{"id":125328,"date":"2023-09-13T08:58:00","date_gmt":"2023-09-13T03:28:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125328"},"modified":"2023-12-06T11:02:26","modified_gmt":"2023-12-06T05:32:26","slug":"rd-sharma-class-9-solutions-chapter-15-exercise-15-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-4\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127677\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.4.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\"><strong>RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/strong>: The PDF is available for free. You can easily download the PDF for <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 15<\/a>&nbsp;Exercise 15.4. These solutions will be very beneficial for your class 9 exams.&nbsp;<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\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-69e718f778a22\" 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-69e718f778a22\"><\/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-class-9-solutions-chapter-15-exercise-15-4\/#download-rd-sharma-class-9-solutions-chapter-15-exercise-154\" title=\"Download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4\">Download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/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-class-9-solutions-chapter-15-exercise-15-4\/#access-rd-sharma-class-9-solutions-chapter-15-exercise-154\" title=\"Access RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4\">Access RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-4\/#question-2-in-the-figure-o-is-the-centre-of-the-circle-find-%e2%88%a0bac-solution-in-the-circle-with-the-centre-o-%e2%88%a0aob-80%c2%b0-and-%e2%88%a0aoc-110%c2%b0-%e2%88%b4-%e2%88%a0boc-%e2%88%a0aob-%e2%88%a0aoc-80%c2%b0-110%c2%b0-190%c2%b0-%e2%88%b4-reflex-%e2%88%a0boc-360%c2%b0-%e2%80%93-190%c2%b0-170%c2%b0-now-arc-bec-subtends-%e2%88%a0boc-at-the-centre-and-%e2%88%a0bac-at-the-remaining-part-of-the-circle-%e2%88%b4-%e2%88%a0boc-2%e2%88%a0bac-%e2%87%92-170%c2%b0-2%e2%88%a0bac-%e2%87%92-%e2%88%a0bac-170%e2%88%982-85%c2%b0-%e2%88%b4-%e2%88%a0bac-85%c2%b0\" title=\"Question 2. In the figure, O is the centre of the circle. Find \u2220BAC.  Solution: In the circle with the centre O \u2220AOB = 80\u00b0 and \u2220AOC =110\u00b0 \u2234 \u2220BOC = \u2220AOB + \u2220AOC = 80\u00b0+ 110\u00b0= 190\u00b0 \u2234 Reflex \u2220BOC = 360\u00b0 \u2013 190\u00b0 = 170\u00b0 Now arc BEC subtends \u2220BOC at the centre and \u2220BAC at the remaining part of the circle.  \u2234 \u2220BOC = 2\u2220BAC \u21d2 170\u00b0 = 2\u2220BAC \u21d2 \u2220BAC =&nbsp;170\u22182&nbsp;= 85\u00b0 \u2234 \u2220BAC = 85\u00b0\">Question 2. In the figure, O is the centre of the circle. Find \u2220BAC.  Solution: In the circle with the centre O \u2220AOB = 80\u00b0 and \u2220AOC =110\u00b0 \u2234 \u2220BOC = \u2220AOB + \u2220AOC = 80\u00b0+ 110\u00b0= 190\u00b0 \u2234 Reflex \u2220BOC = 360\u00b0 \u2013 190\u00b0 = 170\u00b0 Now arc BEC subtends \u2220BOC at the centre and \u2220BAC at the remaining part of the circle.  \u2234 \u2220BOC = 2\u2220BAC \u21d2 170\u00b0 = 2\u2220BAC \u21d2 \u2220BAC =&nbsp;170\u22182&nbsp;= 85\u00b0 \u2234 \u2220BAC = 85\u00b0<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-4\/#faq-rd-sharma-class-9-solutions-chapter-15-exercise-154\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4\">FAQ: RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/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-class-9-solutions-chapter-15-exercise-15-4\/#can-i-open-rd-sharma-class-9-solutions-chapter-15-exercise-154-pdf-on-my-smartphone\" title=\"Can I open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF on my smartphone?\">Can I open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF on my smartphone?<\/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-class-9-solutions-chapter-15-exercise-15-4\/#can-i-download-rd-sharma-class-9-solutions-chapter-15-exercise-154-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF free?<\/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-class-9-solutions-chapter-15-exercise-15-4\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-15-exercise-154\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4?\">What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4?<\/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-class-9-solutions-chapter-15-exercise-15-4\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-15-exercise-154\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\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\/2021\/09\/RD_Sharma_Class_9_Chapter_15_Ex_15_4.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD_Sharma_Class_9_Chapter_15_Ex_15_4.pdf\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-15-exercise-154\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1.<\/strong><br><strong>In the figure, O is the centre of the circle. If \u2220APB = 50\u00b0, find \u2220AOB and \u2220OAB.<\/strong><br><img class=\"alignnone size-full wp-image-67489\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q1.1.png\" sizes=\"(max-width: 191px) 100vw, 191px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q1.1.png 191w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q1.1-150x150.png 150w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q1.1-100x100.png 100w\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q1.1\" width=\"191\" height=\"190\"><br><strong>Solution:<\/strong><br>Arc AB, subtends \u2220AOB at the centre and \u2220APB at the remaining part of the circle<br>\u2234\u2220AOB = 2\u2220APB = 2 x 50\u00b0 = 100\u00b0<br>Join AB<br><img class=\"alignnone size-full wp-image-67490\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q1.2.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q1.2\" width=\"196\" height=\"186\"><br>\u2206AOB is an isosceles triangle in which<br>OA = OB<br>\u2234 \u2220OAB = \u2220OBA But \u2220AOB = 100\u00b0<br>\u2234\u2220OAB + \u2220OBA = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br>\u21d2 2\u2220OAB = 80\u00b0<br>80\u00b0<br>\u2234\u2220OAB =&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\">80<\/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; = 40\u00b0<\/p>\n<h4><span class=\"ez-toc-section\" id=\"question-2-in-the-figure-o-is-the-centre-of-the-circle-find-%e2%88%a0bac-solution-in-the-circle-with-the-centre-o-%e2%88%a0aob-80%c2%b0-and-%e2%88%a0aoc-110%c2%b0-%e2%88%b4-%e2%88%a0boc-%e2%88%a0aob-%e2%88%a0aoc-80%c2%b0-110%c2%b0-190%c2%b0-%e2%88%b4-reflex-%e2%88%a0boc-360%c2%b0-%e2%80%93-190%c2%b0-170%c2%b0-now-arc-bec-subtends-%e2%88%a0boc-at-the-centre-and-%e2%88%a0bac-at-the-remaining-part-of-the-circle-%e2%88%b4-%e2%88%a0boc-2%e2%88%a0bac-%e2%87%92-170%c2%b0-2%e2%88%a0bac-%e2%87%92-%e2%88%a0bac-170%e2%88%982-85%c2%b0-%e2%88%b4-%e2%88%a0bac-85%c2%b0\"><\/span><strong>Question 2.<\/strong><br><strong>In the figure, O is the centre of the circle. Find \u2220BAC.<\/strong><br><img class=\"alignnone size-full wp-image-67491\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q2.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q2.1\" width=\"216\" height=\"190\"><br><strong>Solution:<\/strong><br>In the circle with the centre O<br>\u2220AOB = 80\u00b0 and \u2220AOC =110\u00b0<br>\u2234 \u2220BOC = \u2220AOB + \u2220AOC<br>= 80\u00b0+ 110\u00b0= 190\u00b0<br>\u2234 Reflex \u2220BOC = 360\u00b0 \u2013 190\u00b0 = 170\u00b0<br>Now arc BEC subtends \u2220BOC at the centre and \u2220BAC at the remaining part of the circle.<br><img class=\"alignnone size-full wp-image-67492\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q2.2.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q2.2\" width=\"212\" height=\"203\"><br>\u2234 \u2220BOC = 2\u2220BAC<br>\u21d2 170\u00b0 = 2\u2220BAC<br>\u21d2 \u2220BAC =&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=\"msubsup\"><span id=\"MathJax-Span-16\" class=\"texatom\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"mn\">170<\/span><\/span><\/span><span id=\"MathJax-Span-19\" class=\"texatom\"><span id=\"MathJax-Span-20\" class=\"mrow\"><span id=\"MathJax-Span-21\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-22\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 85\u00b0<br>\u2234 \u2220BAC = 85\u00b0<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>Question 3.<\/strong><br><strong>If O is the centre of the circle, find the value of x in each of the following figures:<\/strong><br><img class=\"alignnone size-full wp-image-67493\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.1.png\" sizes=\"(max-width: 607px) 100vw, 607px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.1.png 607w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.1-300x129.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q3.1\" width=\"607\" height=\"262\"><\/p>\n<p><br><img class=\"alignnone size-full wp-image-67494\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.2.png\" sizes=\"(max-width: 569px) 100vw, 569px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.2.png 569w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.2-300x243.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q3.2\" width=\"569\" height=\"460\"><br><img class=\"alignnone size-full wp-image-67495\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.3.png\" sizes=\"(max-width: 554px) 100vw, 554px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.3.png 554w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q3.3-300x126.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q3.3\" width=\"554\" height=\"232\"><br><strong>Solution:<\/strong><br><strong>(i)<\/strong>&nbsp;A circle with centre O<br>\u2220AOC = 135\u00b0<br>But \u2220AOC + \u2220COB = 180\u00b0 (Linear pair)<br>\u21d2 135\u00b0 + \u2220COB = 180\u00b0<br>\u21d2 \u2220COB = 180\u00b0- 135\u00b0 = 45\u00b0<br>Now arc BC subtends \u2220BOC at the centre and \u2220BPC at the remaining part of the circle<br>\u2234 \u2220BOC = 2\u2220BPC<br>\u21d2 \u2220BPC =&nbsp;<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-23\" class=\"math\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"mfrac\"><span id=\"MathJax-Span-26\" class=\"mn\">1<\/span><span id=\"MathJax-Span-27\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220BOC =&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=\"mn\">1<\/span><span id=\"MathJax-Span-32\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 45\u00b0 =&nbsp;<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-33\" class=\"math\"><span id=\"MathJax-Span-34\" class=\"mrow\"><span id=\"MathJax-Span-35\" class=\"mfrac\"><span id=\"MathJax-Span-36\" class=\"msubsup\"><span id=\"MathJax-Span-37\" class=\"texatom\"><span id=\"MathJax-Span-38\" class=\"mrow\"><span id=\"MathJax-Span-39\" class=\"mn\">45<\/span><\/span><\/span><span id=\"MathJax-Span-40\" class=\"texatom\"><span id=\"MathJax-Span-41\" class=\"mrow\"><span id=\"MathJax-Span-42\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-43\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br>\u2234 \u2220BPC = 22&nbsp;<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-44\" class=\"math\"><span id=\"MathJax-Span-45\" class=\"mrow\"><span id=\"MathJax-Span-46\" class=\"mfrac\"><span id=\"MathJax-Span-47\" class=\"mn\">1<\/span><span id=\"MathJax-Span-48\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0 or x = 22&nbsp;<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-49\" class=\"math\"><span id=\"MathJax-Span-50\" class=\"mrow\"><span id=\"MathJax-Span-51\" class=\"mfrac\"><span id=\"MathJax-Span-52\" class=\"mn\">1<\/span><span id=\"MathJax-Span-53\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0<br><strong>(ii)<\/strong>&nbsp;\u2235 CD and AB are the diameters of the circle with centre O<br>\u2220ABC = 40\u00b0<br>But in \u2206OBC,<br>OB = OC (Radii of the circle)<br>\u2220OCB = \u2220OBC \u2013 40\u00b0<br>Now in ABCD,<br>\u2220ODB + \u2220OCB + \u2220CBD = 180\u00b0 (Angles of a triangle)<br>\u21d2 x + 40\u00b0 + 90\u00b0 = 180\u00b0<br>\u21d2 x + 130\u00b0 = 180\u00b0<br>\u21d2 x = 180\u00b0 \u2013 130\u00b0 = 50\u00b0<br>\u2234 x = 50\u00b0<br><strong>(iii)<\/strong>&nbsp;In circle with centre O,<br>\u2220AOC = 120\u00b0, AB is produced to D<br>\u2235 \u2220AOC = 120\u00b0<br>and \u2220AOC + convex \u2220AOC = 360\u00b0<br>\u21d2 120\u00b0 + convex \u2220AOC = 360\u00b0<br>\u2234 Convex \u2220AOC = 360\u00b0 \u2013 120\u00b0 = 240\u00b0<br>\u2234 Arc APC Subtends \u2220AOC at the centre and \u2220ABC at the remaining part of the circle<br>\u2234 \u2220ABC =&nbsp;<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-54\" class=\"math\"><span id=\"MathJax-Span-55\" class=\"mrow\"><span id=\"MathJax-Span-56\" class=\"mfrac\"><span id=\"MathJax-Span-57\" class=\"mn\">1<\/span><span id=\"MathJax-Span-58\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOC =&nbsp;<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-59\" class=\"math\"><span id=\"MathJax-Span-60\" class=\"mrow\"><span id=\"MathJax-Span-61\" class=\"mfrac\"><span id=\"MathJax-Span-62\" class=\"mn\">1<\/span><span id=\"MathJax-Span-63\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x 240\u00b0 = 120\u00b0<br>But \u2220ABC + \u2220CBD = 180\u00b0 (Linear pair)<br>\u21d2 120\u00b0 + x = 180\u00b0<br>\u21d2 x = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<br>\u2234 x = 60\u00b0<br><strong>(iv)<\/strong>&nbsp;A circle with centre O and \u2220CBD = 65\u00b0<br>But \u2220ABC + \u2220CBD = 180\u00b0 (Linear pair)<br>\u21d2 \u2220ABC + 65\u00b0 = 180\u00b0<br>\u21d2 \u2220ABC = 180\u00b0-65\u00b0= 115\u00b0<br>Now arc AEC subtends \u2220x at the centre and \u2220ABC at the remaining part of the circle<br>\u2234 \u2220AOC = 2\u2220ABC<br>\u21d2 x = 2 x 115\u00b0 = 230\u00b0<br>\u2234 x = 230\u00b0<br><strong>(v)<\/strong>&nbsp;In circle with centre O<br>AB is chord of the circle, \u2220OAB = 35\u00b0<br>In \u2206OAB,<br>OA = OB (Radii of the circle)<br>\u2220OBA = \u2220OAB = 35\u00b0<br>But in \u2206OAB,<br>\u2220OAB + \u2220OBA + \u2220AOB = 180\u00b0 (Angles of a triangle)<br>\u21d2 35\u00b0 + 35\u00b0 + \u2220AOB = 180\u00b0<br>\u21d2 70\u00b0 + \u2220AOB = 180\u00b0<br>\u21d2 \u2220AOB = 180\u00b0-70\u00b0= 110\u00b0<br>\u2234 Convex \u2220AOB = 360\u00b0 -110\u00b0 = 250\u00b0<br>But arc AB subtends \u2220AOB at the centre and \u2220ACB at the remaining part of the circle.<br>\u2234\u2220ACB =&nbsp;<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-64\" class=\"math\"><span id=\"MathJax-Span-65\" class=\"mrow\"><span id=\"MathJax-Span-66\" class=\"mfrac\"><span id=\"MathJax-Span-67\" class=\"mn\">1<\/span><span id=\"MathJax-Span-68\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-69\" class=\"math\"><span id=\"MathJax-Span-70\" class=\"mrow\"><span id=\"MathJax-Span-71\" class=\"mfrac\"><span id=\"MathJax-Span-72\" class=\"mn\">1<\/span><span id=\"MathJax-Span-73\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 250\u00b0 = 125\u00b0<br>\u2234 x= 125\u00b0<br><strong>(vi)<\/strong>&nbsp;In the circle with centre O,<br>BOC is its diameter, \u2220AOB = 60\u00b0<br>Arc AB subtends \u2220AOB at the centre of the circle and \u2220ACB at the remaining part of the circle<br>\u2234 \u2220ACB =&nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-74\" class=\"math\"><span id=\"MathJax-Span-75\" class=\"mrow\"><span id=\"MathJax-Span-76\" class=\"mfrac\"><span id=\"MathJax-Span-77\" class=\"mn\">1<\/span><span id=\"MathJax-Span-78\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220AOB<br>=&nbsp;<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-79\" class=\"math\"><span id=\"MathJax-Span-80\" class=\"mrow\"><span id=\"MathJax-Span-81\" class=\"mfrac\"><span id=\"MathJax-Span-82\" class=\"mn\">1<\/span><span id=\"MathJax-Span-83\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 60\u00b0 = 30\u00b0<br>But in \u2206OAC,<br>OC = OA (Radii of the circle)<br>\u2234 \u2220OAC = \u2220OCA = \u2220ACB<br>\u21d2 x = 30\u00b0<br><strong>(vii)<\/strong>&nbsp;In the circle, \u2220BAC and \u2220BDC are in the same segment<br>\u2234 \u2220BDC = \u2220BAC = 50\u00b0<br>Now in ABCD,<br>\u2220DBC + \u2220BCD + \u2220BDC = 180\u00b0 (Angles of a triangle)<br>\u21d2 70\u00b0 + x + 50\u00b0 = 180\u00b0<br>\u21d2 x + 120\u00b0 = 180\u00b0 \u21d2 x = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<br>\u2234 x = 60\u00b0<br><strong>(viii)&nbsp;<\/strong>In a circle with centre O,<br>\u2220OBD = 40\u00b0<br>AB and CD are the diameters of the circle<br>\u2220DBA and \u2220ACD are in the same segment<br>\u2234 \u2220ACD = \u2220DBA = 40\u00b0<br>In AOAC, OA = OC (Radii of the circle)<br>\u2234 \u2220OAC = \u2220OCA = 40\u00b0<br>and \u2220OAC + \u2220OCA + \u2220AOC = 180\u00b0 (Angles in a triangle)<br>\u21d2 40\u00b0 + 40\u00b0 + x = 180\u00b0<br>\u21d2 x + 80\u00b0 = 180\u00b0 \u21d2 x = 180\u00b0 \u2013 80\u00b0 = 100\u00b0<br>\u2234 x = 100\u00b0<br><strong>(ix)<\/strong>&nbsp;In the circle, ABCD is a cyclic quadrilateral \u2220ADB = 32\u00b0, \u2220DAC = 28\u00b0 and \u2220ABD = 50\u00b0<br>\u2220ABD and \u2220ACD are in the same segment of a circle<br>\u2234 \u2220ABD = \u2220ACD \u21d2 \u2220ACD = 50\u00b0<br>Similarly, \u2220ADB = \u2220ACB<br>\u21d2 \u2220ACB = 32\u00b0<br>Now, \u2220DCB = \u2220ACD + \u2220ACB<br>= 50\u00b0 + 32\u00b0 = 82\u00b0<br>\u2234 x = 82\u00b0<br><strong>(x)<\/strong>&nbsp;In a circle,<br>\u2220BAC = 35\u00b0, \u2220CBD = 65\u00b0<br>\u2220BAC and \u2220BDC are in the same segment<br>\u2234 \u2220BAC = \u2220BDC = 35\u00b0<br>In \u2206BCD,<br>\u2220BDC + \u2220BCD + \u2220CBD = 180\u00b0 (Angles in a triangle)<br>\u21d2 35\u00b0 + x + 65\u00b0 = 180\u00b0<br>\u21d2 x + 100\u00b0 = 180\u00b0<br>\u21d2 x = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br>\u2234 x = 80\u00b0<br><strong>(xi)<\/strong>&nbsp;In the circle,<br>\u2220ABD and \u2220ACD are in the same segment of a circle<br>\u2234 \u2220ABD = \u2220ACD = 40\u00b0<br>Now in \u2206CPD,<br>\u2220CPD + \u2220PCD + \u2220PDC = 180\u00b0 (Angles of a triangle)<br>110\u00b0 + 40\u00b0 + x = 180\u00b0<br>\u21d2 x + 150\u00b0 = 180\u00b0<br>\u2234 x= 180\u00b0- 150\u00b0 = 30\u00b0<br><strong>(xii)<\/strong>&nbsp;In the circle, two diameters AC and BD intersect each other at O<br>\u2220BAC = 50\u00b0<br>In \u2206OAB,<br>OA = OB (Radii of the circle)<br>\u2234 \u2220OBA = \u2220OAB = 52\u00b0<br>\u21d2 \u2220ABD = 52\u00b0<br>But \u2220ABD and \u2220ACD are in the same segment of the circle<br>\u2234 \u2220ABD = \u2220ACD \u21d2 52\u00b0 = x<br>\u2234 x = 52\u00b0<\/p>\n<p><strong>Question 4.<\/strong><br><strong>O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that \u2220BOD = \u2220A.<\/strong><br><strong>Solution:<\/strong><br><strong>Given<\/strong>&nbsp;<strong>:<\/strong>&nbsp;O is the circumcentre of \u2206ABC.<br>OD \u22a5 BC<br>OB is joined<br><strong>To prove :<\/strong>&nbsp;\u2220BOD = \u2220A<br><strong>Construction:<\/strong>&nbsp;Join OC.<br><img class=\"alignnone size-full wp-image-67496\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q4.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q4.1\" width=\"227\" height=\"208\"><br><strong>Proof :<\/strong>&nbsp;Arc BC subtends \u2220BOC at the centre and \u2220BAC at the remaining part of the circle<br>\u2234 \u2220BOC = 2\u2220A \u2026(i)<br>In right \u2206OBD and \u2206OCD Side OD = OD (Common)<br>Hyp. OB = OC (Radii of the circle)<br>\u2234 \u2206OBD \u2245 \u2206OCD (RHS criterion)<br>\u2234 \u2220BOD = \u2220COD =&nbsp;<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-84\" class=\"math\"><span id=\"MathJax-Span-85\" class=\"mrow\"><span id=\"MathJax-Span-86\" class=\"mfrac\"><span id=\"MathJax-Span-87\" class=\"mn\">1<\/span><span id=\"MathJax-Span-88\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220BOC<br>\u21d2 \u2220BOC = 2\u2220BOD \u2026(ii)<br>From (i) and (ii)<br>2\u2220BOD = 2\u2220A<br>\u2234\u2220BOD = \u2220A<\/p>\n<p><strong>Question 5.<\/strong><br><strong>In the figure, O is the centre of the circle, BO is the bisector of \u2220ABC. Show that AB = BC.<\/strong><br><img class=\"alignnone size-full wp-image-67497\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q5.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q5.1\" width=\"219\" height=\"210\"><br><strong>Solution:<\/strong><br><strong>Given :<\/strong>&nbsp;In the figure, a circle with centre O OB is the bisector of \u2220ABC<br><strong>To prove :<\/strong>&nbsp;AB = BC<br><strong>Construction :<\/strong>&nbsp;Draw OL \u22a5 AB and OM \u22a5 BC<br><img class=\"alignnone size-full wp-image-67498\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q5.2.png\" sizes=\"(max-width: 218px) 100vw, 218px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q5.2.png 218w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q5.2-100x100.png 100w\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q5.2\" width=\"218\" height=\"215\"><br><strong>Proof:<\/strong>&nbsp;In \u2206OLB and \u2206OMB,<br>\u22201 = \u22202 (Given)<br>\u2220L = \u2220M (Each = 90\u00b0)<br>OB = OB (Common)<br>\u2234 \u2206OLB \u2245 \u2206OMB (AAS criterion)<br>\u2234 OL = OM (c.p.c.t.)<br>But these are distance from the centre and chords equidistant from the centre are equal<br>\u2234 Chord BA = BC<br>Hence AB = BC<\/p>\n<p><strong>Question 6.<\/strong><br><strong>In the figure, O and O\u2019 are centres of two circles intersecting at B and C. ACD is a straight line, find x.<\/strong><br><img class=\"alignnone size-full wp-image-67499\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q6.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q6.1\" width=\"250\" height=\"173\"><br><strong>Solution:<\/strong><br>In the figure, two circles with centres O and O\u2019 intersect each other at B and C.<br>ACD is a line, \u2220AOB = 130\u00b0<br>Arc AB subtends \u2220AOB at the centre O and \u2220ACB at the remaining part of the circle.<br><img class=\"alignnone size-full wp-image-67500\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q6.2.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q6.2\" width=\"243\" height=\"179\"><br>\u2234 \u2220ACB =<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-89\" class=\"math\"><span id=\"MathJax-Span-90\" class=\"mrow\"><span id=\"MathJax-Span-91\" class=\"mfrac\"><span id=\"MathJax-Span-92\" class=\"mn\">1<\/span><span id=\"MathJax-Span-93\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB<br>=&nbsp;<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-94\" class=\"math\"><span id=\"MathJax-Span-95\" class=\"mrow\"><span id=\"MathJax-Span-96\" class=\"mfrac\"><span id=\"MathJax-Span-97\" class=\"mn\">1<\/span><span id=\"MathJax-Span-98\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 130\u00b0 = 65\u00b0<br>But \u2220ACB + \u2220BCD = 180\u00b0 (Linear pair)<br>\u21d2 65\u00b0 + \u2220BCD = 180\u00b0<br>\u21d2 \u2220BCD = 180\u00b0-65\u00b0= 115\u00b0<br>Now, arc BD subtends reflex \u2220BO\u2019D at the centre and \u2220BCD at the remaining part of the circle<br>\u2234 \u2220BO\u2019D = 2\u2220BCD = 2 x 115\u00b0 = 230\u00b0<br>But \u2220BO\u2019D + reflex \u2220BO\u2019D = 360\u00b0 (Angles at a point)<br>\u21d2 x + 230\u00b0 = 360\u00b0<br>\u21d2 x = 360\u00b0 -230\u00b0= 130\u00b0<br>Hence x = 130\u00b0<\/p>\n<div class=\"code-block code-block-4\">\n<div id=\"div-gpt-ad-1699162041532-0\" data-google-query-id=\"CIefyfD56oIDFY5GnQkdPTYEkA\">\n<div id=\"google_ads_iframe_\/22987339172\/insta_BANNER_300x250_051123_3_0__container__\">&nbsp;<\/div>\n<\/div>\n<\/div>\n<p><strong>Question 7.<\/strong><br><strong>In the figure, if \u2220ACB = 40\u00b0, \u2220DPB = 120\u00b0, find \u2220CBD.<\/strong><br><img class=\"alignnone size-full wp-image-67501\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q7.1.png\" sizes=\"(max-width: 197px) 100vw, 197px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q7.1.png 197w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q7.1-150x150.png 150w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q7.1-100x100.png 100w\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q7.1\" width=\"197\" height=\"196\"><br><strong>Solution:<\/strong><br>Arc AB subtend \u2220ACB and \u2220ADB in the same segment of a circle<br>\u2234 \u2220ACB = \u2220ADB = 40\u00b0<br>In \u2206PDB,<br>\u2220DPB + \u2220PBD + \u2220ADB = 180\u00b0 (Sum of angles of a triangle)<br>\u21d2 120\u00b0 + \u2220PBD + 40\u00b0 = 180\u00b0<br>\u21d2 160\u00b0 + \u2220PBD = 180\u00b0<br>\u21d2 \u2220PBD = 180\u00b0 \u2013 160\u00b0 = 20\u00b0<br>\u21d2 \u2220CBD = 20\u00b0<\/p>\n<p><strong>Question 8.<\/strong><br><strong>A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.<\/strong><br><strong>Solution:<\/strong><br>A circle with centre O, a chord AB = radius of the circle C and D are points on the minor and major arcs of the circle<br><img class=\"alignnone size-full wp-image-67502\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q8.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q8.1\" width=\"206\" height=\"217\"><br>\u2234 \u2220ACB and \u2220ADB are formed Now in \u2206AOB,<br>OA = OB = AB (\u2235 AB = radii of the circle)<br>\u2234 \u2206AOB is an equilateral triangle,<br>\u2234 \u2220AOB = 60\u00b0<br>Now arc AB subtends \u2220AOB at the centre and \u2220ADB at the remainder part of the circle.<br>\u2234 \u2220ADB =&nbsp;<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-99\" class=\"math\"><span id=\"MathJax-Span-100\" class=\"mrow\"><span id=\"MathJax-Span-101\" class=\"mfrac\"><span id=\"MathJax-Span-102\" class=\"mn\">1<\/span><span id=\"MathJax-Span-103\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220AOB =&nbsp;<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-104\" class=\"math\"><span id=\"MathJax-Span-105\" class=\"mrow\"><span id=\"MathJax-Span-106\" class=\"mfrac\"><span id=\"MathJax-Span-107\" class=\"mn\">1<\/span><span id=\"MathJax-Span-108\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>x 60\u00b0 = 30\u00b0<br>Now ACBD is a cyclic quadrilateral,<br>\u2234 \u2220ADB + \u2220ACB = 180\u00b0 (Sum of opposite angles of the cyclic quad.)<br>\u21d2 30\u00b0 + \u2220ACB = 180\u00b0<br>\u21d2 \u2220ACB = 180\u00b0 \u2013 30\u00b0 = 150\u00b0<br>\u2234 \u2220ACB = 150\u00b0<br>Hence angles are 150\u00b0 and 30\u00b0<\/p>\n<p><strong>Question 9.<\/strong><br><strong>In the figure, it is given that O is the centre of the circle and \u2220AOC = 150\u00b0. Find \u2220ABC.<\/strong><br><img class=\"alignnone size-full wp-image-67503\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q9.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q9.1\" width=\"207\" height=\"189\"><br><strong>Solution:<\/strong><br>In circle with centre O and \u2220AOC = 150\u00b0<br>But \u2220AOC + reflex \u2220AOC = 360\u00b0<br>\u2234 150\u00b0 + reflex \u2220AOC = 360\u00b0<br>\u21d2 Reflex \u2220AOC = 360\u00b0 \u2013 150\u00b0 = 210\u00b0<br>Now arc AEC subtends \u2220AOC at the centre and \u2220ABC at the remaining part of the circle.<br><img class=\"alignnone size-full wp-image-67504\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q9.2.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q9.2\" width=\"215\" height=\"200\"><br>Reflex \u2220AOC = 2\u2220ABC<br>\u21d2 210\u00b0 = 2\u2220ABC<br>\u2234 \u2220ABC =&nbsp;<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-109\" class=\"math\"><span id=\"MathJax-Span-110\" class=\"mrow\"><span id=\"MathJax-Span-111\" class=\"mfrac\"><span id=\"MathJax-Span-112\" class=\"msubsup\"><span id=\"MathJax-Span-113\" class=\"texatom\"><span id=\"MathJax-Span-114\" class=\"mrow\"><span id=\"MathJax-Span-115\" class=\"mn\">210<\/span><\/span><\/span><span id=\"MathJax-Span-116\" class=\"texatom\"><span id=\"MathJax-Span-117\" class=\"mrow\"><span id=\"MathJax-Span-118\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-119\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 105\u00b0<\/p>\n<p><strong>Question 10.<\/strong><br><strong>In the figure, O is the centre of the circle, prove that \u2220x = \u2220y + \u2220z.<\/strong><br><strong>Solution:<\/strong><br><strong>Given :<\/strong>&nbsp;In circle, O is centre<br><img class=\"alignnone size-full wp-image-67505\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q10.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q10.1\" width=\"210\" height=\"242\"><br><strong>To prove :<\/strong>&nbsp;\u2220x = \u2220y + \u2220z<br><strong>Proof :<\/strong>&nbsp;\u2235 \u22203 and \u22204 are in the same segment of the circle<br>\u2234 \u22203 = \u22204 \u2026(i)<br>\u2235 Arc AB subtends \u2220AOB at the centre and \u22203 at the remaining part of the circle<br>\u2234 \u2220x = 2\u22203 = \u22203 + \u22203 = \u22203 + \u22204 (\u2235 \u22203 = \u22204) \u2026(ii)<br>In \u2206ACE,<br>Ext. \u2220y = \u22203 + \u22201<br>(Ext. is equal to sum of its interior opposite angles)<br>\u21d2 \u22203 \u2013 \u2220y \u2013 \u22201 \u2026(ii)<br>From (i) and (ii),<br>\u2220x = \u2220y \u2013 \u22201 + \u22204 \u2026(iii)<br>Similarly in \u2206ADF,<br>Ext. \u22204 = \u22201 + \u2220z \u2026(iv)<br>From (iii) and (iv)<br>\u2220x = \u2220y-\u2220l + (\u22201 + \u2220z)<br>= \u2220y \u2013 \u22201 + \u22201 + \u2220z = \u2220y + \u2220z<br>Hence \u2220x = \u2220y + \u2220z<\/p>\n<p><strong>Question 11.<\/strong><br><strong>In the figure, O is the centre of a circle and PQ is the diameter. If \u2220ROS = 40\u00b0, find \u2220RTS.<\/strong><br><img class=\"alignnone size-full wp-image-67506\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q11.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q11.1\" width=\"241\" height=\"170\"><br><strong>Solution:<\/strong><br>In the figure, O is the centre of the circle,<br>PQ is the diameter and \u2220ROS = 40\u00b0<br>Now we have to find \u2220RTS<br>Arc RS subtends \u2220ROS at the centre and \u2220RQS at the remaining part of the circle<br><img class=\"alignnone size-full wp-image-67507\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-15-Areas-of-Parallelograms-and-Triangles-Ex-15.4-Q11.2.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 Q11.2\" width=\"249\" height=\"184\"><br>\u2234 \u2220RQS =&nbsp;<span id=\"MathJax-Element-20-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-120\" class=\"math\"><span id=\"MathJax-Span-121\" class=\"mrow\"><span id=\"MathJax-Span-122\" class=\"mfrac\"><span id=\"MathJax-Span-123\" class=\"mn\">1<\/span><span id=\"MathJax-Span-124\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220ROS<br>=&nbsp;<span id=\"MathJax-Element-21-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-125\" class=\"math\"><span id=\"MathJax-Span-126\" class=\"mrow\"><span id=\"MathJax-Span-127\" class=\"mfrac\"><span id=\"MathJax-Span-128\" class=\"mn\">1<\/span><span id=\"MathJax-Span-129\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;x 40\u00b0 = 20\u00b0<br>\u2235 \u2220PRQ = 90\u00b0 (Angle in a semi-circle)<br>\u2234 \u2220QRT = 180\u00b0 \u2013 90\u00b0 = 90\u00b0 (\u2235 PRT is a straight line)<br>Now in \u2206RQT,<br>\u2220RQT + \u2220QRT + \u2220RTQ = 180\u00b0 (Angles of a triangle)<br>\u21d2 20\u00b0 + 90\u00b0 + \u2220RTQ = 180\u00b0<br>\u21d2 \u2220RTQ = 180\u00b0 \u2013 20\u00b0 \u2013 90\u00b0 = 70\u00b0 or \u2220RTS = 70\u00b0<br>Hence \u2220RTS = 70\u00b0<\/p>\n<p>We have included all the information regarding <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/span><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">. If you have any queries feel free to ask in the comment section.&nbsp;<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-15-exercise-154\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631088226897\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-class-9-solutions-chapter-15-exercise-154-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088343664\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-class-9-solutions-chapter-15-exercise-154-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088344826\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-15-exercise-154\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practising RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088348118\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4: The PDF is available for free. You can easily download the PDF for RD Sharma Class 9 Solutions Chapter 15&nbsp;Exercise 15.4. These solutions will be very beneficial for your class 9 exams.&nbsp; RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 15 &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-4\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 15 Exercise 15.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127677,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76823],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125328"}],"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\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125328"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125328\/revisions"}],"predecessor-version":[{"id":517698,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125328\/revisions\/517698"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127677"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125328"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125328"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125328"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}