class=\"msubsup\"><span id=\"MathJax-Span-147\" class=\"texatom\"><span id=\"MathJax-Span-148\" class=\"mrow\"><span id=\"MathJax-Span-149\" class=\"mn\">105<\/span><\/span><\/span><span id=\"MathJax-Span-150\" class=\"texatom\"><span id=\"MathJax-Span-151\" class=\"mrow\"><span id=\"MathJax-Span-152\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-153\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 35\u00b0<br>\u2234 x = 35\u00b0<\/p>\n<p>Question 2.<br>In the figure, write all pairs of adjacent angles and all the linear pairs.<\/p>\n<p>Solution:<br>In the given figure,<br>(i) \u2220AOD, \u2220COD; \u2220BOC, \u2220COD; \u2220AOD, \u2220BOD and \u2220AOC, \u2220BOC are the pairs of adjacent angles.<br>(ii) \u2220AOD, \u2220BOD and \u2220AOC, \u2220BOC are the pairs of linear pairs.<\/p>\n<p>Question 3.<br>In the figure, find x, further find \u2220BOC, \u2220COD and \u2220AOD.<\/p>\n<p>Solution:<br>In the figure,<br>AOB is a straight line<br>\u2234 \u2220AOD + \u2220DOB = 180\u00b0 (Linear pair)<br>\u21d2 \u2220AOD + \u2220DOC + \u2220COB = 180\u00b0<br>\u21d2 x+ 10\u00b0 + x + x + 20 = 180\u00b0<br>\u21d2 3x + 30\u00b0 = 180\u00b0<br>\u21d2 3x= 180\u00b0 -30\u00b0= 150\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-154\" class=\"math\"><span id=\"MathJax-Span-155\" class=\"mrow\"><span id=\"MathJax-Span-156\" class=\"mfrac\"><span id=\"MathJax-Span-157\" class=\"msubsup\"><span id=\"MathJax-Span-158\" class=\"texatom\"><span id=\"MathJax-Span-159\" class=\"mrow\"><span id=\"MathJax-Span-160\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-161\" class=\"texatom\"><span id=\"MathJax-Span-162\" class=\"mrow\"><span id=\"MathJax-Span-163\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-164\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 50\u00b0<br>\u2234 x = 50\u00b0<br>Now \u2220BOC =x + 20\u00b0 = 50\u00b0 + 20\u00b0 = 70\u00b0<br>\u2220COD = x = 50\u00b0<br>and \u2220AOD = x + 10\u00b0 = 50\u00b0 + 10\u00b0 = 60\u00b0<\/p>\n<p>Question 4.<br>In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that \u2220AOB + \u2220BOC + \u2220COD + \u2220DOE + \u2220EOA = 360\u00b0.<\/p>\n<p>Solution:<br>Produce AO to F such that AOF is a straight line<br>Now \u2220AOB + \u2220BOF = 180\u00b0 (Linear pair)<\/p>\n<p>\u21d2 \u2220AOB + \u2220BOC + \u2220COF = 180\u00b0 \u2026(i)<br>Similarly, \u2220AOE + \u2220EOF = 180\u00b0<br>\u21d2 \u2220AOE + \u2220EOD + \u2220DOF = 180\u00b0 \u2026(ii)<br>Adding (i) and (ii)<br>\u2220AOB + \u2220BOC + \u2220COF + \u2220DOF + \u2220EOD + \u2220AOE = 180\u00b0 + 180\u00b0<br>\u21d2 \u2220AOB + \u2220BOC + \u2220COD + \u2220DOE + \u2220EOA = 360\u00b0<br>Hence proved.<\/p>\n<p>Question 5.<br>In the figure, \u2220AOC and \u2220BOC form a linear pair. If a \u2013 2b = 30\u00b0, find a and b.<\/p>\n<p>Solution:<br>\u2220AOC + \u2220BOC = 180\u00b0 (Linear pair)<br>\u21d2 a + 6 = 180\u00b0 \u2026(i)<br>and a \u2013 2b = 30\u00b0 \u2026(ii)<br>Subtracting (ii) from (i), 3b = 150\u00b0<br>\u21d2 b =&nbsp;<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-165\" class=\"math\"><span id=\"MathJax-Span-166\" class=\"mrow\"><span id=\"MathJax-Span-167\" class=\"mfrac\"><span id=\"MathJax-Span-168\" class=\"msubsup\"><span id=\"MathJax-Span-169\" class=\"texatom\"><span id=\"MathJax-Span-170\" class=\"mrow\"><span id=\"MathJax-Span-171\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-172\" class=\"texatom\"><span id=\"MathJax-Span-173\" class=\"mrow\"><span id=\"MathJax-Span-174\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-175\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 50\u00b0<br>and a + 50\u00b0 = 180\u00b0<br>\u21d2 a= 180\u00b0-50\u00b0= 130\u00b0<br>\u2234 a = 130\u00b0, b = 50\u00b0<\/p>\n<p>Question 6.<br>How many pairs of adjacent angles are formed when two lines intersect in a point?<br>Solution:<br>If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.<br>\u2220AOC, \u2220BOC; \u2220BOC, \u2220BOD; \u2220BOD, \u2220AOD and \u2220AOD, \u2220AOC<\/p>\n<p>Question 7.<br>How many pairs of adjacent angles, in all, can you name in the figure.<\/p>\n<p>Solution:<br>In the given figure 10 pairs of adjacent angles are formed as given below:<br>\u2220AOB, \u2220BOC; \u2220AOB, \u2220BOD; \u2220AOC, \u2220COD; \u2220AOD, \u2220BOE; \u2220AOB, \u2220BOE; \u2220AOC, \u2220COF; \u2220BOC, \u2220COD; \u2220BOC, \u2220COE; \u2220COD, \u2220DOE and \u2220BOD, \u2220DOE<\/p>\n<p>Question 8.<br>In the figure, determine the value of x.<\/p>\n<p>Solution:<br>In the figure<br>\u2220AOC + \u2220COB + \u2220BOD +\u2220AOD = 360 (angles at a point )<\/p>\n<p>\u21d2 3x + 3x + x + 150\u00b0 = 360\u00b0<br>\u21d2 7x \u2013 360\u00b0 \u2013 150\u00b0 = 210\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-18-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=\"msubsup\"><span id=\"MathJax-Span-180\" class=\"texatom\"><span id=\"MathJax-Span-181\" class=\"mrow\"><span id=\"MathJax-Span-182\" class=\"mn\">210<\/span><\/span><\/span><span id=\"MathJax-Span-183\" class=\"texatom\"><span id=\"MathJax-Span-184\" class=\"mrow\"><span id=\"MathJax-Span-185\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-186\" class=\"mn\">7<\/span><\/span><\/span><\/span><\/span>&nbsp;= 30\u00b0<br>\u2234 x = 30\u00b0<\/p>\n<p>Question 9.<br>In the figure, AOC is a line, find x.<\/p>\n<p>Solution:<br>In the figure,<br>\u2220AOB + \u2220BOC = 180\u00b0 (Linear pair)<br>\u21d2 70\u00b0 + 2x = 180\u00b0<br>\u21d2 2x = 180\u00b0 \u2013 70\u00b0<br>\u21d2 2x = 110\u00b0\u21d2x =&nbsp;<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-187\" class=\"math\"><span id=\"MathJax-Span-188\" class=\"mrow\"><span id=\"MathJax-Span-189\" class=\"mfrac\"><span id=\"MathJax-Span-190\" class=\"msubsup\"><span id=\"MathJax-Span-191\" class=\"texatom\"><span id=\"MathJax-Span-192\" class=\"mrow\"><span id=\"MathJax-Span-193\" class=\"mn\">110<\/span><\/span><\/span><span id=\"MathJax-Span-194\" class=\"texatom\"><span id=\"MathJax-Span-195\" class=\"mrow\"><span id=\"MathJax-Span-196\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-197\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 55\u00b0<br>\u2234 x = 55\u00b0<\/p>\n<p>Question 10.<br>In the figure, POS is a line, find x.<\/p>\n<p>Solution:<br>In the figure, POS is a line<br>\u2234 \u2220POQ + \u2220QOS = 180\u00b0 (Linear pair)<br>\u21d2 \u2220POQ + \u2220QOR + \u2220ROS = 180\u00b0<br>\u21d2 60\u00b0 + 4x + 40\u00b0 = 180\u00b0<br>\u21d2 4x + 100\u00b0 \u2013 180\u00b0<br>\u21d2 4x = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-20-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-198\" class=\"math\"><span id=\"MathJax-Span-199\" class=\"mrow\"><span id=\"MathJax-Span-200\" class=\"mfrac\"><span id=\"MathJax-Span-201\" class=\"msubsup\"><span id=\"MathJax-Span-202\" class=\"texatom\"><span id=\"MathJax-Span-203\" class=\"mrow\"><span id=\"MathJax-Span-204\" class=\"mn\">80<\/span><\/span><\/span><span id=\"MathJax-Span-205\" class=\"texatom\"><span id=\"MathJax-Span-206\" class=\"mrow\"><span id=\"MathJax-Span-207\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-208\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;=20\u00b0<br>\u2234 x = 20\u00b0<\/p>\n<p>Question 11.<br>In the figure, ACB is a line such that \u2220DCA = 5x and \u2220DCB = 4x. Find the value of x.<\/p>\n<p>Solution:<br>ACB is a line, \u2220DCA = 5x and \u2220DCB = 4x<br>\u2220ACD + \u2220DCB = 180\u00b0 (Linear pair)<br>\u21d2 5x + 4x = 180\u00b0 \u21d2 9x = 180\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-21-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-209\" class=\"math\"><span id=\"MathJax-Span-210\" class=\"mrow\"><span id=\"MathJax-Span-211\" class=\"mfrac\"><span id=\"MathJax-Span-212\" class=\"msubsup\"><span id=\"MathJax-Span-213\" class=\"texatom\"><span id=\"MathJax-Span-214\" class=\"mrow\"><span id=\"MathJax-Span-215\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-216\" class=\"texatom\"><span id=\"MathJax-Span-217\" class=\"mrow\"><span id=\"MathJax-Span-218\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-219\" class=\"mn\">9<\/span><\/span><\/span><\/span><\/span>&nbsp;= 20\u00b0<br>\u2234 x = 20\u00b0<br>\u2234 \u2220ACD = 5x = 5 x 20\u00b0 = 100\u00b0 and \u2220DCB = 4x = 4 x 20\u00b0 = 80\u00b0<\/p>\n<p>Question 12.<br>Given \u2220POR = 3x and \u2220QOR = 2x + 10\u00b0, find the value ofx for which POQ will be a line.<\/p>\n<p>Solution:<br>\u2220POR = 3x and \u2220QOR = 2x + 10\u00b0<br>If POQ is a line, then<br>\u2220POR + \u2220QOR = 180\u00b0 (Linear pair)<br>\u21d2 3x + 2x + 10\u00b0 = 180\u00b0<br>\u21d2 5x = 180\u00b0 \u2013 10\u00b0 = 170\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-22-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-220\" class=\"math\"><span id=\"MathJax-Span-221\" class=\"mrow\"><span id=\"MathJax-Span-222\" class=\"mfrac\"><span id=\"MathJax-Span-223\" class=\"msubsup\"><span id=\"MathJax-Span-224\" class=\"texatom\"><span id=\"MathJax-Span-225\" class=\"mrow\"><span id=\"MathJax-Span-226\" class=\"mn\">170<\/span><\/span><\/span><span id=\"MathJax-Span-227\" class=\"texatom\"><span id=\"MathJax-Span-228\" class=\"mrow\"><span id=\"MathJax-Span-229\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-230\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp;= 34\u00b0<br>\u2234 x = 34\u00b0<\/p>\n<p>Question 13.<br>What value ofy would make AOB, a line in the figure, if \u2220AOC = 4y and \u2220BOC = (6y + 30).<\/p>\n<p>Solution:<br>In the figure,<br>AOB is a line if<br>\u2220AOC + \u2220BOC = 180\u00b0<br>\u21d2 6y + 30\u00b0 + 4y= 180\u00b0<br>\u21d2 10y= 180\u00b0-30\u00b0= 150\u00b0<br>150\u00b0<br>\u21d2 y =&nbsp;<span id=\"MathJax-Element-23-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-231\" class=\"math\"><span id=\"MathJax-Span-232\" class=\"mrow\"><span id=\"MathJax-Span-233\" class=\"mfrac\"><span id=\"MathJax-Span-234\" class=\"msubsup\"><span id=\"MathJax-Span-235\" class=\"texatom\"><span id=\"MathJax-Span-236\" class=\"mrow\"><span id=\"MathJax-Span-237\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-238\" class=\"texatom\"><span id=\"MathJax-Span-239\" class=\"mrow\"><span id=\"MathJax-Span-240\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-241\" class=\"mn\">10<\/span><\/span><\/span><\/span><\/span>&nbsp;= 15\u00b0<br>\u2234 y = 15\u00b0<\/p>\n<p>Question 14.<br>In the figure, OP, OQ, OR and OS are four rays. Prove that: \u2220POQ + \u2220QOR + \u2220SOR + \u2220POS = 360\u00b0 [NCERT]<\/p>\n<p>Solution:<br>In the figure, OP, OQ, OR and OS are the rays from O<br>To prove : \u2220POQ + \u2220QOR + \u2220SOR + \u2220POS = 360\u00b0<br>Construction : Produce PO to E<\/p>\n<p>Proof: \u2220POQ + \u2220QOE = 180\u00b0 (Linear pair)<br>Similarly, \u2220EOS + \u2220POS = 180\u00b0<br>Adding we get,<br>\u2220POQ + \u2220QOR + \u2220ROE + \u2220EOS + \u2220POS = 180\u00b0 + 180\u00b0 ,<br>\u21d2 \u2220POQ + \u2220QOR + \u2220ROS + \u2220POS = 360\u00b0 Hence \u2220POQ + \u2220QOR + \u2220SOR + \u2220POS = 360\u00b0<\/p>\n<p>Question 15.<br>In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of \u2220POS and \u2220SOQ respectively. If \u2220POS = x, find \u2220ROT. [NCERT]<\/p>\n<p>Solution:<br>Ray OR stands on a line POQ forming \u2220POS and \u2220QOS<\/p>\n<p>OR and OT the angle bisects of \u2220POS and \u2220QOS respectively. \u2220POS = x<br>But \u2220POS + \u2220QOS = 180\u00b0 (Linear pair)<br>\u21d2 x + \u2220QOS = 180\u00b0<br>\u21d2 \u2220QOS = 180\u00b0 \u2013 x<br>\u2235 OR and OT are the bisectors of angle<\/p>\n<p>Question 16.<br>In the figure, lines PQ and RS intersect each other at point O. If \u2220POR : \u2220ROQ = 5 : 7. Find all the angles. [NCERT]<\/p>\n<p>Solution:<br>Lines PQ and PR, intersect each other at O<br>\u2235 Vertically opposite angles are equal<br>\u2234 \u2220POR = \u2220QOS and \u2220ROQ = \u2220POS<br>\u2220POR : \u2220ROQ = 5:7<br>Let \u2220POR = 5x and \u2220ROQ = 7x<br>But \u2220POR + \u2220ROQ = 180\u00b0 (Linear pair)<br>\u2234 5x + 7x = 180\u00b0 \u21d2 12x = 180\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-24-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-242\" class=\"math\"><span id=\"MathJax-Span-243\" class=\"mrow\"><span id=\"MathJax-Span-244\" class=\"mfrac\"><span id=\"MathJax-Span-245\" class=\"msubsup\"><span id=\"MathJax-Span-246\" class=\"texatom\"><span id=\"MathJax-Span-247\" class=\"mrow\"><span id=\"MathJax-Span-248\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-249\" class=\"texatom\"><span id=\"MathJax-Span-250\" class=\"mrow\"><span id=\"MathJax-Span-251\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-252\" class=\"mn\">12<\/span><\/span><\/span><\/span><\/span>&nbsp;= 15\u00b0<br>\u2234 \u2220POR = 5x = 5 x 15\u00b0 = 75\u00b0<br>and \u2220ROQ = 7x = 7 x 15\u00b0 = 105\u00b0<br>But \u2220QOS = POR = 75\u00b0 (Vertically opposite angles)<br>and \u2220POS = \u2220ROQ = 105\u00b0<\/p>\n<p>Question 17.<br>In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.<\/p>\n<p>Solution:<br>In the figure,<br>\u2220AOC + \u2220BOC = 180\u00b0 (Linear pair)<br>\u21d2 a + b =180\u00b0 \u2026(i)<br>But a = b +&nbsp;<span id=\"MathJax-Element-25-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-253\" class=\"math\"><span id=\"MathJax-Span-254\" class=\"mrow\"><span id=\"MathJax-Span-255\" class=\"mfrac\"><span id=\"MathJax-Span-256\" class=\"mn\">1<\/span><span id=\"MathJax-Span-257\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;x 90\u00b0 = b + 30\u00b0<br>\u21d2 a \u2013 b = 30\u00b0 \u2026(ii)<br>Adding (i) and (ii)<br>210\u00b0<br>2a = 210\u00b0 \u21d2 a =&nbsp;<span id=\"MathJax-Element-26-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-258\" class=\"math\"><span id=\"MathJax-Span-259\" class=\"mrow\"><span id=\"MathJax-Span-260\" class=\"mfrac\"><span id=\"MathJax-Span-261\" class=\"msubsup\"><span id=\"MathJax-Span-262\" class=\"texatom\"><span id=\"MathJax-Span-263\" class=\"mrow\"><span id=\"MathJax-Span-264\" class=\"mn\">210<\/span><\/span><\/span><span id=\"MathJax-Span-265\" class=\"texatom\"><span id=\"MathJax-Span-266\" class=\"mrow\"><span id=\"MathJax-Span-267\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-268\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 105\u00b0<br>and 105\u00b0 + b = 180\u00b0<br>\u21d2 b = 180\u00b0 \u2013 105\u00b0<br>\u2234 b = 75\u00b0<br>Hence a = 105\u00b0, b = 75\u00b0<\/p>\n<p>Question 18.<br>In the figure, \u2220AOF and \u2220FOG form a linear pair.<br>\u2220EOB = \u2220FOC = 90\u00b0 and \u2220DOC = \u2220FOG = \u2220AOB = 30\u00b0<\/p>\n<p>(i) Find the measures of \u2220FOE, \u2220COB and \u2220DOE.<br>(ii) Name all the right angles.<br>(iii) Name three pairs of adjacent complementary angles.<br>(iv) Name three pairs of adjacent supplementary angles.<br>(v) Name three pairs of adjacent angles.<br>Solution:<br>In the figure,<br>\u2220AOF and \u2220FOG form a linear pair<br>\u2220EOB = \u2220FOC = 90\u00b0<br>\u2220DOC = \u2220FOG = \u2220AOB = 30\u00b0<\/p>\n<p>(i) \u2220BOE = 90\u00b0, \u2220AOB = 30\u00b0<br>But \u2220BOE + \u2220AOB + \u2220EOG = 180\u00b0<br>\u21d2 30\u00b0 + 90\u00b0 + \u2220EOG = 180\u00b0<br>\u2234\u2220EOG = 180\u00b0 \u2013 30\u00b0 \u2013 90\u00b0 = 60\u00b0<br>But \u2220FOG = 30\u00b0<br>\u2234 \u2220FOE = 60\u00b0 \u2013 30\u00b0 = 30\u00b0<br>\u2220COD = 30\u00b0, \u2220COF = 90\u00b0<br>\u2234 \u2220DOF = 90\u00b0 \u2013 30\u00b0 = 60\u00b0<br>\u2220DOE = \u2220DOF \u2013 \u2220EOF<br>= 60\u00b0 \u2013 30\u00b0 = 30\u00b0<br>\u2220BOC = BOE \u2013 \u2220COE<br>= 90\u00b0 \u2013 30\u00b0 \u2013 30\u00b0 = 90\u00b0 \u2013 60\u00b0 = 30\u00b0<br>(ii) Right angles are,<br>\u2220AOD = 30\u00b0 + 30\u00b0 + 30\u00b0 = 90\u00b0<br>\u2220BOE = 30\u00b0 + 30\u00b0 + 30\u00b0 = 90\u00b0<br>\u2220COF = 30\u00b0 + 30\u00b0 + 30\u00b0 = 90\u00b0<br>and \u2220DOG = 30\u00b0 + 30\u00b0 + 30\u00b0 = 90\u00b0<br>(iii) Pairs of adjacent complementaiy angles are \u2220AOB, \u2220BOD; \u2220AOC, \u2220COD; \u2220BOC, \u2220COE<br>(iv) Pairs of adjacent supplementary angles are \u2220AOB, \u2220BOG; \u2220AOC, \u2220COG and \u2220AOD, \u2220DOG<br>(v) Pairs of adjacent angles are \u2220BOC, \u2220COD; \u2220COD, \u2220DOE and \u2220DOE, \u2220EOF.<\/p>\n<p>Question 19.<br>In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that \u2220ROS =&nbsp;<span id=\"MathJax-Element-27-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=\"mn\">1<\/span><span id=\"MathJax-Span-273\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>(\u2220QOS \u2013 \u2220POS). [NCERT]<\/p>\n<p>Solution:<br>Ray OR \u22a5 ROQ. OS is another ray lying between OP and OR.<\/p>\n<p>To prove : \u2220ROS =&nbsp;<span id=\"MathJax-Element-28-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-274\" class=\"math\"><span id=\"MathJax-Span-275\" class=\"mrow\"><span id=\"MathJax-Span-276\" class=\"mfrac\"><span id=\"MathJax-Span-277\" class=\"mn\">1<\/span><span id=\"MathJax-Span-278\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>(\u2220QOS \u2013 \u2220POS)<br>Proof : \u2235 RO \u22a5 POQ<br>\u2234 \u2220POR = 90\u00b0<br>\u21d2 \u2220POS + \u2220ROS = 90\u00b0 \u2026(i)<br>\u21d2 \u2220ROS = 90\u00b0 \u2013 \u2220POS<br>But \u2220POS + \u2220QOS = 180\u00b0 (Linear pair)<br>= 2(\u2220POS + \u2220ROS) [From (i)]<br>\u2220POS + \u2220QOS = 2\u2220ROS + 2\u2220POS<br>\u21d2 2\u2220ROS = \u2220POS + \u2220QOS \u2013 2\u2220POS<br>= \u2220QOS \u2013 \u2220POS<br>\u2234 ROS =&nbsp;<span id=\"MathJax-Element-29-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-279\" class=\"math\"><span id=\"MathJax-Span-280\" class=\"mrow\"><span id=\"MathJax-Span-281\" class=\"mfrac\"><span id=\"MathJax-Span-282\" class=\"mn\">1<\/span><span id=\"MathJax-Span-283\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(\u2220QOS \u2013 \u2220POS)<\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-maths-chapter-10-congruent-triangles-rd-sharma-solutions-ex-103\"><\/span>Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions Ex 10.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In the figure, lines l<sub>1<\/sub>&nbsp;and l<sub>2<\/sub>&nbsp;intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and n.<br><img src=\"https:\/\/farm2.staticflickr.com\/1941\/43813800120_8baa5667b1_o.png\" alt=\"RD Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"234\" height=\"115\"><br>Solution:<br>Two lines l<sub>1<\/sub>&nbsp;and l<sub>2<\/sub>&nbsp;intersect each other at O \u2220x = 45\u00b0<br>\u2235 \u2220z = \u2220x (Vertically opposite angles)<br>= 45\u00b0<br>But x + y = 180\u00b0 (Linear pair)<br>\u21d245\u00b0 + y= 180\u00b0<br>\u21d2 y= 180\u00b0-45\u00b0= 135\u00b0<br>But u = y (Vertically opposite angles)<br>\u2234 u = 135\u00b0<br>Hence y = 135\u00b0, z = 45\u00b0 and u = 135\u00b0<\/p>\n<p>Question 2.<br>In the figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.<br><img src=\"https:\/\/farm2.staticflickr.com\/1974\/43813799960_9e16f11ea0_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles\" width=\"221\" height=\"251\"><br>Solution:<br>Three lines AB, CD and EF intersect at O<br>\u2220BOD = 90\u00b0, \u2220DOF = 50\u00b0<br>\u2235 AB is a line<br>\u2234 \u2220BOD + \u2220DOF + FOA = 180\u00b0<br>\u21d2 90\u00b0 + 50\u00b0 + u = 180\u00b0<br>\u21d2 140\u00b0 + w = 180\u00b0<br>\u2234 u= 180\u00b0- 140\u00b0 = 40\u00b0<br>But x = u (Vertically opposite angles)<br>\u2234 x = 40\u00b0<br>Similarly, y = 50\u00b0 and z = 90\u00b0<br>Hence x = 40\u00b0, y = 50\u00b0, z = 90\u00b0 and u = 40\u00b0<\/p>\n<p>Question 3.<br>In the figure, find the values of x, y and z.<br><img src=\"https:\/\/farm2.staticflickr.com\/1945\/43813799590_d1e64f7d2f_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles\" width=\"222\" height=\"99\"><br>Solution:<br>Two lines l<sub>1<\/sub>&nbsp;and l<sub>2<\/sub>&nbsp;intersect each other at O<br>\u2234 Vertically opposite angles are equal,<br>\u2234 y = 25\u00b0 and x = z<br>Now 25\u00b0 + x = 180\u00b0 (Linear pair)<br>\u21d2 x= 180\u00b0-25\u00b0= 155\u00b0<br>\u2234 z = x = 155\u00b0<br>Hence x = 155\u00b0, y = 25\u00b0, z = 155\u00b0<\/p>\n<p>Question 4.<br>In the figure, find the value of x.<br><img src=\"https:\/\/farm2.staticflickr.com\/1959\/43813799430_6cccc212c6_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 10 Congruent Triangles\" width=\"204\" height=\"187\"><br>Solution:<br>\u2235 EF and CD intersect each other at O<br>\u2234 Vertically opposite angles are equal,<br>\u2234 \u22201 = 2x<br>AB is a line<br>3x + \u22201 + 5x = 180\u00b0 (Angles on the same side of a line)<br>\u21d2 3x + 2x + 5x = 180\u00b0<br>\u21d2 10x = 180\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-30-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-284\" class=\"math\"><span id=\"MathJax-Span-285\" class=\"mrow\"><span id=\"MathJax-Span-286\" class=\"mfrac\"><span id=\"MathJax-Span-287\" class=\"msubsup\"><span id=\"MathJax-Span-288\" class=\"texatom\"><span id=\"MathJax-Span-289\" class=\"mrow\"><span id=\"MathJax-Span-290\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-291\" class=\"texatom\"><span id=\"MathJax-Span-292\" class=\"mrow\"><span id=\"MathJax-Span-293\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-294\" class=\"mn\">10<\/span><\/span><\/span><\/span><\/span>&nbsp; = 18\u00b0<br>Hence x = 18\u00b0<\/p>\n<p>Question 5.<br>If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right angle.<br>Solution:<br>Given : Two lines AB and CD intersect each other at O. \u2220AOC = 90\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1909\/44717744295_bbcd123312_o.png\" alt=\"Congruent Triangles Class 9 RD Sharma Solutions\" width=\"258\" height=\"243\"><br>To prove: \u2220AOD = \u2220BOC = \u2220BOD = 90\u00b0<br>Proof : \u2235 AB and CD intersect each other at O<br>\u2234 \u2220AOC = \u2220BOD and \u2220BOC = \u2220AOD (Vertically opposite angles)<br>\u2234 But \u2220AOC = 90\u00b0<br>\u2234 \u2220BOD = 90\u00b0<br>\u2234 \u2220AOC + \u2220BOC = 180\u00b0 (Linear pair)<br>\u21d2 90\u00b0 + \u2220BOC = 180\u00b0<br>\u2234 \u2220BOC = 180\u00b0 -90\u00b0 = 90\u00b0<br>\u2234 \u2220AOD = \u2220BOC = 90\u00b0<br>\u2234 \u2220AOD = \u2220BOC = \u2220BOD = 90\u00b0<\/p>\n<p>Question 6.<br>In the figure, rays AB and CD intersect at O.<br>(i) Determine y when x = 60\u00b0<br>(ii) Determine x when y = 40\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1947\/44717743895_d7e56a3ecb_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 10 Congruent Triangles\" width=\"218\" height=\"165\"><br>Solution:<br>In the figure,<br>AB is a line<br>\u2234 2x + y = 180\u00b0 (Linear pair)<br>(i) If x = 60\u00b0, then<br>2 x 60\u00b0 + y = 180\u00b0<br>\u21d2 120\u00b0 +y= 180\u00b0<br>\u2234 y= 180\u00b0- 120\u00b0 = 60\u00b0<br>(ii) If y = 40\u00b0, then<br>2x + 40\u00b0 = 180\u00b0<br>\u21d2 2x = 180\u00b0 \u2013 40\u00b0 = 140\u00b0<br>\u21d2 x=&nbsp;<span id=\"MathJax-Element-31-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-295\" class=\"math\"><span id=\"MathJax-Span-296\" class=\"mrow\"><span id=\"MathJax-Span-297\" class=\"mfrac\"><span id=\"MathJax-Span-298\" class=\"msubsup\"><span id=\"MathJax-Span-299\" class=\"texatom\"><span id=\"MathJax-Span-300\" class=\"mrow\"><span id=\"MathJax-Span-301\" class=\"mn\">140<\/span><\/span><\/span><span id=\"MathJax-Span-302\" class=\"texatom\"><span id=\"MathJax-Span-303\" class=\"mrow\"><span id=\"MathJax-Span-304\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-305\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; =70\u00b0<br>\u2234 x = 70\u00b0<\/p>\n<p>Question 7.<br>In the figure, lines AB, CD and EF intersect at O. Find the measures of \u2220AOC, \u2220COF, \u2220DOE and \u2220BOF.<br><img src=\"https:\/\/farm2.staticflickr.com\/1907\/44717743735_10a1d32610_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles\" width=\"244\" height=\"148\"><br>Solution:<br>Three lines AB, CD and EF intersect each other at O<br>\u2220AOE = 40\u00b0 and \u2220BOD = 35\u00b0<br>(i) \u2220AOC = \u2220BOD (Vertically opposite angles)<br>= 35\u00b0<br>AB is a line<br>\u2234 \u2220AOE + \u2220DOE + \u2220BOD = 180\u00b0<br>\u21d2 40\u00b0 + \u2220DOE + 35\u00b0 = 180\u00b0<br>\u21d2 75\u00b0 + \u2220DOE = 180\u00b0<br>\u21d2 \u2220DOE = 180\u00b0-75\u00b0 = 105\u00b0<br>But \u2220COF = \u2220DOE (Vertically opposite angles)<br>\u2234 \u2220COF = 105\u00b0<br>Similarly, \u2220BOF = \u2220AOE (Vertically opposite angles)<br>\u21d2 \u2220BOF = 40\u00b0<br>Hence \u2220AOC = 35\u00b0, \u2220COF = 105\u00b0, \u2220DOE = 105\u00b0 and \u2220BOF = 40\u00b0<\/p>\n<p>Question 8.<br>AB, CD and EF are three concurrent lines passing through the point O such that OF bisects \u2220BOD. If \u2220BOF = 35\u00b0, find \u2220BOC and \u2220AOD.<br>Solution:<br>AB, CD and EF intersect at O. Such that OF is the bisector of<br>\u2220BOD \u2220BOF = 35\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1968\/43813797980_6b1db997d0_o.png\" alt=\"Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions\" width=\"284\" height=\"178\"><br>\u2235 OF bisects \u2220BOD,<br>\u2234 \u2220DOF = \u2220BOF = 35\u00b0 (Vertically opposite angles)<br>\u2234 \u2220BOD = 35\u00b0 + 35\u00b0 = 70\u00b0<br>But \u2220BOC + \u2220BOD = 180\u00b0 (Linear pair)<br>\u21d2 \u2220BOC + 70\u00b0 = 180\u00b0<br>\u21d2 \u2220BOC = 180\u00b0-70\u00b0= 110\u00b0<br>But \u2220AOD = \u2220BOC (Vertically opposite angles)<br>= 110\u00b0<br>Hence \u2220BOC = 110\u00b0 and \u2220AOD =110\u00b0<\/p>\n<p>Question 9.<br>In the figure, lines AB and CD intersect at O. If \u2220AOC + \u2220BOE = 70\u00b0 and \u2220BOD = 40\u00b0, find \u2220BOE and reflex \u2220COE.<br><img src=\"https:\/\/farm2.staticflickr.com\/1967\/43813797600_c933a1c9fc_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 10 Congruent Triangles\" width=\"280\" height=\"172\"><br>Solution:<br>In the figure, AB and CD intersect each other at O<br>\u2220AOC + \u2220BOE = 70\u00b0<br>\u2220BOD = 40\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1935\/43813797430_493ee79231_o.png\" alt=\"RD Sharma Class 9 Book Chapter 10 Congruent Triangles\" width=\"284\" height=\"177\"><br>AB is a line<br>\u2234 \u2220AOC + \u2220BOE + \u2220COE = 180\u00b0 (Angles on one side of a line)<br>\u21d2 70\u00b0 + \u2220COE = 180\u00b0<br>\u21d2 \u2220COE = 180\u00b0-70\u00b0= 110\u00b0<br>and \u2220AOC = \u2220BOD (Vertically opposite angles)<br>\u21d2 \u2220AOC = 40\u00b0<br>\u2234 \u2220BOE = 70\u00b0 \u2013 40\u00b0 = 30\u00b0<br>and reflex \u2220COE = 360\u00b0 \u2013 \u2220COE<br>= 360\u00b0- 110\u00b0 = 250\u00b0<\/p>\n<p>Question 10.<br>Which of the following statements are true (T) and which are false (F)?<br>(i) Angles forming a linear pair are supplementary.<br>(ii) If two adjacent angles are equal, then each angle measures 90\u00b0.<br>(iii) Angles forming a linear pair can both be acute angles.<br>(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90\u00b0.<br>Solution:<br>(i) True.<br>(ii) False. It can be possible if they are a linear pair.<br>(iii) False. In a linear pair, if one is acute, then the other will be obtuse.<br>(iv) True.<\/p>\n<p>Question 11.<br>Fill in the blanks so as to make the following statements true:<br>(i) If one angle of a linear pair is acute, then its other angle will be \u2026\u2026.. .<br>(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is \u2026\u2026\u2026 .<br>(iii) If the sum of two adjacent angles is 180\u00b0, then the \u2026\u2026 arms of the two angles are opposite rays.<br>Solution:<br>(i) If one angle of a linear pair is acute, then its other angle will be obtuse.<br>(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is 180\u00b0.<br>(iii) If the sum of two adjacent angles is 180\u00b0, then the uncommon arms of the two angles are opposite rays.<\/p>\n<p>Question 12.<br>Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.<br>Solution:<br>Given : Lines AB and CD intersect each other at O.<br>OE and OF are the bisectors of \u2220AOC and \u2220BOD respectively<br><img src=\"https:\/\/farm2.staticflickr.com\/1930\/43813797160_ec4430ddba_o.png\" alt=\"Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"282\" height=\"175\"><br>To prove : OE and OF are in the same line<br>Proof : \u2235 \u2220AOC = \u2220BOD (Vertically opposite angles)<br>\u2235 OE and OF are the bisectors of \u2220AOC and \u2220BOD<br>\u2234 \u22201 = \u22202 and \u22203 = \u22204<br>\u21d2 \u22201 = \u22202 =&nbsp;<span id=\"MathJax-Element-32-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-306\" class=\"math\"><span id=\"MathJax-Span-307\" class=\"mrow\"><span id=\"MathJax-Span-308\" class=\"mfrac\"><span id=\"MathJax-Span-309\" class=\"mn\">1<\/span><span id=\"MathJax-Span-310\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220AOC and<br>\u22203 = \u22204 =&nbsp;<span id=\"MathJax-Element-33-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-311\" class=\"math\"><span id=\"MathJax-Span-312\" class=\"mrow\"><span id=\"MathJax-Span-313\" class=\"mfrac\"><span id=\"MathJax-Span-314\" class=\"mn\">1<\/span><span id=\"MathJax-Span-315\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220BOD<br>\u2234 \u22201 = \u22202 = \u22203 = \u22204<br>\u2235 AOB is a line<br>\u2234 \u2220BOD + \u2220AOD = 180\u00b0 (Linear pair)<br>\u21d2 \u22203 + \u22204 + \u2220AOD = 180\u00b0<br>\u21d2 \u22203 + \u22201 + \u2220AOD = 180\u00b0 (\u2235 \u22201 = \u22204)<br>\u2234 EOF is a straight line<\/p>\n<p>Question 13.<br>If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.<br>Solution:<br>Given : AB and CD intersect each other at O. OE is the bisector of \u2220AOD and EO is produced to F.<br><img src=\"https:\/\/farm2.staticflickr.com\/1903\/43813796950_a4e1eb075f_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles\" width=\"285\" height=\"230\"><br>To prove : OF is the bisector of \u2220BOC<br>Proof : \u2235 AB and CD intersect each other at O<br>\u2234 \u2220AOD = \u2220BOC (Vertically opposite angles)<br>\u2235OE is the bisector of \u2220AOD<br>\u2234 \u22201 = \u22202<br>\u2235 AB and EF intersect each other at O<br>\u2234\u22201 = \u22204 (Vertically opposite angles) Similarly, CD and EF intersect each other at O<br>\u2234 \u22202 = \u22203<br>But \u22201 = \u22202<br>\u2234 \u22203 = \u22204<br>OF is the bisector of \u2220BOC<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-maths-book-questions-chapter-10-congruent-triangles-ex-104\"><\/span>RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles Ex 10.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In the figure, AB || CD and \u22201 and \u22202 are in the ratio 3 : 2. Determine all angles from 1 to 8.<\/p>\n<p>Solution:<br>AB || CD and l is transversal \u22201 : \u22202 = 3 : 2<br>Let \u22201 = 3x<br>Then \u22202 = 2x<br>But \u22201 + \u22202 = 180\u00b0 (Linear pair)<br>\u2234 3x + 2x = 180\u00b0 \u21d2 5x = 180\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-34-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-316\" class=\"math\"><span id=\"MathJax-Span-317\" class=\"mrow\"><span id=\"MathJax-Span-318\" class=\"mfrac\"><span id=\"MathJax-Span-319\" class=\"msubsup\"><span id=\"MathJax-Span-320\" class=\"texatom\"><span id=\"MathJax-Span-321\" class=\"mrow\"><span id=\"MathJax-Span-322\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-323\" class=\"texatom\"><span id=\"MathJax-Span-324\" class=\"mrow\"><span id=\"MathJax-Span-325\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-326\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp; = 36\u00b0<br>\u2234 \u22201 = 3x = 3 x 36\u00b0 = 108\u00b0<br>\u22202 = 2x = 2 x 36\u00b0 = 72\u00b0<br>Now \u22201 = \u22203 and \u22202 = \u22204 (Vertically opposite angles)<br>\u2234 \u22203 = 108\u00b0 and \u22204 = 72\u00b0<br>\u22201 = \u22205 and \u22202 = \u22206 (Corresponding angles)<br>\u2234 \u22205 = 108\u00b0, \u22206 = 72\u00b0<br>Similarly, \u22204 = \u22208 and<br>\u22203 = \u22207<br>\u2234 \u22208 = 72\u00b0 and \u22207 = 108\u00b0<br>Hence, \u22201 = 108\u00b0, \u22202= 72\u00b0<br>\u22203 = 108\u00b0, \u22204 = 72\u00b0<br>\u22205 = 108\u00b0, \u22206 = 72\u00b0<br>\u22207 = 108\u00b0, \u22208 = 12\u00b0<\/p>\n<p>Question 2.<br>In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find \u2220l, \u22202 and \u22203.<\/p>\n<p>Solution:<br>l || m || n and p is then transversal which intersects then at X, Y and Z respectively \u22204 = 120\u00b0<br>\u22202 = \u22204 (Alternate angles)<br>\u2234 \u22202 = 120\u00b0<br>But \u22203 + \u22204 = 180\u00b0 (Linear pair)<br>\u21d2 \u22203 + 120\u00b0 = 180\u00b0<br>\u21d2 \u22203 = 180\u00b0 \u2013 120\u00b0<br>\u2234 \u22203 = 60\u00b0<br>But \u2220l = \u22203 (Corresponding angles)<br>\u2234 \u2220l = 60\u00b0<br>Hence \u2220l = 60\u00b0, \u22202 = 120\u00b0, \u22203 = 60\u00b0<\/p>\n<p>Question 3.<br>In the figure, if AB || CD and CD || EF, find \u2220ACE.<\/p>\n<p>Solution:<br>Given : In the figure, AB || CD and CD || EF<br>\u2220BAC = 70\u00b0, \u2220CEF = 130\u00b0<br>\u2235 EF || CD<br>\u2234 \u2220ECD + \u2220CEF = 180\u00b0 (Co-interior angles)<br>\u21d2 \u2220ECD + 130\u00b0 = 180\u00b0<br>\u2234 \u2220ECD = 180\u00b0 \u2013 130\u00b0 = 50\u00b0<br>\u2235 BA || CD<br>\u2234 \u2220BAC = \u2220ACD (Alternate angles)<br>\u2234 \u2220ACD = 70\u00b0 (\u2235 \u2220BAC = 70\u00b0)<br>\u2235 \u2220ACE = \u2220ACD \u2013 \u2220ECD = 70\u00b0 \u2013 50\u00b0 = 20\u00b0<\/p>\n<p>Question 4.<br>In the figure, state which lines are parallel and why.<\/p>\n<p>Solution:<br>In the figure,<br>\u2235 \u2220ACD = \u2220CDE = 100\u00b0<br>But they are alternate angles<br>\u2234 AC || DE<\/p>\n<p>Question 5.<br>In the figure, if l || m,n|| p and \u22201 = 85\u00b0, find \u22202.<\/p>\n<p>Solution:<br>In the figure, l || m, n|| p and \u22201 = 85\u00b0<br>\u2235 n || p<br>\u2234 \u22201 = \u22203 (Corresponding anlges)<br>But \u22201 = 85\u00b0<br>\u2234 \u22203 = 85\u00b0<br>\u2235 m || 1<br>\u22203 + \u22202 = 180\u00b0 (Sum of co-interior angles)<br>\u21d2 85\u00b0 + \u22202 = 180\u00b0<br>\u21d2 \u22202 = 180\u00b0 \u2013 85\u00b0 = 95\u00b0<\/p>\n<p>Question 6.<br>If two straight lines are perpendicular to the same line, prove that they are parallel to each other.<br>Solution:<\/p>\n<p>Question 7.<br>Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.<br>Solution:<br>In ||gm ABCD,<\/p>\n<p>\u2220A and \u2220B are unequal<br>and \u2220A : \u2220B = 2 : 3<br>Let \u2220A = 2x, then<br>\u2220B = 3x<br>But \u2220A + \u2220B = 180\u00b0 (Co-interior angles)<br>\u2234 2x + 3x = 180\u00b0<br>\u21d2 5x = 180\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-35-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-327\" class=\"math\"><span id=\"MathJax-Span-328\" class=\"mrow\"><span id=\"MathJax-Span-329\" class=\"mfrac\"><span id=\"MathJax-Span-330\" class=\"msubsup\"><span id=\"MathJax-Span-331\" class=\"texatom\"><span id=\"MathJax-Span-332\" class=\"mrow\"><span id=\"MathJax-Span-333\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-334\" class=\"texatom\"><span id=\"MathJax-Span-335\" class=\"mrow\"><span id=\"MathJax-Span-336\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-337\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp; = 36\u00b0<br>\u2234 \u2220A = 2x = 2 x 36\u00b0 = 72\u00b0<br>\u2220B = 3x = 3 x 36\u00b0 = 108\u00b0<br>But \u2220A = \u2220C and \u2220B = \u2220D (Opposite angles of a ||gm)<br>\u2234 \u2220C = 72\u00b0 and \u2220D = 108\u00b0<br>Hence \u2220A = 72\u00b0, \u2220B = 108\u00b0, \u2220C = 72\u00b0, \u2220D = 108\u00b0<\/p>\n<p>Question 8.<br>In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?<br>Solution:<br>AB \u22a5 line l and CD \u22a5 line l<\/p>\n<p>\u2234 \u2220B = 90\u00b0 and \u2220D = 90\u00b0<br>\u2234 \u2220B = \u2220D<br>But there are corresponding angles<br>\u2234 AB || CD<\/p>\n<p>Question 9.<br>In the figure, \u22201 = 60\u00b0 and \u22202 = (<span id=\"MathJax-Element-36-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-338\" class=\"math\"><span id=\"MathJax-Span-339\" class=\"mrow\"><span id=\"MathJax-Span-340\" class=\"mfrac\"><span id=\"MathJax-Span-341\" class=\"mn\">2<\/span><span id=\"MathJax-Span-342\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>)3 a right angle. Prove that l || m.<\/p>\n<p>Solution:<br>In the figure, a transversal n intersects two lines l and m<br>\u22201 = 60\u00b0 and<br>\u22202 =&nbsp;<span id=\"MathJax-Element-37-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-343\" class=\"math\"><span id=\"MathJax-Span-344\" class=\"mrow\"><span id=\"MathJax-Span-345\" class=\"mfrac\"><span id=\"MathJax-Span-346\" class=\"mn\">2<\/span><span id=\"MathJax-Span-347\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;rd of a right angle 2<br>=&nbsp;<span id=\"MathJax-Element-38-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-348\" class=\"math\"><span id=\"MathJax-Span-349\" class=\"mrow\"><span id=\"MathJax-Span-350\" class=\"mfrac\"><span id=\"MathJax-Span-351\" class=\"mn\">2<\/span><span id=\"MathJax-Span-352\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;x 90\u00b0 = 60\u00b0<br>\u2234 \u22201 = \u22202<br>But there are corresponding angles<br>\u2234 l || m<\/p>\n<p>Question 10.<br>In the figure, if l || m || n and \u22201 = 60\u00b0, find \u22202.<\/p>\n<p>Solution:<br>In the figure,<\/p>\n<p>l || m || n and a transversal p, intersects them at P, Q and R respectively<br>\u22201 = 60\u00b0<br>\u2234 \u22201 = \u22203 (Corresponding angles)<br>\u2234 \u22203 = 60\u00b0<br>But \u22203 + \u22204 = 180\u00b0 (Linear pair)<br>60\u00b0 + \u22204 = 180\u00b0 \u21d2 \u22204 = 180\u00b0 \u2013 60\u00b0<br>\u2234 \u22204 = 120\u00b0<br>But \u22202 = \u22204 (Alternate angles)<br>\u2234 \u22202 = 120\u00b0<\/p>\n<p>Question 11.<br>Prove that the straight lines perpendicular to the same straight line are parallel to one another.<br>Solution:<br>Given : l is a line, AB \u22a5 l and CD \u22a5 l<\/p>\n<p>Question 12.<br>The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60\u00b0, find the other angles.<br>Solution:<br>In quadrilateral ABCD, AB || DC and AD || BC and \u2220A = 60\u00b0<\/p>\n<p>\u2235 AD || BC and AB || DC<br>\u2234 ABCD is a parallelogram<br>\u2234 \u2220A + \u2220B = 180\u00b0 (Co-interior angles)<br>60\u00b0 + \u2220B = 180\u00b0<br>\u21d2 \u2220B = 180\u00b0-60\u00b0= 120\u00b0<br>But \u2220A = \u2220C and \u2220B = \u2220D (Opposite angles of a ||gm)<br>\u2234 \u2220C = 60\u00b0 and \u2220D = 120\u00b0<br>Hence \u2220B = 120\u00b0, \u2220C = 60\u00b0 and \u2220D = 120\u00b0<\/p>\n<p>Question 13.<br>Two lines AB and CD intersect at O. If \u2220AOC + \u2220COB + \u2220BOD = 270\u00b0, find the measure of \u2220AOC, \u2220COB, \u2220BOD and \u2220DOA.<br>Solution:<br>Two lines AB and CD intersect at O<br>and \u2220AOC + \u2220COB + \u2220BOD = 270\u00b0<br>But \u2220AOC + \u2220COB + \u2220BOD + \u2220DOA = 360\u00b0 (Angles at a point)<\/p>\n<p>\u2234 270\u00b0 + \u2220DOA = 360\u00b0<br>\u21d2 \u2220DOA = 360\u00b0 \u2013 270\u00b0 = 90\u00b0<br>But \u2220DOA = \u2220BOC (Vertically opposite angles)<br>\u2234 \u2220BOC = 90\u00b0<br>But \u2220DOA + \u2220BOD = 180\u00b0 (Linear pair)<br>\u21d2 90\u00b0 + \u2220BOD = 180\u00b0<br>\u2234 \u2220BOD= 180\u00b0-90\u00b0 = 90\u00b0 ,<br>But \u2220BOD = \u2220AOC (Vertically opposite angles)<br>\u2234 \u2220AOC = 90\u00b0<br>Hence \u2220AOC = 90\u00b0,<br>\u2220COB = 90\u00b0,<br>\u2220BOD = 90\u00b0 and \u2220DOA = 90\u00b0<\/p>\n<p>Question 14.<br>In the figure, p is a transversal to lines m and n, \u22202 = 120\u00b0 and \u22205 = 60\u00b0. Prove that m || n.<\/p>\n<p>Solution:<br>Given : p is a transversal to the lines m and n<br>Forming \u2220l, \u22202, \u22203, \u22204, \u22205, \u22206, \u22207 and \u22208<br>\u22202 = 120\u00b0, and \u22205 = 60\u00b0<br>To prove : m || n<br>Proof : \u22202 + \u22203 = 180\u00b0 (Linear pair)<br>\u21d2 120\u00b0+ \u22203 = 180\u00b0<br>\u21d2 \u22203 = 180\u00b0- 120\u00b0 = 60\u00b0<br>But \u22205 = 60\u00b0<br>\u2234 \u22203 = \u22205<br>But there are alternate angles<br>\u2234 m || n<\/p>\n<p>Question 15.<br>In the figure, transversal l, intersects two lines m and n, \u22204 = 110\u00b0 and \u22207 = 65\u00b0. Is m || n?<\/p>\n<p>Solution:<br>A transversal l, intersects two lines m and n, forming \u22201, \u22202, \u22203, \u22204, \u22205, \u22206, \u22207 and \u22208<br>\u22204 = 110\u00b0 and \u22207 = 65\u00b0<br>To prove : Whether m || n or not<br>Proof : \u22204 = 110\u00b0 and \u22207 = 65\u00b0<br>\u22207 = \u22205 (Vertically opposite angles)<br>\u2234 \u22205 = 65\u00b0<br>Now \u22204 + \u22205 = 110\u00b0 + 65\u00b0 = 175\u00b0<br>\u2235 Sum of co-interior angles \u22204 and \u22205 is not 180\u00b0.<br>\u2234 m is not parallel to n<\/p>\n<p>Question 16.<br>Which pair of lines in the figure are parallel? Give reasons.<\/p>\n<p>Solution:<br>Given : In the figure, \u2220A = 115\u00b0, \u2220B = 65\u00b0, \u2220C = 115\u00b0 and \u2220D = 65\u00b0<br>\u2235 \u2220A + \u2220B = 115\u00b0+ 65\u00b0= 180\u00b0<br>But these are co-interior angles,<br>\u2234 AD || BC<br>Similarly, \u2220A + \u2220D = 115\u00b0 + 65\u00b0 = 180\u00b0<br>\u2234 AB || DC<\/p>\n<p>Question 17.<br>If l, m, n are three lines such that l ||m and n \u22a5 l, prove that n \u22a5 m.<br>Solution:<br>Given : l, m, n are three lines such that l || m and n \u22a5 l<\/p>\n<p>To prove : n \u22a5 m<br>Proof : \u2235 l || m and n is the transversal.<br>\u2234 \u2220l = \u22202 (Corresponding angles)<br>But \u22201 = 90\u00b0 (\u2235 n\u22a5l)<br>\u2234 \u22202 = 90\u00b0<br>\u2234 n \u22a5 m<\/p>\n<p>Question 18.<br>Which of the following statements are true (T) and which are false (F)? Give reasons.<br>(i) If two lines are intersected by a transversal, then corresponding angles are equal.<br>(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.<br>(iii) Two lines perpendicular to the same line are perpendicular to each other.<br>(iv) Two lines parallel to the same line are parallel to each other.<br>(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.<br>Solution:<br>(i) False. Because if lines are parallel, then it is possible.<br>(ii) True.<br>(iii) False. Not perpendicular but parallel to each other.<br>(iv) True.<br>(v) False. Sum of interior angles on the same side is 180\u00b0 not are equal.<\/p>\n<p>Question 19.<br>Fill in the blanks in each of the following to make the statement true:<br>(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are \u2026\u2026..<br>(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are \u2026\u2026.<br>(iii) Two lines perpendicular to the same line are \u2026\u2026\u2026 to each other.<br>(iv) Two lines parallel to the same line are \u2026\u2026\u2026 to each other.<br>(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are \u2026\u2026.<br>(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180\u00b0, then the lines are \u2026\u2026.<br>Solution:<br>(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.<br>(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.<br>(iii) Two lines perpendicular to the same line are parallel to each other.<br>(iv) Two lines parallel to the same line are parallel to each other.<br>(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.<br>(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180\u00b0, then the lines are parallel.<\/p>\n<p>Question 20.<br>In the figure, AB || CD || EF and GH || KL. Find \u2220HKL.<\/p>\n<p>Solution:<br>In the figure, AB || CD || EF and KL || HG Produce LK and GH<\/p>\n<p>\u2235 AB || CD and HK is transversal<br>\u2234 \u22201 = 25\u00b0 (Alternate angles)<br>\u22203 = 60\u00b0 (Corresponding angles)<br>and \u22203 = \u22204 (Corresponding angles)<br>= 60\u00b0<br>But \u22204 + \u22205 = 180\u00b0 (Linear pair)<br>\u21d2 60\u00b0 + \u22205 = 180\u00b0<br>\u21d2 \u22205 = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<br>\u2234 \u2220HKL = \u22201 + \u22205 = 25\u00b0 + 120\u00b0 = 145\u00b0<\/p>\n<p>Question 21.<br>In the figure, show that AB || EF.<\/p>\n<p>Solution:<br>Given : In the figure, AB || EF<br>\u2220BAC = 57\u00b0, \u2220ACE = 22\u00b0<br>\u2220ECD = 35\u00b0 and \u2220CEF =145\u00b0<br>To prove : AB || EF,<br>Proof : \u2220ECD + \u2220CEF = 35\u00b0 + 145\u00b0<br>= 180\u00b0<br>But these are co-interior angles<br>\u2234 EF || CD<br>But AB || CD<br>\u2234 AB || EF<\/p>\n<p>Question 22.<br>In the figure, PQ || AB and PR || BC. If \u2220QPR = 102\u00b0. Determine \u2220ABC. Give reasons.<\/p>\n<p>Solution:<br>In the figure, PQ || AB and PR || BC<br>\u2220QPR = 102\u00b0<br>Produce BA to meet PR at D<\/p>\n<p>\u2235 PQ || AB or DB<br>\u2234 \u2220QPR = \u2220ADR (Corresponding angles)<br>\u2234\u2220ADR = 102\u00b0 or \u2220BDR = 102\u00b0<br>\u2235 PR || BC<br>\u2234 \u2220BDR + \u2220DBC = 180\u00b0<br>(Sum of co-interior angles) \u21d2 102\u00b0 + \u2220DBC = 180\u00b0<br>\u21d2 \u2220DBC = 180\u00b0 \u2013 102\u00b0 = 78\u00b0<br>\u21d2 \u2220ABC = 78\u00b0<\/p>\n<p>Question 23.<br>Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.<br>Solution:<br>Given : In two angles \u2220ABC and \u2220DEF AB \u22a5 DE and BC \u22a5 EF<br>To prove: \u2220ABC + \u2220DEF = 180\u00b0 or \u2220ABC = \u2220DEF<br>Construction : Produce the sides DE and EF of \u2220DEF, to meet the sides of \u2220ABC at H and G.<\/p>\n<p>Proof: In figure (i) BGEH is a quadrilateral<br>\u2220BHE = 90\u00b0 and \u2220BGE = 90\u00b0<br>But sum of angles of a quadrilateral is 360\u00b0<br>\u2234 \u2220HBG + \u2220HEG = 360\u00b0 \u2013 (90\u00b0 + 90\u00b0)<br>= 360\u00b0 \u2013 180\u00b0= 180\u00b0<br>\u2234 \u2220ABC and \u2220DEF are supplementary<br>In figure (if) in quadrilateral BGEH,<br>\u2220BHE = 90\u00b0 and \u2220HEG = 90\u00b0<br>\u2234 \u2220HBG + \u2220HEG = 360\u00b0 \u2013 (90\u00b0 + 90\u00b0)<br>= 360\u00b0- 180\u00b0 = 180\u00b0 \u2026(i)<br>But \u2220HEF + \u2220HEG = 180\u00b0 \u2026(ii) (Linear pair)<br>From (i) and (ii)<br>\u2234 \u2220HEF = \u2220HBG<br>\u21d2 \u2220DEF = \u2220ABC<br>Hence \u2220ABC and \u2220DEF are equal or supplementary<\/p>\n<p>Question 24.<br>In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that \u2220ABP + \u2220CDP = \u2220DPB.<\/p>\n<p>Solution:<br>Given : In the figure, AB || CD<br>P is a point between AB and CD PD<br>and PB are joined<br>To prove : \u2220APB + \u2220CDP = \u2220DPB<br>Construction : Through P, draw PQ || AB or CD<\/p>\n<p>Proof: \u2235 AB || PQ<br>\u2234 \u2220ABP = BPQ \u2026(i) (Alternate angles)<br>Similarly,<br>CD || PQ<br>\u2234 \u2220CDP = \u2220DPQ \u2026(ii)<br>(Alternate angles)<br>Adding (i) and (ii)<br>\u2220ABP + \u2220CDP = \u2220BPQ + \u2220DPQ<br>Hence \u2220ABP + \u2220CDP = \u2220DPB<\/p>\n<p>Question 25.<br>In the figure, AB || CD and P is any point shown in the figure. Prove that:<br>\u2220ABP + \u2220BPD + \u2220CDP = 360\u00b0<\/p>\n<p>Solution:<br>Given : AB || CD and P is any point as shown in the figure<br>To prove : \u2220ABP + \u2220BPD + \u2220CDP = 360\u00b0<br>Construction : Through P, draw PQ || AB and CD<\/p>\n<p>Proof : \u2235 AB || PQ<br>\u2234 \u2220ABP+ \u2220BPQ= 180\u00b0 \u2026\u2026(i) (Sum of co-interior angles)<br>Similarly, CD || PQ<br>\u2234 \u2220QPD + \u2220CDP = 180\u00b0 \u2026(ii)<br>Adding (i) and (ii)<br>\u2220ABP + \u2220BPQ + \u2220QPD + \u2220CDP<br>= 180\u00b0+ 180\u00b0 = 360\u00b0<br>\u21d2 \u2220ABP + \u2220BPD + \u2220CDP = 360\u00b0<\/p>\n<p>Question 26.<br>In the figure, arms BA and BC of \u2220ABC are respectively parallel to arms ED and EF of \u2220DEF. Prove that \u2220ABC = \u2220DEF.<\/p>\n<p>Solution:<br>Given : In \u2220ABC and \u2220DEF. Their arms are parallel such that BA || ED and BC || EF<br>To prove : \u2220ABC = \u2220DEF<br>Construction : Produce BC to meet DE at G<br>Proof: AB || DE<br>\u2234 \u2220ABC = \u2220DGH\u2026(i) (Corresponding angles)<\/p>\n<p>BC or BH || EF<br>\u2234 \u2220DGH = \u2220DEF (ii) (Corresponding angles)<br>From (i) and (ii)<br>\u2220ABC = \u2220DEF<\/p>\n<p>Question 27.<br>In the figure, arms BA and BC of \u2220ABC are respectively parallel to arms ED and EF of \u2220DEF. Prove that \u2220ABC + \u2220DEF = 180\u00b0.<\/p>\n<p>Solution:<br>Given: In \u2220ABC = \u2220DEF<br>BA || ED and BC || EF<br>To prove: \u2220ABC = \u2220DEF = 180\u00b0<br>Construction : Produce BC to H intersecting ED at G<\/p>\n<p>Proof: \u2235 AB || ED<br>\u2234 \u2220ABC = \u2220EGH \u2026(i) (Corresponding angles)<br>\u2235 BC or BH || EF<br>\u2220EGH || \u2220DEF = 180\u00b0 (Sum of co-interior angles)<br>\u21d2 \u2220ABC + \u2220DEF = 180\u00b0 [From (i)]<br>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-10-congruent-triangles-vsaqs\"><\/span>RD Sharma Class 9 Solution Chapter 10 Congruent Triangles VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Define complementary angles.<br>Solution:<br>Two angles whose sum is 90\u00b0, are called complementary angles.<\/p>\n<p>Question 2.<br>Define supplementary angles.<br>Solution:<br>Two angles whose sum is 180\u00b0, are called supplementary angles.<\/p>\n<p>Question 3.<br>Define adjacent angles.<br>Solution:<br>Two angles which have common vertex and one arm common are called adjacent angles.<\/p>\n<p>Question 4.<br>The complement of an acute angles is\u2026\u2026.<br>Solution:<br>The complement of an acute angles is an acute angle.<\/p>\n<p>Question 5.<br>The supplement of an acute angles is\u2026\u2026\u2026<br>Solution:<br>The supplement of an acute angles is a obtuse angle.<\/p>\n<p>Question 6.<br>The supplement of a right angle is\u2026\u2026.<br>Solution:<br>The supplement of a right angle is a right angle.<\/p>\n<p>Question 7.<br>Write the complement of an angle of measure x\u00b0.<br>Solution:<br>The complement of x\u00b0 is (90\u00b0 \u2013 x)\u00b0<\/p>\n<p>Question 8.<br>Write the supplement of an angle of measure 2y\u00b0.<br>Solution:<br>The supplement of 2y\u00b0 is (180\u00b0 \u2013 2y)\u00b0<\/p>\n<p>Question 9.<br>If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.<br>Solution:<br>Total measure of angle around a point = 360\u00b0<br>Number of spokes = 6<br>\u2234 Angle between the two adjacent spokes =&nbsp;<span id=\"MathJax-Element-39-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-353\" class=\"math\"><span id=\"MathJax-Span-354\" class=\"mrow\"><span id=\"MathJax-Span-355\" class=\"mfrac\"><span id=\"MathJax-Span-356\" class=\"msubsup\"><span id=\"MathJax-Span-357\" class=\"texatom\"><span id=\"MathJax-Span-358\" class=\"mrow\"><span id=\"MathJax-Span-359\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-360\" class=\"texatom\"><span id=\"MathJax-Span-361\" class=\"mrow\"><span id=\"MathJax-Span-362\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-363\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>&nbsp;= 60\u00b0<\/p>\n<p>Question 10.<br>An angle is equal to its supplement. Determine its measure.<br>Solution:<br>Let required angle = x\u00b0<br>Then its supplement angle = 180\u00b0 \u2013 x<br>x = 180\u00b0 \u2013 x<br>\u21d2 x + x = 180\u00b0<br>\u21d2&nbsp; 2x = 180\u00b0 \u21d2&nbsp; x =&nbsp;<span id=\"MathJax-Element-40-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-364\" class=\"math\"><span id=\"MathJax-Span-365\" class=\"mrow\"><span id=\"MathJax-Span-366\" class=\"mfrac\"><span id=\"MathJax-Span-367\" class=\"msubsup\"><span id=\"MathJax-Span-368\" class=\"texatom\"><span id=\"MathJax-Span-369\" class=\"mrow\"><span id=\"MathJax-Span-370\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-371\" class=\"texatom\"><span id=\"MathJax-Span-372\" class=\"mrow\"><span id=\"MathJax-Span-373\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-374\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 90\u00b0<br>\u2234 Required angle = 90\u00b0<\/p>\n<p>Question 11.<br>An angle is equal to five times its complement. Determine its measure.<br>Solution:<br>Let required measure of angle = x\u00b0<br>\u2234&nbsp; Its complement angle = 90\u00b0 \u2013 x<br>\u2234&nbsp; x = 5(90\u00b0 \u2013 x)<br>\u21d2&nbsp; x = 450\u00b0 \u2013 5x<br>\u21d2&nbsp; x + 5x = 450\u00b0<br>\u21d2&nbsp; 6x = 450\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-41-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-375\" class=\"math\"><span id=\"MathJax-Span-376\" class=\"mrow\"><span id=\"MathJax-Span-377\" class=\"mfrac\"><span id=\"MathJax-Span-378\" class=\"msubsup\"><span id=\"MathJax-Span-379\" class=\"texatom\"><span id=\"MathJax-Span-380\" class=\"mrow\"><span id=\"MathJax-Span-381\" class=\"mn\">450<\/span><\/span><\/span><span id=\"MathJax-Span-382\" class=\"texatom\"><span id=\"MathJax-Span-383\" class=\"mrow\"><span id=\"MathJax-Span-384\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-385\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>&nbsp;= 75\u00b0<br>\u2234 Required angle = 75\u00b0<\/p>\n<p>Question 12.<br>How many pairs of adjacent angles are formed when two lines intersect in a point?<br>Solution:<br>If two lines AB and CD intersect at a point O, then pairs of two adjacent angles are, \u2220AOC and \u2220COB, \u2220COB and \u2220BOD, \u2220BOD and DOA, \u2220DOA and \u2220ZAOC<br>i.e, 4 pairs<br><img src=\"https:\/\/farm2.staticflickr.com\/1956\/44717750205_f0721aa81b_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 10 Congruent Triangles VSAQS\" width=\"232\" height=\"181\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-10-congruent-triangles-mcqs\"><\/span>RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Mark the correct alternative in each of the following:<br>Question 1.<br>One angle is equal to three times its supplement. The measure of the angle is<br>(a) 130\u00b0<br>(b) 135\u00b0<br>(c) 90\u00b0<br>(d) 120\u00b0<br>Solution:<br>Let required angle = x<br>Then its supplement = (180\u00b0 \u2013 x)<br>x = 3(180\u00b0 \u2013 x) = 540\u00b0 \u2013 3x<br>\u21d2 x + 3x = 540\u00b0<br>\u21d2 4x = 540\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-42-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-386\" class=\"math\"><span id=\"MathJax-Span-387\" class=\"mrow\"><span id=\"MathJax-Span-388\" class=\"mfrac\"><span id=\"MathJax-Span-389\" class=\"msubsup\"><span id=\"MathJax-Span-390\" class=\"texatom\"><span id=\"MathJax-Span-391\" class=\"mrow\"><span id=\"MathJax-Span-392\" class=\"mn\">540<\/span><\/span><\/span><span id=\"MathJax-Span-393\" class=\"texatom\"><span id=\"MathJax-Span-394\" class=\"mrow\"><span id=\"MathJax-Span-395\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-396\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp; = 135\u00b0<br>\u2234 Required angle = 135\u00b0 (b)<\/p>\n<p>Question 2.<br>Two straight lines AB and CD intersect one another at the point O. If \u2220AOC + \u2220COB + \u2220BOD = 274\u00b0, then \u2220AOD =<br>(a) 86\u00b0<br>(b) 90\u00b0<br>(c) 94\u00b0<br>(d) 137\u00b0<br>Solution:<br>Sum of angles at a point O = 360\u00b0<br>Sum of three angles \u2220AOC + \u2220COB + \u2220BOD = 274\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1907\/45580877072_ec256e9315_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles\" width=\"206\" height=\"171\"><br>\u2234 Fourth angle \u2220AOD = 360\u00b0 \u2013 274\u00b0<br>= 86\u00b0 (a)<\/p>\n<p>Question 3.<br>Two straight lines AB and CD cut each other at O. If \u2220BOD = 63\u00b0, then \u2220BOC =<br>(a) 63\u00b0<br>(b) 117\u00b0<br>(c) 17\u00b0<br>(d) 153\u00b0<br>Solution:<br>CD is a line<br>\u2234 \u2220BOD + \u2220BOC = 180\u00b0 (Linear pair)<br><img src=\"https:\/\/farm2.staticflickr.com\/1959\/45580876982_a34b44355d_o.png\" alt=\"Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions\" width=\"147\" height=\"178\"><br>\u21d2 63\u00b0 + \u2220BOC = 180\u00b0<br>\u21d2 \u2220BOC = 180\u00b0 \u2013 63\u00b0<br>\u2234 \u2220BOC =117\u00b0 (b)<\/p>\n<p>Question 4.<br>Consider the following statements:<br>When two straight lines intersect:<br>(i) adjacent angles are complementary<br>(ii) adjacent angles are supplementary<br>(iii) opposite angles are equal<br>(iv) opposite angles are supplementary Of these statements<br>(a) (i) and (iii) are correct<br>(b) (ii) and (iii) are correct<br>(c) (i) and (iv) are correct<br>(d) (ii) and (iv) are correct<br>Solution:<br>Only (ii) and (iii) arc true. (b)<\/p>\n<p>Question 5.<br>Given \u2220POR = 3x and \u2220QOR = 2x + 10\u00b0. If POQ is a striaght line, then the value of x is<br>(a) 30\u00b0<br>(b) 34\u00b0<br>(c) 36\u00b0<br>(d) none of these<br>Solution:<br>\u2235 POQ is a straight line<br>\u2234 \u2220POR + \u2220QOR = 180\u00b0 (Linear pair)<br>\u21d2 3x + 2x + 10\u00b0 = 180\u00b0<br>\u21d2 5x = 180 \u2013 10\u00b0 = 170\u00b0<br>\u2234 x =&nbsp;<span id=\"MathJax-Element-43-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=\"msubsup\"><span id=\"MathJax-Span-401\" class=\"texatom\"><span id=\"MathJax-Span-402\" class=\"mrow\"><span id=\"MathJax-Span-403\" class=\"mn\">170<\/span><\/span><\/span><span id=\"MathJax-Span-404\" class=\"texatom\"><span id=\"MathJax-Span-405\" class=\"mrow\"><span id=\"MathJax-Span-406\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-407\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp; = 34\u00b0 (b)<br><img src=\"https:\/\/farm2.staticflickr.com\/1941\/45580876752_c130cab833_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 10 Congruent Triangles\" width=\"201\" height=\"151\"><\/p>\n<p>Question 6.<br>In the figure, AOB is a straight line. If \u2220AOC + \u2220BOD = 85\u00b0, then \u2220COD =<br>(a) 85\u00b0<br>(b) 90\u00b0<br>(c) 95\u00b0<br>(d) 100\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1961\/45580876652_6a99f3cf59_o.png\" alt=\"RD Sharma Class 9 Book Chapter 10 Congruent Triangles\" width=\"250\" height=\"173\"><br>Solution:<br>AOB is a straight line,<br>OC and OD are rays on it<br>and \u2220AOC + \u2220BOD = 85\u00b0<br>But \u2220AOC + \u2220BOD + \u2220COD = 180\u00b0<br>\u21d2 85\u00b0 + \u2220COD = 180\u00b0<br>\u2220COD = 180\u00b0 \u2013 85\u00b0 = 95\u00b0 (c)<\/p>\n<p>Question 7.<br>In the figure, the value of y is<br>(a) 20\u00b0<br>(b) 30\u00b0<br>(c) 45\u00b0<br>(d) 60\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1970\/45580876592_e2677faa33_o.png\" alt=\"Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"205\" height=\"216\"><br>Solution:<br>In the figure,<br><img src=\"https:\/\/farm2.staticflickr.com\/1922\/45580876462_9734209360_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles\" width=\"239\" height=\"225\"><br>y = x (Vertically opposite angles)<br>\u22201 = 3x<br>\u22202 = 3x<br>\u2234 2(x + 3x + 2x) = 360\u00b0 (Angles at a point)<br>2x + 6x + 4x = 360\u00b0<br>12x = 360\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-44-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-408\" class=\"math\"><span id=\"MathJax-Span-409\" class=\"mrow\"><span id=\"MathJax-Span-410\" class=\"mfrac\"><span id=\"MathJax-Span-411\" class=\"msubsup\"><span id=\"MathJax-Span-412\" class=\"texatom\"><span id=\"MathJax-Span-413\" class=\"mrow\"><span id=\"MathJax-Span-414\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-415\" class=\"texatom\"><span id=\"MathJax-Span-416\" class=\"mrow\"><span id=\"MathJax-Span-417\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-418\" class=\"mn\">12<\/span><\/span><\/span><\/span><\/span>&nbsp; = 30\u00b0<br>\u2234 y = x = 30\u00b0 (b)<\/p>\n<p>Question 8.<br>In the figure, the value of x is<br>(a) 12<br>(b) 15<br>(c) 20<br>(d) 30<br><img src=\"https:\/\/farm2.staticflickr.com\/1950\/45580876132_0297750bae_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 10 Congruent Triangles\" width=\"207\" height=\"265\"><br>Solution:<br>\u22201 = 3x+ 10 (Vertically opposite angles)<br>But x + \u22201 + \u22202 = 180\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45580875912_6af1021a28_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"216\" height=\"272\"><br>\u21d2 x + 3x + 10\u00b0 + 90\u00b0 = 180\u00b0<br>\u21d2 4x = 180\u00b0 \u2013 10\u00b0 \u2013 90\u00b0 = 80\u00b0<br>x =&nbsp;<span id=\"MathJax-Element-45-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-419\" class=\"math\"><span id=\"MathJax-Span-420\" class=\"mrow\"><span id=\"MathJax-Span-421\" class=\"mfrac\"><span id=\"MathJax-Span-422\" class=\"msubsup\"><span id=\"MathJax-Span-423\" class=\"texatom\"><span id=\"MathJax-Span-424\" class=\"mrow\"><span id=\"MathJax-Span-425\" class=\"mn\">80<\/span><\/span><\/span><span id=\"MathJax-Span-426\" class=\"texatom\"><span id=\"MathJax-Span-427\" class=\"mrow\"><span id=\"MathJax-Span-428\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-429\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;= 20&nbsp; &nbsp;(c)<\/p>\n<p>Question 9.<br>In the figure, which of the following statements must be true?<br>(i) a + b = d + c<br>(ii) a + c + e = 180\u00b0<br>(iii) b + f= c + e<br>(a) (i) only<br>(b) (ii) only<br>(c) (iii) only<br>(d) (ii) and (iii) only<br><img src=\"https:\/\/farm2.staticflickr.com\/1927\/45580875642_32633417fd_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 10 Congruent Triangles\" width=\"172\" height=\"239\"><br>Solution:<br>In the figure,<br>(i) a + b = d + c<br>a\u00b0 = d\u00b0<br>b\u00b0 = e\u00b0<br>c\u00b0= f\u00b0<br>(ii) a + b + e = 180\u00b0<br>a + e + c = 180\u00b0<br>\u21d2 a + c + e = 180\u00b0<br>(iii) b + f= e + c<br>\u2234 (ii) and (iii) are true statements (d)<\/p>\n<p>Question 10.<br>If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is<br>(a) 54\u00b0<br>(b) 120\u00b0<br>(c) 108\u00b0<br>(d) 136\u00b0<br>Solution:<br>In figure, l || m and p is transversal<br><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44717756405_54611f28cd_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 10 Congruent Triangles\" width=\"277\" height=\"319\"><br>=&nbsp;<span id=\"MathJax-Element-46-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-430\" class=\"math\"><span id=\"MathJax-Span-431\" class=\"mrow\"><span id=\"MathJax-Span-432\" class=\"mfrac\"><span id=\"MathJax-Span-433\" class=\"mn\">3<\/span><span id=\"MathJax-Span-434\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp;x 180\u00b0 = 108\u00b0 (c)<\/p>\n<p>Question 11.<br>In the figure, if AB || CD, then the value of x is<br>(a) 20\u00b0<br>(b) 30\u00b0<br>(c) 45\u00b0<br>(d) 60\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1908\/45580875392_6b8120e2b5_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"235\" height=\"274\"><br>Solution:<br>In the figure, AB || CD,<br>and \/ is transversal<br>\u22201 = x (Vertically opposite angles)<br>and 120\u00b0 + x + \u22201 = 180\u00b0 (Co-interior angles)<br><img src=\"https:\/\/farm2.staticflickr.com\/1925\/44717756295_5d86635936_o.png\" alt=\"RD Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"367\" height=\"390\"><\/p>\n<p>Question 12.<br>Two lines AB and CD intersect at O. If \u2220AOC + \u2220COB + \u2220BOD = 270\u00b0, then \u2220AOC =<br>(a) 70\u00b0<br>(b) 80\u00b0<br>(c) 90\u00b0<br>(d) 180\u00b0<br>Solution:<br>Two lines AB and CD intersect at O<br><img src=\"https:\/\/farm2.staticflickr.com\/1921\/44717755795_f389a9b261_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles\" width=\"169\" height=\"205\"><br>\u2220AOC + \u2220COB + \u2220BOD = 270\u00b0 \u2026(i)<br>But \u2220AOC + \u2220COB + \u2220BOD + \u2220DOA = 360\u00b0 \u2026(ii)<br>Subtracting (i) from (ii),<br>\u2220DOA = 360\u00b0 \u2013 270\u00b0 = 90\u00b0<br>But \u2220DOA + \u2220AOC = 180\u00b0<br>\u2234 \u2220AOC = 180\u00b0 \u2013 90\u00b0 = 90\u00b0 (c)<\/p>\n<p>Question 13.<br>In the figure, PQ || RS, \u2220AEF = 95\u00b0, \u2220BHS = 110\u00b0 and \u2220ABC = x\u00b0. Then the value of x is<br>(a) 15\u00b0<br>(b) 25\u00b0<br>(c) 70\u00b0<br>(d) 35\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1922\/44717755475_083f08e1e6_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles\" width=\"247\" height=\"245\"><br>Solution:<br>In the figure,<br><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44717755205_8555ec768b_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 10 Congruent Triangles\" width=\"246\" height=\"232\"><br>PQ || RS, \u2220AEF = 95\u00b0<br>\u2220BHS = 110\u00b0, \u2220ABC = x<br>\u2235 PQ || RS,<br>\u2234 \u2220AEF = \u22201 = 95\u00b0 (Corresponding anlges)<br>But \u22201 + \u22202 = 180\u00b0 (Linear pair)<br>\u21d2 \u22202 = 180\u00b0 \u2013 \u22201 = 180\u00b0 \u2013 95\u00b0 = 85\u00b0<br>In \u2206AGH,<br>Ext. \u2220BHS = \u22202 +x<br>\u21d2 110\u00b0 = 85\u00b0 + x<br>\u21d2 x= 110\u00b0-85\u00b0 = 25\u00b0 (b)<\/p>\n<p>Question 14.<br>In the figure, if l1 || l2, what is the value of x?<br>(a) 90\u00b0<br>(b) 85\u00b0<br>(c) 75\u00b0<br>(d) 70\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1902\/44717754925_a809ae36a4_o.png\" alt=\"Congruent Triangles Class 9 RD Sharma Solutions\" width=\"283\" height=\"168\"><br>Solution:<br>In the figure,<br><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44717754635_de675d8839_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 10 Congruent Triangles\" width=\"350\" height=\"196\"><br>\u22201 = 58\u00b0 (Vertically opposite angles)<br>Similarly, \u22202 = 37\u00b0<br>\u2235 l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub>, EF is transversal<br>\u2220GEF + EFD = 180\u00b0 (Co-interior angles)<br>\u21d2 \u22202 + \u2220l +x = 180\u00b0<br>\u21d2 37\u00b0 + 58\u00b0 + x = 180\u00b0<br>\u21d2 95\u00b0 + x= 180\u00b0<br>x = 180\u00b0-95\u00b0 = 85\u00b0 (b)<\/p>\n<p>Question 15.<br>In the figure, if l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub>, what is x + y in terms of w and z?<br>(a) 180-w + z<br>(b) 180\u00b0 + w- z<br>(c) 180 -w- z<br>(d) 180 + w + z<br><img src=\"https:\/\/farm2.staticflickr.com\/1976\/45580873902_09eed0b22c_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles width=\" height=\"227\"><br>Solution:<br>In the figure, l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub><br><img src=\"https:\/\/farm2.staticflickr.com\/1916\/45580873662_4b6149f240_o.png\" alt=\"Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions\" width=\"246\" height=\"231\"><br>p and q are transversals<br>\u2234 w + x = 180\u00b0 \u21d2 x = 180\u00b0 \u2013 w (Co-interior angle)<br>z = y (Alternate angles)<br>\u2234 x + y = 180\u00b0 \u2013 w + z (a)<\/p>\n<p>Question 16.<br>In the figure, if l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub>, what is the value of y?<br>(a) 100<br>(b) 120<br>(c) 135<br>(d) 150<br><img src=\"https:\/\/farm2.staticflickr.com\/1931\/45580873402_31e148fe4c_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 10 Congruent Triangles\" width=\"258\" height=\"143\"><br>Solution:<br>In the figure, l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub>&nbsp;and l<sub>3<\/sub>&nbsp;is the transversal<br><img src=\"https:\/\/farm2.staticflickr.com\/1962\/44717753975_641504b9b9_o.png\" alt=\"RD Sharma Class 9 Book Chapter 10 Congruent Triangles\" width=\"358\" height=\"503\"><\/p>\n<p>Question 17.<br>In the figure, if l1 || l2 and l3 || l4 what is y in terms of x?<br>(a) 90 + x<br>(b) 90 + 2x<br>(c) 90 \u2013&nbsp;<span id=\"MathJax-Element-47-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-435\" class=\"math\"><span id=\"MathJax-Span-436\" class=\"mrow\"><span id=\"MathJax-Span-437\" class=\"mfrac\"><span id=\"MathJax-Span-438\" class=\"mi\">x<\/span><span id=\"MathJax-Span-439\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br>(d) 90 \u2013 2x<br><img src=\"https:\/\/farm2.staticflickr.com\/1932\/45580873182_1140d81811_o.png\" alt=\"Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"275\" height=\"257\"><br>Solution:<br>In the figure,<br><img src=\"https:\/\/farm2.staticflickr.com\/1923\/45580872792_6ceb9cdfc0_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles\" width=\"294\" height=\"257\"><br>l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub>&nbsp;and l<sub>3<\/sub>&nbsp;|| l<sub>4<\/sub>&nbsp;and m is the angle bisector<br>\u2234 \u22202 = \u22203 = y<br>\u2235 l<sub>1<\/sub>&nbsp;|| l<sub>2<\/sub><br>\u22201 = x (Corresponding angles)<br>\u2235 l<sub>3<\/sub>&nbsp;|| l<sub>4<\/sub><br>\u2234 \u22201 + (\u22202 + \u22203) = 180\u00b0 (Co-interior angles)<br>\u21d2 x + 2y= 180\u00b0<br>\u21d2 2y= 180\u00b0-x<br>\u21d2 y =&nbsp;<span id=\"MathJax-Element-48-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-440\" class=\"math\"><span id=\"MathJax-Span-441\" class=\"mrow\"><span id=\"MathJax-Span-442\" class=\"mfrac\"><span id=\"MathJax-Span-443\" class=\"mrow\"><span id=\"MathJax-Span-444\" class=\"msubsup\"><span id=\"MathJax-Span-445\" class=\"texatom\"><span id=\"MathJax-Span-446\" class=\"mrow\"><span id=\"MathJax-Span-447\" class=\"mn\">540<\/span><\/span><\/span><span id=\"MathJax-Span-448\" class=\"texatom\"><span id=\"MathJax-Span-449\" class=\"mrow\"><span id=\"MathJax-Span-450\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-451\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-452\" class=\"mi\">x<\/span><\/span><span id=\"MathJax-Span-453\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span><br>= 90\u00b0 \u2013&nbsp;<span id=\"MathJax-Element-49-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-454\" class=\"math\"><span id=\"MathJax-Span-455\" class=\"mrow\"><span id=\"MathJax-Span-456\" class=\"mfrac\"><span id=\"MathJax-Span-457\" class=\"mi\">x<\/span><span id=\"MathJax-Span-458\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(c)<\/p>\n<p>Question 18.<br>In the figure, if 11| m, what is the value of x?<br>(a) 60<br>(b) 50<br>(c) 45<br>d) 30<br><img src=\"https:\/\/farm2.staticflickr.com\/1909\/45580872532_1a36935f2b_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 10 Congruent Triangles\" width=\"313\" height=\"216\"><br>Solution:<br>In the figure, l || m and n is the transversal<br><img src=\"https:\/\/farm2.staticflickr.com\/1913\/45580872502_3a2d2ddf5a_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 10 Congruent Triangles \" width=\"351\" height=\"413\"><br>\u21d2 y = 25\u00b0<br>But 2y + 25\u00b0 = x+ 15\u00b0<br>(Vertically opposite angles) \u21d2 x = 2y + 25\u00b0 \u2013 15\u00b0 = 2y+ 10\u00b0<br>= 2 x 25\u00b0+10\u00b0 = 50\u00b0+10\u00b0 = 60\u00b0 (a)<\/p>\n<p>Question 19.<br>In the figure, if AB || HF and DE || FG, then the measure of \u2220FDE is<br>(a) 108\u00b0<br>(b) 80\u00b0<br>(c) 100\u00b0<br>(d) 90\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1941\/45580872322_35f6d5b7e1_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"275\" height=\"183\"><br>Solution:<br>In the figure,<br>AB || HF, DE || FG<br><img src=\"https:\/\/farm2.staticflickr.com\/1921\/45580872112_aa3f70b007_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles\" height=\"184\"><br>\u2234 HF || AB<br>\u22201 =28\u00b0 (Corresponding angles)<br>But \u22201 + \u2220FDE + 72\u00b0 \u2013 180\u00b0 (Angles of a straight line)<br>\u21d2 28\u00b0 + \u2220FDE + 72\u00b0 = 180\u00b0<br>\u21d2 \u2220FDE + 100\u00b0 = 180\u00b0<br>\u21d2 \u2220FDE = 180\u00b0 \u2013 100 = 80\u00b0 (b)<\/p>\n<p>Question 20.<br>In the figure, if lines l and m are parallel, then x =<br>(a) 20\u00b0<br>(b) 45\u00b0<br>(c) 65\u00b0<br>(d) 85\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1916\/45580871902_89537f0d9a_o.png\" alt=\"Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions\" width=\"229\" height=\"157\"><br>Solution:<br>In the figure, l || m<br><img src=\"https:\/\/farm2.staticflickr.com\/1972\/45580871662_96e7dd3d2f_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 10 Congruent Triangles\" width=\"262\" height=\"172\"><br>\u2234 \u22201 =65\u00b0 (Corresponding angles)<br>In \u2206BCD,<br>Ext. \u22201 = x + 20\u00b0<br>\u21d2 65\u00b0 = x + 20\u00b0<br>\u21d2 x = 65\u00b0 \u2013 20\u00b0<br>\u21d2 x = 45\u00b0 (b)<\/p>\n<p>Question 21.<br>In the figure, if AB || CD, then x =<br>(a) 100\u00b0<br>(b) 105\u00b0<br>(c) 110\u00b0<br>(d) 115\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1964\/45580871452_7a16c7a98c_o.png\" alt=\"RD Sharma Class 9 Book Chapter 10 Congruent Triangles\" width=\"249\" height=\"143\"><br>Solution:<br>In the figure, AB || CD<br>Through P, draw PQ || AB or CD<br><img src=\"https:\/\/farm2.staticflickr.com\/1923\/44717751765_6cba3bdf40_o.png\" alt=\"Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"274\" height=\"160\"><br>\u2220A + \u22201 = 180\u00b0 (Co-interior angles)<br>\u21d2 132\u00b0 + \u22201 = 180\u00b0<br>\u21d2 \u22201 = 180\u00b0- 132\u00b0 = 48\u00b0<br>\u2234 \u22202 = 148\u00b0 \u2013 \u22201 = 148\u00b0 \u2013 48\u00b0 = 100\u00b0<br>\u2235 DQ || CP<br>\u2234 \u22202 = x (Corresponding angles)<br>\u2234 x = 100\u00b0 (a)<\/p>\n<p>Question 22.<br>In tlie figure, if lines l and in are parallel lines, then x =<br>(a) 70\u00b0<br>(b) 100\u00b0<br>(c) 40\u00b0<br>(d) 30\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1901\/45580871252_67975fd0cd_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles\" width=\"197\" height=\"242\"><br>Solution:<br>In the figure, l || m<br><img src=\"https:\/\/farm2.staticflickr.com\/1962\/45580871152_d3b481cf9a_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 10 Congruent Triangles\" width=\"212\" height=\"261\"><br>\u2220l =70\u00b0 (Corresponding angles)<br>In \u2206DEF,<br>Ext. \u2220l = x + 30\u00b0<br>\u21d2 70\u00b0 = x + 30\u00b0<br>\u21d2 x = 70\u00b0 \u2013 30\u00b0 = 40\u00b0 (c)<\/p>\n<p>Question 23.<br>In the figure, if l || m, then x =<br>(a) 105\u00b0<br>(b) 65\u00b0<br>(c) 40\u00b0<br>(d) 25\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1972\/44717751455_75c2995614_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"235\" height=\"187\"><br>Solution:<br>In the figure,<br>l || m and n is the transversal<br><img src=\"https:\/\/farm2.staticflickr.com\/1947\/45580870952_d4776cc368_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 10 Congruent Triangles\" width=\"258\" height=\"206\"><br>\u22201 = 65\u00b0 (Alternate angles)<br>In \u2206GHF,<br>Ext. x = \u22201 + 40\u00b0 = 65\u00b0 + 40\u00b0<br>\u21d2 x = 105\u00b0<br>\u2234 x = 105\u00b0 (a)<\/p>\n<p>Question 24.<br>In the figure, if lines l and m are parallel, then the value of x is<br>(a) 35\u00b0<br>(b) 55\u00b0<br>(c) 65\u00b0<br>(d) 75\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44717751295_4bbb6fd5df_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 10 Congruent Triangles\" width=\"225\" height=\"152\"><br>Solution:<br>In the figure, l || m<br>and PQ is the transversal<br><img src=\"https:\/\/farm2.staticflickr.com\/1969\/45580870742_4740f7542c_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"242\" height=\"182\"><br>\u22201 = 90\u00b0<br>In \u2206EFG,<br>Ext. \u2220G = \u2220E + \u2220F<br>\u21d2 125\u00b0 = x + \u22201 = x + 90\u00b0<br>\u21d2 x = 125\u00b0 \u2013 90\u00b0 = 35\u00b0 (a)<\/p>\n<p>Question 25.<br>Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is<br>(a) 45\u00b0<br>(b) 30\u00b0<br>(c) 36\u00b0<br>(d) none of these<br>Solution:<br>Let first angle = x<br>Then its complementary angle = 90\u00b0 \u2013 x<br>\u2234 2x = 3(90\u00b0 \u2013 x)<br>\u21d2 2x = 270\u00b0 \u2013 3x<br>\u21d2 2x + 3x = 270\u00b0<br>\u21d2 5x = 270\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-50-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-459\" class=\"math\"><span id=\"MathJax-Span-460\" class=\"mrow\"><span id=\"MathJax-Span-461\" class=\"mfrac\"><span id=\"MathJax-Span-462\" class=\"msubsup\"><span id=\"MathJax-Span-463\" class=\"texatom\"><span id=\"MathJax-Span-464\" class=\"mrow\"><span id=\"MathJax-Span-465\" class=\"mn\">270<\/span><\/span><\/span><span id=\"MathJax-Span-466\" class=\"texatom\"><span id=\"MathJax-Span-467\" class=\"mrow\"><span id=\"MathJax-Span-468\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-469\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp; = 54\u00b0<br>\u2234 second angle = 90\u00b0 \u2013 54\u00b0 = 36\u00b0<br>\u2234 smaller angle = 36\u00b0 (c)<\/p>\n<p>Question 26.<br><img src=\"https:\/\/farm2.staticflickr.com\/1913\/43813805800_0626ffdb3a_o.png\" alt=\"RD Sharma Class 9 Chapter 10 Congruent Triangles\" width=\"351\" height=\"269\"><br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1924\/45580870632_6a7a2b0006_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles\" width=\"354\" height=\"470\"><\/p>\n<p>Question 27.<br>In the figure, AB || CD || EF and GH || KL.<br>The measure of \u2220HKL is<br>(a) 85\u00b0<br>(b) 135\u00b0<br>(c) 145\u00b0<br>(d) 215\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1952\/44717750805_4849633b58_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles\" width=\"260\" height=\"214\"><br>Solution:<br>In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.<br><img src=\"https:\/\/farm2.staticflickr.com\/1940\/43813805330_cef63fc899_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 10 Congruent Triangles\" width=\"247\" height=\"238\"><br>\u2235 AB || CD<br>\u2234 \u22201 = 25\u00b0 (Alternate angle)<br>and GH || KL<br>\u2234 \u22204 = 60\u00b0 (Corresponding angles)<br>\u22205 = \u22204 = 60\u00b0 (Vertically opposite angle)<br>\u22205 + \u22202 = 180\u00b0 (Co-interior anlges)<br>\u2234 \u21d2 60\u00b0 + \u22202 = 180\u00b0<br>\u22202 = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<br>Now \u2220HKL = \u22201 + \u22202 = 25\u00b0 + 120\u00b0<br>= 145\u00b0 (c)<\/p>\n<p>Question 28.<br>AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of \u2220FEB. If \u2220LEB = 35\u00b0, then \u2220CFQ will be<br>(a) 55\u00b0<br>(b) 70\u00b0<br>(c) 110\u00b0<br>(d) 130\u00b0<br>Solution:<br>AB || CD and PQ is the transversal EL is the bisector of \u2220FEB and \u2220LEB = 35\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44717750575_18a7de62ed_o.png\" alt=\"Congruent Triangles Class 9 RD Sharma Solutions\" width=\"259\" height=\"211\"><br>\u2234 \u2220FEB = 2 x 35\u00b0 = 70\u00b0<br>\u2235 AB || CD<br>\u2234 \u2220FEB + \u2220EFD = 180\u00b0<br>(Co-interior angles)<br>70\u00b0 + \u2220EFD = 180\u00b0<br>\u2234 \u2220EFD = 180\u00b0-70\u00b0= 110\u00b0<br>But \u2220CFQ = \u2220EFD<br>(Vertically opposite angles)<br>\u2234 \u2220CFQ =110\u00b0 (c)<\/p>\n<p>Question 29.<br>In the figure, if line segment AB is parallel to the line segment CD, what is the value of y?<br>(a) 12<br>(b) 15<br>(c) 18<br>(d) 20<br><img src=\"https:\/\/farm2.staticflickr.com\/1952\/43813805120_24205e136f_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 10 Congruent Triangles\" width=\"261\" height=\"148\"><br>Solution:<br>In the figure, AB || CD<br>BD is transversal<br>\u2234 \u2220ABD + \u2220BDC = 180\u00b0 (Co-interior angles)<br>\u21d2y + 2y+y + 5y = 180\u00b0<br>\u21d2 9y = 180\u00b0 \u21d2 y =&nbsp;<span id=\"MathJax-Element-51-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-470\" class=\"math\"><span id=\"MathJax-Span-471\" class=\"mrow\"><span id=\"MathJax-Span-472\" class=\"mfrac\"><span id=\"MathJax-Span-473\" class=\"msubsup\"><span id=\"MathJax-Span-474\" class=\"texatom\"><span id=\"MathJax-Span-475\" class=\"mrow\"><span id=\"MathJax-Span-476\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-477\" class=\"texatom\"><span id=\"MathJax-Span-478\" class=\"mrow\"><span id=\"MathJax-Span-479\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-480\" class=\"mn\">9<\/span><\/span><\/span><\/span><\/span>&nbsp; = 20\u00b0 (d)<\/p>\n<p>Question 30.<br>In the figure, if CP || DQ, then the measure of x is<br>(a) 130\u00b0<br>(b) 105\u00b0<br>(c) 175\u00b0<br>(d) 125\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1974\/44717750355_3545af13b9_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles\" width=\"318\" height=\"188\"><br>Solution:<br>In the figure, CP || DQ<br>BA is transversal<br>Produce PC to meet BA at D<br><img src=\"https:\/\/farm2.staticflickr.com\/1970\/43813804930_8db421e7ae_o.png\" alt=\"Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions\" width=\"320\" height=\"199\"><br>\u2235 QB || PD<br>\u2234 \u2220D = 105\u00b0 (Corresponding angles)<br>In \u2206ADC,<br>Ext. \u2220ACP = \u2220CDA + \u2220DAC<br>\u21d2 x = \u22201 + 25\u00b0<br>= 105\u00b0 + 25\u00b0 = 130\u00b0 (a)<\/p>\n<h3><span class=\"ez-toc-section\" id=\"important-concepts-from-rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\"><\/span>Important concepts from RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li>Congruence of Line segments<\/li>\n<li>Congruence of Angles<\/li>\n<li>Congruence of Triangles<\/li>\n<li>Congruence Relation<\/li>\n<li>Some inequality Relations in a Triangle<\/li>\n<\/ul>\n<p>This is the complete blog on RD Sharma Solution Class 9 Maths Chapter 9. If you have any doubts regarding the <a href=\"https:\/\/cbse.nic.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-10-congruent-triangle\"><\/span>FAQs on RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangle<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-1630673380443\" 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-10\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 10?<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-1630673483047\" 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-10\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 10?<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-1630673497147\" 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-10-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 10\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 10 Congruent Triangles: Start practisng for your Class 9 Maths exam with the RD Sharma Solutions Class 9 Maths Chapter 9. All the solutions are designed by subject matter experts and that too as per the latest CBSE syllabus.&nbsp; Download RD Sharma Solutions Class 9 Maths &#8211; Chapter &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 10 &#8211; Congruent Triangles (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 10 &#8211; Congruent Triangles (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124434,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3081,3037,3048],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/61717"}],"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=61717"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/61717\/revisions"}],"predecessor-version":[{"id":520872,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/61717\/revisions\/520872"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124434"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=61717"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=61717"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=61717"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}