{"id":62900,"date":"2023-03-30T04:02:00","date_gmt":"2023-03-29T22:32:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62900"},"modified":"2023-11-28T10:11:29","modified_gmt":"2023-11-28T04:41:29","slug":"rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 14 &#8211; Quadrilaterals (Updated for 2024)"},"content":{"rendered":"\n<p><span style=\"font-weight: 400;\"><strong>RD Sharma Solutions Class 9 Maths Chapter 14 &#8211; Quadrilaterals: Quadrilateral:<\/strong> In this blog, we will be dealing with the important aspects of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 14 which will guide you to bring changes in your methods of preparation. The parts that we will be concentrating on are the benefits of following Chapter 14 Solutions along with the exercise-wise explanations. <\/span><\/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-69d10e35f2b0c\" 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-69d10e35f2b0c\"><\/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-14-quadrilaterals\/#download-rd-sharma-class-9-solutions-chapter-14-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 14 PDF\">Download RD Sharma Class 9 Solutions Chapter 14 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-14-quadrilaterals\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-14\" title=\"Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 14\u00a0\">Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 14\u00a0<\/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-14-quadrilaterals\/#access-answers-of-rd-sharma-solutions-class-9-maths-chapter-14\" title=\"Access answers of RD Sharma Solutions Class 9 Maths Chapter 14\u00a0\">Access answers of RD Sharma Solutions Class 9 Maths Chapter 14\u00a0<\/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-14-quadrilaterals\/#rd-sharma-class-9-solution-chapter-14-quadrilaterals-ex-141\" title=\"RD Sharma Class 9 Solution Chapter 14 Quadrilaterals Ex 14.1\">RD Sharma Class 9 Solution Chapter 14 Quadrilaterals Ex 14.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/#exercise-142\" title=\"Exercise 14.2\">Exercise 14.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/#exercise-143\" title=\"Exercise 14.3\">Exercise 14.3<\/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-14-quadrilaterals\/#exercise-144\" title=\"Exercise 14.4\">Exercise 14.4<\/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-14-quadrilaterals\/#exercise-vsaqs\" title=\"Exercise VSAQs\">Exercise VSAQs<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/#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-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/#rd-sharma-class-9-chapter-14-exercise-14a\" title=\"RD Sharma Class 9 Chapter 14 Exercise 14A\">RD Sharma Class 9 Chapter 14 Exercise 14A<\/a><\/li><li class='ez-toc-page-1 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-14-quadrilaterals\/#rd-sharma-class-9-chapter-14-exercise-14b\" title=\"RD Sharma Class 9 Chapter 14 Exercise 14B\">RD Sharma Class 9 Chapter 14 Exercise 14B<\/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-14-quadrilaterals\/#rd-sharma-class-9-chapter-14-exercise-14c\" title=\"RD Sharma Class 9 Chapter 14 Exercise 14C\">RD Sharma Class 9 Chapter 14 Exercise 14C<\/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-14-quadrilaterals\/#rd-sharma-class-9-chapter-14-exercise-14d\" title=\"RD Sharma Class 9 Chapter 14 Exercise 14D\">RD Sharma Class 9 Chapter 14 Exercise 14D<\/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-14-quadrilaterals\/#important-concepts-from-rd-sharma-solutions-class-9-maths\" title=\"Important concepts from RD Sharma Solutions Class 9 Maths\">Important concepts from RD Sharma Solutions Class 9 Maths<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-14\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 14\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 14<\/a><ul class='ez-toc-list-level-3'><li class='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-14-quadrilaterals\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-14-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 14\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 14\u00a0PDF offline?<\/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-14-quadrilaterals\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-14\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 14?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 14?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-14\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 14?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 14?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-14-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 14 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/14-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 14<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/14-1.pdf\", \"#example1\");<\/script><\/p>\n<h2 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-14\"><\/span><strong>Exercise-Wise RD Sharma Solutions Class 9 Maths Chapter 14\u00a0<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-14-class-9-maths-exercise-14-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 14 Exercise 14A<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-14-class-9-maths-exercise-14-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 14 Exercise 14B<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-14-class-9-maths-exercise-14-3-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 14 Exercise 14C<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-14-class-9-maths-exercise-14-4-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Chapter 14 Exercise 14D<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-9-maths-chapter-14\"><\/span><strong>Access answers of <\/strong><strong>RD Sharma Solutions Class 9 Maths Chapter 14\u00a0<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-14-quadrilaterals-ex-141\"><\/span>RD Sharma Class 9 Solution Chapter 14 Quadrilaterals Ex 14.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: Three angles of a quadrilateral are respectively equal to 110<sup>0<\/sup>, 50<sup>0<\/sup>\u00a0and 40<sup>0<\/sup>. Find its fourth angle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Three angles of a quadrilateral are 110<sup>0<\/sup>, 50<sup>0<\/sup>\u00a0and 40<sup>0<\/sup><\/p>\n<p>Let the fourth angle be \u2018x\u2019<\/p>\n<p>We know, sum of all angles of a quadrilateral = 360<sup>0<\/sup><\/p>\n<p>110<sup>0<\/sup>\u00a0+ 50<sup>0<\/sup>\u00a0+ 40<sup>0<\/sup>\u00a0+ x<sup>0<\/sup>\u00a0= 360<sup>0<\/sup><\/p>\n<p>\u21d2 x = 360<sup>0<\/sup>\u00a0\u2013 200<sup>0<\/sup><\/p>\n<p>\u21d2x = 160<sup>0<\/sup><\/p>\n<p><em>Therefore, the required fourth angle is 160<sup>0<\/sup>.<\/em><\/p>\n<p><strong>Question 2: In a quadrilateral ABCD, the angles A, B, C and D are in the ratio of 1:2:4:5. Find the measure of each angles of the quadrilateral.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the angles of the quadrilaterals are A = x, B = 2x, C = 4x and D = 5x<\/p>\n<p>We know, sum of all angles of a quadrilateral = 360<sup>0<\/sup><\/p>\n<p>A + B + C + D = 360<sup>0<\/sup><\/p>\n<p>x + 2x + 4x + 5x = 360<sup>0<\/sup><\/p>\n<p>12x = 360<sup>0<\/sup><\/p>\n<p>x = 360<sup>0<\/sup>\/12 = 30<sup>0<\/sup><\/p>\n<p><em>Therefore,<\/em><\/p>\n<p><em>A = x = 30<sup>0<\/sup><\/em><\/p>\n<p><em>B = 2x = 60<sup>0<\/sup><\/em><\/p>\n<p><em>C = 4x = 120<sup>0<\/sup><\/em><\/p>\n<p><em>D = 5x = 150<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 3: In a quadrilateral ABCD, CO and DO are the bisectors of \u2220C and \u2220D respectively. Prove that \u2220COD = 1\/2 (\u2220A + \u2220B).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"rd sharma class 9 chapter 14 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/word-image.png\" alt=\"rd sharma class 9 chapter 14 - 1\" \/><\/p>\n<p>In \u0394DOC,<\/p>\n<p>\u2220CDO + \u2220COD + \u2220DCO = 180<sup>0<\/sup>\u00a0[Angle sum property of a triangle]<\/p>\n<p>or 1\/2\u2220CDA + \u2220COD + 1\/2\u2220DCB = 180<sup>0<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>\u2220COD = 180<sup>0<\/sup>\u00a0\u2013 1\/2(\u2220CDA + \u2220DCB) \u2026..(i)<\/p>\n<p>Also<\/p>\n<p>We know, sum of all angles of a quadrilateral = 360<sup>0<\/sup><\/p>\n<p>\u2220CDA + \u2220DCB = 360<sup>0<\/sup>\u00a0\u2013 (\u2220DAB + \u2220CBA) \u2026\u2026(ii)<\/p>\n<p>Substituting (ii) in (i)<\/p>\n<p>\u2220COD = 180<sup>0<\/sup>\u00a0\u2013 1\/2{360<sup>0<\/sup>\u00a0\u2013 (\u2220DAB + \u2220CBA) }<\/p>\n<p>We can also write, \u2220DAB = \u2220A and \u2220CBA = \u2220B<\/p>\n<p>\u2220COD = 180<sup>0\u00a0<\/sup>\u2212 180<sup>0\u00a0<\/sup>+ 1\/2(\u2220A + \u2220B))<\/p>\n<p><em>\u2220COD = 1\/2(\u2220A + \u2220B)<\/em><\/p>\n<p>Hence Proved.<\/p>\n<p><strong>Question 4: The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The angles of a quadrilateral are 3x, 5x, 9x and 13x respectively.<\/p>\n<p>We know, sum of all interior angles of a quadrilateral = 360<sup>0<\/sup><\/p>\n<p>Therefore, 3x + 5x + 9x + 13x = 360<sup>0<\/sup><\/p>\n<p>30x = 360<sup>0<\/sup><\/p>\n<p>or x = 12<sup>0<\/sup><\/p>\n<p><em>Hence, angles measures are<\/em><\/p>\n<p><em>3x = 3(12) = 36<sup>0<\/sup><\/em><\/p>\n<p><em>5x = 5(12) = 60<sup>0<\/sup><\/em><\/p>\n<p><em>9x = 9(12) = 108<sup>0<\/sup><\/em><\/p>\n<p><em>13x = 13(12) = 156<sup>0<\/sup><\/em><\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-142\"><\/span>Exercise 14.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: Two opposite angles of a parallelogram are (3x \u2013 2)<sup>0<\/sup>\u00a0and (50 \u2013 x)<sup>\u00a00<\/sup>. Find the measure of each angle of the parallelogram.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Given<strong>:\u00a0<\/strong>Two opposite angles of a parallelogram are (3x \u2013 2)<sup>0<\/sup>\u00a0and (50 \u2013 x)<sup>\u00a00<\/sup>.<\/p>\n<p>We know, opposite sides of a parallelogram are equal.<\/p>\n<p>(3x \u2013 2)<sup>0<\/sup>\u00a0= (50 \u2013 x)<sup>\u00a00<\/sup><\/p>\n<p>3x + x = 50 + 2<\/p>\n<p>4x = 52<\/p>\n<p>x = 13<\/p>\n<p>Angle x is 13<sup>0<\/sup><\/p>\n<p>Therefore,<\/p>\n<p>(3x-2)<sup>\u00a00<\/sup>\u00a0= (3(13) \u2013 2) = 37<sup>0<\/sup><\/p>\n<p>(50-x)<sup>\u00a00<\/sup>\u00a0= (50 \u2013 13) = 37<sup>0<\/sup><\/p>\n<p>Adjacent angles of a parallelogram are supplementary.<\/p>\n<p>x + 37 = 180<sup>0<\/sup><\/p>\n<p>x = 180<sup>0<\/sup>\u00a0\u2212 37<sup>0\u00a0<\/sup>= 143<sup>0<\/sup><\/p>\n<p><em>Therefore, required angles are : 37<sup>0<\/sup>, 143<sup>0<\/sup>, 37<sup>0<\/sup>\u00a0and 143<sup>0<\/sup>.<\/em><\/p>\n<p><strong>Question 2: If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Let the measure of the angle be x. Therefore, measure of the adjacent angle is 2x\/3.<\/p>\n<p>We know, adjacent angle of a parallelogram is supplementary.<\/p>\n<p>x + 2x\/3 = 180<sup>0<\/sup><\/p>\n<p>3x + 2x = 540<sup>0<\/sup><\/p>\n<p>5x = 540<sup>0<\/sup><\/p>\n<p>or x = 108<sup>0<\/sup><\/p>\n<p>Measure of second angle is 2x\/3 = 2(108<sup>0<\/sup>)\/3 = 72<sup>0<\/sup><\/p>\n<p>Similarly measure of 3<sup>rd<\/sup>\u00a0and 4<sup>th<\/sup>\u00a0angles are 108<sup>0<\/sup>\u00a0and 72<sup>0<\/sup><\/p>\n<p><em>Hence, four angles are 108<sup>0<\/sup>, 72<sup>0<\/sup>, 108<sup>0<\/sup>, 72<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 3: Find the measure of all the angles of a parallelogram, if one angle is 24<sup>0<\/sup>\u00a0less than twice the smallest angle.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Given: One angle of a parallelogram is 24<sup>0<\/sup>\u00a0less than twice the smallest angle.<\/p>\n<p>Let x be the smallest angle, then<\/p>\n<p>x + 2x \u2013 24<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>3x \u2013 24<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>3x = 108<sup>0<\/sup>\u00a0+ 24<sup>0<\/sup><\/p>\n<p>3x = 204<sup>0<\/sup><\/p>\n<p>x = 204<sup>0<\/sup>\/3 = 68<sup>0<\/sup><\/p>\n<p>So, x = 68<sup>0<\/sup><\/p>\n<p>Another angle = 2x \u2013 24<sup>0<\/sup>\u00a0= 2(68<sup>0<\/sup>) \u2013 24<sup>0<\/sup>\u00a0= 112<sup>0<\/sup><\/p>\n<p><em>Hence, four angles are 68<sup>0<\/sup>, 112<sup>0<\/sup>, 68<sup>0<\/sup>, 112<sup>0<\/sup>.<\/em><\/p>\n<p><strong>Question 4: The perimeter of a parallelogram is 22cm. If the longer side measures 6.5cm what is the measure of the shorter side?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let x be the shorter side of a parallelogram.<\/p>\n<p>Perimeter = 22 cm<\/p>\n<p>Longer side = 6.5 cm<\/p>\n<p>Perimeter = Sum of all sides = x + 6.5 + 6.5 + x<\/p>\n<p>22 = 2 ( x + 6.5 )<\/p>\n<p>11 = x + 6.5<\/p>\n<p>or x = 11 \u2013 6.5 = 4.5<\/p>\n<p><em>Therefore, shorter side of a parallelogram is 4.5 cm<\/em><\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-143\"><\/span>Exercise 14.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: In a parallelogram ABCD, determine the sum of angles \u2220C and \u2220D.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In a parallelogram ABCD , \u2220C and \u2220D are consecutive interior angles on the same side of the transversal CD.<\/p>\n<p>So,\u00a0<em>\u2220C + \u2220D = 180<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 2: In a parallelogram ABCD, if \u2220B = 135<sup>0<\/sup>, determine the measures of its other angles.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: In a parallelogram ABCD, if \u2220B = 135<sup>0<\/sup><\/p>\n<p>Here, \u2220A = \u2220C, \u2220B = \u2220D and \u2220A + \u2220B = 180<sup>0<\/sup><\/p>\n<p>\u2220A + 135<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>\u2220A = 45<sup>0<\/sup><\/p>\n<p>Answer:<\/p>\n<p><em>\u2220A = \u2220C = 45<sup>0<\/sup><\/em><\/p>\n<p><em>\u2220B = \u2220D = 135<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 3: ABCD is a square. AC and BD intersect at O. State the measure of \u2220AOB.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know, diagonals of a square bisect each other at right angle.<\/p>\n<p>So,\u00a0<em>\u2220AOB = 90<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 4: ABCD is a rectangle with \u2220ABD = 40<sup>0<\/sup>. Determine \u2220DBC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Each angle of a rectangle = 90<sup>o<\/sup><\/p>\n<p>So, \u2220ABC = 90<sup>0<\/sup><\/p>\n<p>\u2220ABD = 40<sup>0\u00a0<\/sup>(given)<\/p>\n<p>Now, \u2220ABD + \u2220DBC = 90<sup>0<\/sup><\/p>\n<p>40<sup>0<\/sup>\u00a0+ \u2220DBC = 90<sup>0<\/sup><\/p>\n<p>or\u00a0<em>\u2220DBC = 50<sup>0<\/sup>\u00a0.<\/em><\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-144\"><\/span>Exercise 14.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: In a \u0394ABC, D, E and F are, respectively, the mid points of BC, CA and AB. If the lengths of sides AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of \u0394DEF.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AB = 7 cm, BC = 8 cm, AC = 9 cm<\/p>\n<p>In \u2206ABC,<\/p>\n<p>In a \u0394ABC, D, E and F are, respectively, the mid points of BC, CA and AB.<\/p>\n<p>According to Midpoint Theorem:<\/p>\n<p>EF = 1\/2BC, DF = 1\/2 AC and DE = 1\/2 AB<\/p>\n<p>Now, Perimeter of \u2206DEF = DE + EF + DF<\/p>\n<p>= 1\/2 (AB + BC + AC)<\/p>\n<p>= 1\/2 (7 + 8 + 9)<\/p>\n<p>= 12<\/p>\n<p><em>Perimeter of \u0394DEF = 12cm<\/em><\/p>\n<p><strong>Question 2: In a \u0394ABC, \u2220A = 50<sup>0<\/sup>, \u2220B = 60<sup>0<\/sup>\u00a0and \u2220C = 70<sup>0<\/sup>. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 2\" width=\"268\" height=\"187\" \/><\/p>\n<p>In \u0394ABC,<\/p>\n<p>D, E and F are mid points of AB,BC and AC respectively.<\/p>\n<p>In a Quadrilateral DECF:<\/p>\n<p>By Mid-point theorem,<\/p>\n<p>DE \u2225 AC \u21d2 DE = AC\/2<\/p>\n<p>And CF = AC\/2<\/p>\n<p>\u21d2 DE = CF<\/p>\n<p>Therefore, DECF is a parallelogram.<\/p>\n<p>\u2220C = \u2220D = 70<sup>0<\/sup><\/p>\n<p>[Opposite sides of a parallelogram]<\/p>\n<p>Similarly,<\/p>\n<p>ADEF is a parallelogram, \u2220A = \u2220E = 50<sup>0<\/sup><\/p>\n<p>BEFD is a parallelogram, \u2220B = \u2220F = 60<sup>0<\/sup><\/p>\n<p><em>Hence, Angles of \u0394DEF are: \u2220D = 70<sup>0<\/sup>, \u2220E = 50<sup>0<\/sup>, \u2220F = 60<sup>0<\/sup>.<\/em><\/p>\n<p><strong>Question 3: In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-1.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 3\" width=\"336\" height=\"235\" \/><\/p>\n<p>In \u0394ABC,<\/p>\n<p>R and P are mid points of AB and BC<\/p>\n<p>By Mid-point Theorem<\/p>\n<p>RP \u2225 AC \u21d2 RP = AC\/2<\/p>\n<p>In a quadrilateral, ARPQ<\/p>\n<p>RP \u2225 AQ \u21d2 RP = AQ<\/p>\n<p>[A pair of side is parallel and equal]<\/p>\n<p>Therefore, ARPQ is a parallelogram.<\/p>\n<p>Now, AR = AB\/2 = 30\/2 = 15 cm<\/p>\n<p>[AB = 30 cm (Given)]<\/p>\n<p>AR = QP = 15 cm<\/p>\n<p>[ Opposite sides are equal ]<\/p>\n<p>And RP = AC\/2 = 21\/2 = 10.5 cm<\/p>\n<p>[AC = 21 cm (Given)]<\/p>\n<p>RP = AQ = 10.5cm<\/p>\n<p>[ Opposite sides are equal ]<\/p>\n<p>Now,<\/p>\n<p>Perimeter of ARPQ = AR + QP + RP +AQ<\/p>\n<p>= 15 +15 +10.5 +10.5<\/p>\n<p>= 51<\/p>\n<p><em>Perimeter of quadrilateral ARPQ is 51 cm.<\/em><\/p>\n<p><strong>Question 4: In a \u0394ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-2.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 4\" width=\"284\" height=\"187\" \/><\/p>\n<p>In a quadrilateral ABXC,<\/p>\n<p>AD = DX [Given]<\/p>\n<p>BD = DC [Given]<\/p>\n<p>From figure, Diagonals AX and BC bisect each other.<\/p>\n<p><em>ABXC is a parallelogram.<\/em><\/p>\n<p>Hence Proved.<\/p>\n<p><strong>Question 5: In a \u0394ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-3.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 5\" width=\"265\" height=\"222\" \/><\/p>\n<p>In a \u0394ABC<\/p>\n<p>E and F are mid points of AC and AB (Given)<\/p>\n<p>EF \u2225 BC \u21d2 EF = BC\/2 and<\/p>\n<p>[By mid-point theorem]<\/p>\n<p>In \u0394ABP<\/p>\n<p>F is the mid-point of AB, again by mid-point theorem<\/p>\n<p>FQ \u2225 BP<\/p>\n<p>Q is the mid-point of AP<\/p>\n<p><em>AQ = QP<\/em><\/p>\n<p>Hence Proved.<\/p>\n<p><strong>Question 6: In a \u0394ABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML = NL.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-4.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 6\" width=\"316\" height=\"208\" \/><\/p>\n<p>Given that,<\/p>\n<p>In \u0394BLM and \u0394CLN<\/p>\n<p>\u2220BML = \u2220CNL = 90<sup>0<\/sup><\/p>\n<p>BL = CL [L is the mid-point of BC]<\/p>\n<p>\u2220MLB = \u2220NLC [Vertically opposite angle]<\/p>\n<p>By ASA criterion:<\/p>\n<p>\u0394BLM \u2245 \u0394CLN<\/p>\n<p><em>So, LM = LN [By CPCT]<\/em><\/p>\n<p><strong>Question 7: In figure, triangle ABC is a right-angled triangle at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate<\/strong><\/p>\n<p><strong>(i) The length of BC (ii) The area of \u0394ADE.<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-5.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 7\" width=\"367\" height=\"199\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In \u0394ABC,<\/p>\n<p>\u2220B=90<sup>0\u00a0<\/sup>(Given)<\/p>\n<p>AB = 9 cm, AC = 15 cm (Given)<\/p>\n<p>By using Pythagoras theorem<\/p>\n<p>AC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup><\/p>\n<p>\u21d215<sup>2<\/sup>\u00a0= 9<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup><\/p>\n<p>\u21d2BC<sup>2\u00a0<\/sup>= 225 \u2013 81 = 144<\/p>\n<p>or BC = 12<\/p>\n<p>Again,<\/p>\n<p>AD = DB = AB\/2 = 9\/2 = 4.5 cm [D is the mid\u2212point of AB<\/p>\n<p>D and E are mid-points of AB and AC<\/p>\n<p>DE \u2225 BC \u21d2 DE = BC\/2 [By mid\u2212point theorem]<\/p>\n<p>Now,<\/p>\n<p>Area of \u0394ADE = 1\/2 x AD x DE<\/p>\n<p>= 1\/2 x 4.5 x 6<\/p>\n<p>=13.5<\/p>\n<p><em>Area of \u0394ADE is 13.5 cm<sup>2<\/sup><\/em><\/p>\n<p><strong>Question 8: In figure, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-6.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 8\" width=\"361\" height=\"209\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm.<\/p>\n<p>M and N are mid-points of AB and AC<\/p>\n<p>By mid\u2212point theorem, we have<\/p>\n<p>MN\u2225BC \u21d2 MN = BC\/2<\/p>\n<p>or BC = 2MN<\/p>\n<p><em>BC = 6 cm<\/em><\/p>\n<p>[MN = 3 cm given)<\/p>\n<p>Similarly,<\/p>\n<p><em>AC = 2MP = 2 (2.5) = 5 cm<\/em><\/p>\n<p><em>AB = 2 NP = 2 (3.5) = 7 cm<\/em><\/p>\n<p><strong>Question 9: ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of \u0394PQR is double the perimeter of \u0394ABC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-7.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 9\" width=\"396\" height=\"250\" \/><\/p>\n<p>ABCQ and ARBC are parallelograms.<\/p>\n<p>Therefore, BC = AQ and BC = AR<\/p>\n<p>\u21d2AQ = AR<\/p>\n<p>\u21d2A is the mid-point of QR<\/p>\n<p>Similarly B and C are the mid points of PR and PQ respectively.<\/p>\n<p>By mid\u2212point theorem, we have<\/p>\n<p>AB = PQ\/2, BC = QR\/2 and CA = PR\/2<\/p>\n<p>or PQ = 2AB, QR = 2BC and PR = 2CA<\/p>\n<p>\u21d2PQ + QR + RP = 2 (AB + BC + CA)<\/p>\n<p><em>\u21d2 Perimeter of \u0394PQR = 2 (Perimeter of \u0394ABC)<\/em><\/p>\n<p>Hence proved.<\/p>\n<p><strong>Question 10: In figure, BE \u22a5 AC, AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that \u2220PQR = 90<sup>0<\/sup>.<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 14 ex 14.4 question 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-14-ex-14-4-questio-8.png\" alt=\"RD sharma class 9 maths chapter 14 ex 14.4 question 10\" width=\"346\" height=\"218\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>BE\u22a5AC and P, Q and R are respectively mid-point of AH, AB and BC. (Given)<\/p>\n<p>In \u0394ABC, Q and R are mid-points of AB and BC respectively.<\/p>\n<p>By Mid-point theorem:<\/p>\n<p>QR \u2225 AC \u2026..(i)<\/p>\n<p>In \u0394ABH, Q and P are the mid-points of AB and AH respectively<\/p>\n<p>QP \u2225 BH \u2026.(ii)<\/p>\n<p>But, BE\u22a5AC<\/p>\n<p>From (i) and (ii) we have,<\/p>\n<p>QP\u22a5QR<\/p>\n<p><em>\u21d2\u2220PQR = 90<sup>0<\/sup><\/em><\/p>\n<p>Hence Proved.<\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-vsaqs\"><\/span>Exercise VSAQs<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: In a parallelogram ABCD, write the sum of angles A and B.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In parallelogram ABCD, Adjacent angles of a parallelogram are supplementary.<\/p>\n<p>Therefore,\u00a0<em>\u2220A + \u2220B = 180<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 2: In a parallelogram ABCD, if \u2220D = 115<sup>0<\/sup>, then write the measure of \u2220A.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In a parallelogram ABCD,<\/p>\n<p>\u2220D = 115<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>Since, \u2220A and \u2220D are adjacent angles of parallelogram.<\/p>\n<p>We know, Adjacent angles of a parallelogram are supplementary.<\/p>\n<p>\u2220A + \u2220D = 180<sup>0<\/sup><\/p>\n<p>\u2220A = 180<sup>0\u00a0<\/sup>\u2013 115<sup>0\u00a0<\/sup>= 65<sup>0<\/sup><\/p>\n<p><em>Measure of \u2220A is 65<sup>0<\/sup>.<\/em><\/p>\n<p><strong>Question 3: PQRS is a square such that PR and SQ intersect at O. State the measure of \u2220POQ.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>PQRS is a square such that PR and SQ intersect at O. (Given)<\/p>\n<p>We know, diagonals of a square bisects each other at 90 degrees.<\/p>\n<p><em>So, \u2220POQ = 90<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 4: In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that \u2220AOB = 75\u00b0, then write the value of \u2220C + \u2220D.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220AOB = 75<sup>o<\/sup>\u00a0(given)<\/p>\n<p>In a quadrilateral ABCD, bisectors of angles A and B intersect at O, then<\/p>\n<p>\u2220AOB = 1\/2 (\u2220ADC + \u2220ABC)<\/p>\n<p>or \u2220AOB = 1\/2 (\u2220D + \u2220C)<\/p>\n<p>By substituting given values, we get<\/p>\n<p>75<sup>\u00a0o<\/sup>\u00a0= 1\/2 (\u2220D + \u2220C)<\/p>\n<p><em>or \u2220C + \u2220D = 150<sup>\u00a0o<\/sup><\/em><\/p>\n<p><strong>Question 5: The diagonals of a rectangle ABCD meet at O. If \u2220BOC = 44<sup>o<\/sup>, find \u2220OAD.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>ABCD is a rectangle and \u2220BOC = 44<sup>o<\/sup>\u00a0(given)<\/p>\n<p>\u2220AOD = \u2220BOC (vertically opposite angles)<\/p>\n<p>\u2220AOD = \u2220BOC = 44<sup>o<\/sup><\/p>\n<p>\u2220OAD = \u2220ODA (Angles facing same side)<\/p>\n<p>and OD = OA<\/p>\n<p>Since sum of all the angles of a triangle is 180<sup>\u00a0o<\/sup>, then<\/p>\n<p><em>So, \u2220OAD = 1\/2 (180<sup>\u00a0o<\/sup>\u00a0\u2013 44<sup>\u00a0o<\/sup>) = 68<sup>\u00a0o<\/sup><\/em><\/p>\n<p><strong>Question 6: If PQRS is a square, then write the measure of \u2220SRP.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>PQRS is a square.<\/p>\n<p>\u21d2 All side are equal, and each angle is 90<sup>o<\/sup>\u00a0degrees and diagonals bisect the angles.<\/p>\n<p><em>So, \u2220SRP = 1\/2 (90<sup>\u00a0o<\/sup>) = 45<sup>o<\/sup><\/em><\/p>\n<p><strong>Question 7: If ABCD is a rectangle with \u2220BAC = 32<sup>o<\/sup>, find the measure of \u2220DBC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>ABCD is a rectangle and \u2220BAC=32<strong><sup>\u00a0o<\/sup><\/strong>\u00a0(given)<\/p>\n<p>We know, diagonals of a rectangle bisects each other.<\/p>\n<p>AO = BO<\/p>\n<p>\u2220DBA = \u2220BAC = 32<strong><sup>\u00a0o<\/sup><\/strong>\u00a0(Angles facing same side)<\/p>\n<p>Each angle of a rectangle = 90 degrees<\/p>\n<p>So, \u2220DBC + \u2220DBA = 90<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>or \u2220DBC + 32<strong><sup>\u00a0o<\/sup><\/strong>\u00a0= 90<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p><em>or \u2220DBC = 58<strong><sup>\u00a0o<\/sup><\/strong><\/em><\/p>\n<p><strong>Question 8: If ABCD is a rhombus with \u2220ABC = 56<sup>o<\/sup>, find the measure of \u2220ACD.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In a rhombus ABCD,<\/p>\n<p>&lt;ABC = 56<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>So, &lt;BCD = 2 (&lt;ACD) (Diagonals of a rhombus bisect the interior angles)<\/p>\n<p>or &lt;ACD = 1\/2 (&lt;BCD) \u2026..(1)<\/p>\n<p>We know, consecutive angles of a rhombus are supplementary.<\/p>\n<p>\u2220BCD + \u2220ABC = 180<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>\u2220BCD = 180<strong><sup>\u00a0o<\/sup><\/strong>\u00a0\u2013 56<strong><sup>\u00a0o<\/sup><\/strong>\u00a0= 124<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>Equation (1)\u00a0<em>\u21d2 &lt;ACD = 1\/2 x 124<strong><sup>\u00a0o<\/sup><\/strong>\u00a0= 62<strong><sup>\u00a0o<\/sup><\/strong><\/em><\/p>\n<p><strong>Question 9: The perimeter of a parallelogram is 22 cm. If the longer side measure 6.5 cm, what is the measure of shorter side?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Perimeter of a parallelogram = 22 cm. (Given)<\/p>\n<p>Longer side = 6.5 cm<\/p>\n<p>Let x be the shorter side.<\/p>\n<p>Perimeter = 2x + 2\u00d76.5<\/p>\n<p>22 = 2x + 13<\/p>\n<p>2x = 22 \u2013 13 = 9<\/p>\n<p>or x = 4.5<\/p>\n<p><em>Measure of shorter side is 4.5 cm.<\/em><\/p>\n<p><strong>Question 10: If the angles of a quadrilateral are in the ratio 3:5:9:13, then find the measure of the smallest angle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13 (Given)<\/p>\n<p>Let the sides are 3x, 5x, 9x, 13x<\/p>\n<p>We know, sum of all the angles of a quadrilateral = 360<sup>o<\/sup><\/p>\n<p>3x + 5x + 9x + 13x = 360<sup>\u00a0o<\/sup><\/p>\n<p>30 x = 360<sup>\u00a0o<\/sup><\/p>\n<p>x = 12<sup>\u00a0o<\/sup><\/p>\n<p><em>Measure of smallest angle = 3x = 3(12) = 36<sup>o<\/sup>\u00a0.<\/em><\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-important-topics-in-the-exercise\"><\/span><strong>Detailed Exercise-wise Explanation with Important Topics in the Exercise<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-14-exercise-14a\"><\/span>RD Sharma Class 9 Chapter 14 Exercise 14A<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">In <\/span>Chapter 14A RD Sharma Class 9 <span style=\"font-weight: 400;\">the<\/span> <span style=\"font-weight: 400;\">students will be able to examine their performances by attempting <\/span><span style=\"font-weight: 400;\">objective-type questions that are associated with Quadrilaterals. After going through the solutions of RD Sharma Class 9 Chapter 14A the students will be able to understand the applications of <\/span><span style=\"font-weight: 400;\">quadrilaterals angles, the angle sum property of a quadrilateral, and many more<\/span><span style=\"font-weight: 400;\">. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">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.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-14-exercise-14b\"><\/span>RD Sharma Class 9 Chapter 14 Exercise 14B<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">In the<\/span> RD Sharma Class 9 Chapter 14B <span style=\"font-weight: 400;\">the students will learn about the techniques to handle problems based on <\/span><span style=\"font-weight: 400;\">Quadrilaterals. The main areas of <\/span>RD Sharma Solutions Class 9 Maths Chapter 14B <span style=\"font-weight: 400;\">which <\/span><span style=\"font-weight: 400;\">you need to be careful about are various types of quadrilaterals like rectangles, squares, rhombus, and many more. The problems which the students will find over there are all connected with the application of the basic concept. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">By looking at the step-wise explanations the students can attempt all kinds of questions from this chapter with ease. Practicing <\/span>RD Sharma Solutions Class 9 Maths Chapter 14 <span style=\"font-weight: 400;\">is something you cannot ignore to strengthen your preparation style.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-14-exercise-14c\"><\/span>RD Sharma Class 9 Chapter 14 Exercise 14C<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">\u00a0In the RD Sharma Exercise 14C<\/span> <span style=\"font-weight: 400;\">Class, 9 Chapter 14<\/span> <span style=\"font-weight: 400;\">students will get new inputs from the in-depth explanations. The main areas <\/span><span style=\"font-weight: 400;\">the students need to focus on are the conditions for a quadrilateral to be a parallelogram and some important theorems results.<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> If the students concentrate on each part of Chapter 14C RD Sharma Solutions carefully then they will be able to grasp the important concepts of Quadrilaterals.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-14-exercise-14d\"><\/span>RD Sharma Class 9 Chapter 14 Exercise 14D<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">The <\/span>RD Sharma Class 9 Chapter 14D <span style=\"font-weight: 400;\">consists of<\/span> <span style=\"font-weight: 400;\">the elements which are related <\/span><span style=\"font-weight: 400;\">to Quadrilaterals. The students are expected to practice all the problems which are given in <\/span>RD Sharma Solutions Class 9 Chapter 14 Exercise 14B <span style=\"font-weight: 400;\">regularly. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">The experts have presented the answers to the complex problems in <\/span>RD Sharma Solutions Class 9 Chapter 14D <span style=\"font-weight: 400;\">to make things easier for the students. Solving <\/span>RD Sharma Class 9 Chapter 14D <span style=\"font-weight: 400;\">is the key to success and to get good results you must make proper use of the key. <\/span>\u00a0<span style=\"font-weight: 400;\">The essential areas are the <\/span><span style=\"font-weight: 400;\">triangle and theorems results of the triangle<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In the last exercise, here students can check the progress of their performances by practicing <\/span><span style=\"font-weight: 400;\">objective-type questions in connection with Quadrilaterals. When you follow the important parts of the Chapter 14 solutions your confidence gets a boost and you will be automatically prepared to face more challenges, You can test yourself when you reach this part to know where you stand.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"important-concepts-from-rd-sharma-solutions-class-9-maths\"><\/span><span style=\"font-weight: 400;\">Important concepts from RD Sharma Solutions Class 9 Maths<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Quadrilaterals Introduction.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Quadrilaterals angles.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Angle sum property.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Various types of Quadrilaterals.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Conditions for a Quadrilaterals to be a Parallelogram.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-14\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 14<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630752228365\" 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-14-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 14\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 id=\"faq-question-1630752298993\" 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-14\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 14?<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-1630752332328\" 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-14\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 14?<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>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 9 Maths Chapter 14 &#8211; Quadrilaterals: Quadrilateral: In this blog, we will be dealing with the important aspects of RD Sharma Solutions Class 9 Maths Chapter 14 which will guide you to bring changes in your methods of preparation. The parts that we will be concentrating on are the benefits of &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 14 &#8211; Quadrilaterals (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 14 &#8211; Quadrilaterals (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124489,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62900"}],"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=62900"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62900\/revisions"}],"predecessor-version":[{"id":513200,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62900\/revisions\/513200"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124489"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62900"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62900"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62900"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}