{"id":125333,"date":"2023-09-13T08:59:00","date_gmt":"2023-09-13T03:29:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125333"},"modified":"2023-11-20T11:56:20","modified_gmt":"2023-11-20T06:26:20","slug":"rd-sharma-class-9-solutions-chapter-15-exercise-15-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-5\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127679\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.5.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.5-768x432.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 15.5&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 15.5<\/strong>: PDF is available for free. You can easily download PDF Sharma <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> Exercise 15.5. These solutions will be very beneficial for your 9th-grade exams.<\/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-69e70438ce1d1\" 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\" 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studying RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5?<\/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-15-5\/#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-155\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5&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 15.5<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\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\/RD_Sharma_Class_9_Chapter_15_Ex_15_5.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_5.pdf\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-15-exercise-155\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5&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 15.5<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>In the figure, \u2206ABC is an equilateral triangle. Find m \u2220BEC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/45594767832_a4e2d56549_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"187\" height=\"202\" \/><br \/>Solution:<br \/>\u2235 \u2206ABC is an equilateral triangle<br \/>\u2234 A = 60\u00b0<br \/>\u2235 ABEC is a cyclic quadrilateral<br \/>\u2234 \u2220A + \u2220E = 180\u00b0 (Sum of opposite angles)<br \/>\u21d2 60\u00b0 + \u2220E = 180\u00b0<br \/>\u21d2 \u2220E = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<br \/>\u2234 m \u2220BEC = 120\u00b0<\/p>\n<p>Question 2.<br \/>In the figure, \u2206PQR is an isosceles triangle with PQ = PR and m \u2220PQR = 35\u00b0. Find m \u2220QSR and m \u2220QTR.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1913\/43827181760_811766f7f8_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"209\" height=\"194\" \/><br \/>Solution:<br \/>In the figure, \u2206PQR is an isosceles PQ = PR<br \/>\u2220PQR = 35\u00b0<br \/>\u2234 \u2220PRQ = 35\u00b0<\/p>\n<p>But \u2220PQR + \u2220PRQ + \u2220QPR = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 35\u00b0 + 35\u00b0 + \u2220QPR = 180\u00b0<br \/>\u21d2 70\u00b0 + \u2220QPR = 180\u00b0<br \/>\u2234 \u2220QPR = 180\u00b0 \u2013 70\u00b0 = 110\u00b0<br \/>\u2235 \u2220QSR = \u2220QPR (Angle in the same segment of circles)<br \/>\u2234 \u2220QSR = 110\u00b0<\/p>\n<p>But PQTR is a cyclic quadrilateral<br \/>\u2234 \u2220QTR + \u2220QPR = 180\u00b0<br \/>\u21d2 \u2220QTR + 110\u00b0 = 180\u00b0<br \/>\u21d2 \u2220QTR = 180\u00b0 -110\u00b0 = 70\u00b0<br \/>Hence \u2220QTR = 70\u00b0<\/p>\n<p>Question 3.<br \/>In the figure, O is the center of the circle. If \u2220BOD = 160\u00b0, find the values of x and y.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1956\/45594767662_cc25d44231_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"216\" height=\"199\" \/><br \/>Solution:<br \/>In the figure, O is the center of the circle \u2220BOD =160\u00b0<\/p>\n<p>ABCD is the cyclic quadrilateral<br \/>\u2235 Arc BAD subtends \u2220BOD is the angle at the center and \u2220BCD is on the other part of the circle<br \/>\u2234 \u2220BCD =\u00a0<span id=\"MathJax-Element-31-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-181\" class=\"math\"><span id=\"MathJax-Span-182\" class=\"mrow\"><span id=\"MathJax-Span-183\" class=\"mfrac\"><span id=\"MathJax-Span-184\" class=\"mn\">1<\/span><span id=\"MathJax-Span-185\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220BOD<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-32-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-186\" class=\"math\"><span id=\"MathJax-Span-187\" class=\"mrow\"><span id=\"MathJax-Span-188\" class=\"mfrac\"><span id=\"MathJax-Span-189\" class=\"mn\">1<\/span><span id=\"MathJax-Span-190\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 160\u00b0 = 80\u00b0<br \/>\u2235 ABCD is a cyclic quadrilateral,<br \/>\u2234 \u2220A + \u2220C = 180\u00b0<br \/>\u21d2 y + x = 180\u00b0<br \/>\u21d2 y + 80\u00b0 = 180\u00b0<br \/>\u21d2 y =180\u00b0- 80\u00b0 = 100\u00b0<br \/>\u2234 x = 80\u00b0, y = 100\u00b0<\/p>\n<p>Question 4.<br \/>In the figure, ABCD is a cyclic quadrilateral. If \u2220BCD = 100\u00b0 and \u2220ABD = 70\u00b0, find \u2220ADB.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/43827181520_7b252c43a0_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"185\" height=\"168\" \/><br \/>Solution:<br \/>In a circle, ABCD is a cyclic quadrilateral \u2220BCD = 100\u00b0 and \u2220ABD = 70\u00b0<br \/>\u2235 ABCD is a cyclic quadrilateral,<br \/>\u2234 \u2220A + \u2220C = 180\u00b0 (Sum of opposite angles)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1921\/45594767552_91fbb2143e_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"211\" height=\"174\" \/><\/p>\n<p>\u21d2 \u2220A + 100\u00b0= 180\u00b0<br \/>\u2220A = 180\u00b0- 100\u00b0 = 80\u00b0<br \/>Now in \u2206ABD,<br \/>\u2220A + \u2220ABD + \u2220ADB = 180\u00b0<br \/>\u21d2 80\u00b0 + 70\u00b0 + \u2220ADB = 180\u00b0<br \/>\u21d2 150\u00b0 +\u2220ADB = 180\u00b0<br \/>\u2234 \u2220ADB = 180\u00b0- 150\u00b0 = 30\u00b0<br \/>Hence \u2220ADB = 30\u00b0<\/p>\n<p>Question 5.<br \/>If ABCD is a cyclic quadrilateral in which AD || BC. Prove that \u2220B = \u2220C.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45594767402_baa55fa0f8_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"187\" height=\"160\" \/><\/p>\n<p>Solution:<br \/>Given: ABCD is a cyclic quadrilateral in which AD || BC<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1952\/45594767362_965824b123_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"189\" height=\"159\" \/><br \/>To prove: \u2220B = \u2220C<br \/>Proof: \u2235 AD || BC<br \/>\u2234 \u2220A + \u2220B = 180\u00b0<br \/>(Sum of interior angles)<br \/>But \u2220A + \u2220C = 180\u00b0<br \/>(Opposite angles of the cyclic quadrilateral)<br \/>\u2234 \u2220A + \u2220B = \u2220A + \u2220C<br \/>\u21d2 \u2220B = \u2220C<br \/>Hence \u2220B = \u2220C<\/p>\n<p>Question 6.<br \/>In the figure, O is the center of the circle. Find \u2220CBD.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1961\/43827180870_4c0208fa9a_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"185\" height=\"217\" \/><br \/>Solution:<br \/>Arc AC subtends \u2220AOC at the center and \u2220APC at the remaining part of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/45594767232_ebd191ebcf_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"186\" height=\"209\" \/><br \/>\u2234 \u2220APC =\u00a0<span id=\"MathJax-Element-33-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-191\" class=\"math\"><span id=\"MathJax-Span-192\" class=\"mrow\"><span id=\"MathJax-Span-193\" class=\"mfrac\"><span id=\"MathJax-Span-194\" class=\"mn\">1<\/span><span id=\"MathJax-Span-195\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220AOC<br \/>=\u00a0<span id=\"MathJax-Element-34-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-196\" class=\"math\"><span id=\"MathJax-Span-197\" class=\"mrow\"><span id=\"MathJax-Span-198\" class=\"mfrac\"><span id=\"MathJax-Span-199\" class=\"mn\">1<\/span><span id=\"MathJax-Span-200\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 100\u00b0 = 50\u00b0<br \/>\u2235 APCB is aa. cyclic quadrilateral,<\/p>\n<p>\u2234 \u2220APC + \u2220ABC = 180\u00b0<br \/>\u21d2 50\u00b0 + \u2220ABC = 180\u00b0 \u21d2 \u2220ABC =180\u00b0- 50\u00b0<br \/>\u2234 \u2220ABC =130\u00b0<\/p>\n<p>But \u2220ABC + \u2220CBD = 180\u00b0 (Linear pair)<br \/>\u21d2 130\u00b0 + \u2220CBD = 180\u00b0<br \/>\u21d2 \u2220CBD = 180\u00b0- 130\u00b0 = 50\u00b0<br \/>\u2234 \u2220CBD = 50\u00b0<\/p>\n<p>Question 7.<br \/>In the figure, AB and CD are the diameters of a circle with center O. If \u2220OBD = 50\u00b0, find \u2220AOC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1907\/43827180680_63b3cfe859_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"185\" height=\"166\" \/><br \/>Solution:<br \/>Two diameters AB and CD intersect each other at O. AC, CB, and BD are joined<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/45594767032_2949852485_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"161\" height=\"171\" \/><\/p>\n<p>\u2220DBA = 50\u00b0<br \/>\u2220DBA and \u2220DCA are in the same segment<br \/>\u2234 \u2220DBA = \u2220DCA = 50\u00b0<br \/>In \u2206OAC, OA = OC (Radii of the circle)<br \/>\u2234 \u2220OAC = \u2220OCA = \u2220DCA = 50\u00b0<br \/>and \u2220OAC + \u2220OCA + \u2220AOC = 180\u00b0 (Sum of angles of a triangle)<\/p>\n<p>\u21d2 50\u00b0 + 50\u00b0 + \u2220AOC = 180\u00b0<br \/>\u21d2 100\u00b0 + \u2220AOC = 180\u00b0<br \/>\u21d2 \u2220AOC = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br \/>Hence \u2220AOC = 80\u00b0<\/p>\n<p>Question 8.<br \/>On a semi-circle with AB as diameter, a point C is taken so that m (\u2220CAB) = 30\u00b0. Find m (\u2220ACB) and m (\u2220ABC).<br \/>Solution:<br \/>A semicircle with AB as the diameter<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/45594766952_065f833871_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"233\" height=\"124\" \/><\/p>\n<p>\u2220 CAB = 30\u00b0<br \/>\u2220ACB = 90\u00b0 (Angle in a semi-circle)<br \/>But \u2220CAB + \u2220ACB + \u2220ABC = 180\u00b0<br \/>\u21d2 30\u00b0 + 90\u00b0 + \u2220ABC \u2013 180\u00b0<br \/>\u21d2 120\u00b0 + \u2220ABC = 180\u00b0<br \/>\u2234 \u2220ABC = 180\u00b0- 120\u00b0 = 60\u00b0<br \/>Hence m \u2220ACB = 90\u00b0<br \/>and m \u2220ABC = 60\u00b0<\/p>\n<p>Question 9.<br \/>In a cyclic quadrilateral ABCD, if AB || CD and \u2220B = 70\u00b0, find the remaining angles.<br \/>Solution:<br \/>In a cyclic quadrilateral ABCD, AB || CD and \u2220B = 70\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1901\/43827180300_6c746fb55c_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"244\" height=\"199\" \/><\/p>\n<p>\u2235 ABCD is a cyclic quadrilateral<br \/>\u2234 \u2220B + \u2220D = 180\u00b0<br \/>\u21d2 70\u00b0 + \u2220D = 180\u00b0<br \/>\u21d2 \u2220D = 180\u00b0-70\u00b0 = 110\u00b0<br \/>\u2235 AB || CD<\/p>\n<p>\u2234 \u2220A + \u2220D = 180\u00b0 (Sum of interior angles)<br \/>\u2220A+ 110\u00b0= 180\u00b0<br \/>\u21d2 \u2220A= 180\u00b0- 110\u00b0 = 70\u00b0<br \/>Similarly, \u2220B + \u2220C = 180\u00b0<br \/>\u21d2 70\u00b0 + \u2220C- 180\u00b0 \u2018<br \/>\u21d2 \u2220C = 180\u00b0-70\u00b0= 110\u00b0<br \/>\u2234 \u2220A = 70\u00b0, \u2220C = 110\u00b0, \u2220D = 110\u00b0<\/p>\n<p>Question 10.<br \/>In a cyclic quadrilateral ABCD, if m \u2220A = 3(m \u2220C). Find m \u2220A.<br \/>Solution:<br \/>In cyclic quadrilateral ABCD, m \u2220A = 3(m \u2220C)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/43827180110_9f8da2997f_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"218\" height=\"190\" \/><\/p>\n<p>\u2235 ABCD is a cyclic quadrilateral,<br \/>\u2234 \u2220A + \u2220C = 180\u00b0<br \/>\u21d2 3 \u2220C + \u2220C = 180\u00b0 \u21d2 4\u2220C = 180\u00b0<br \/>\u21d2 \u2220C =\u00a0<span id=\"MathJax-Element-35-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-201\" class=\"math\"><span id=\"MathJax-Span-202\" class=\"mrow\"><span id=\"MathJax-Span-203\" class=\"mfrac\"><span id=\"MathJax-Span-204\" class=\"msubsup\"><span id=\"MathJax-Span-205\" class=\"texatom\"><span id=\"MathJax-Span-206\" class=\"mrow\"><span id=\"MathJax-Span-207\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-208\" class=\"texatom\"><span id=\"MathJax-Span-209\" class=\"mrow\"><span id=\"MathJax-Span-210\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-211\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0 = 45\u00b0<br \/>\u2234 \u2220A = 3 x 45\u00b0= 135\u00b0<br \/>Hence m \u2220A =135\u00b0<\/p>\n<p>Question 11.<br \/>In the figure, O is the center of the circle, and \u2220DAB = 50\u00b0. Calculate the values of x and y.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/43827180020_249a137f05_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"197\" height=\"164\" \/><br \/>Solution:<\/p>\n<p>In the figure, O is the center of the circle \u2220DAB = 50\u00b0<br \/>\u2235 ABCD is a cyclic quadrilateral<br \/>\u2234 \u2220A + \u2220C = 180\u00b0<br \/>\u21d2 50\u00b0 + y = 180\u00b0<br \/>\u21d2 y = 180\u00b0 \u2013 50\u00b0 = 130\u00b0<\/p>\n<p>In \u2206OAB, OA = OB (Radii of the circle)<br \/>\u2234 \u2220A = \u2220OBA = 50\u00b0<br \/>\u2234 Ext. \u2220DOB = \u2220A + \u2220OBA<br \/>x = 50\u00b0 + 50\u00b0 = 100\u00b0<br \/>\u2234 x= 100\u00b0, y= 130\u00b0<\/p>\n<p>Question 12.<br \/>In the figure, if \u2220BAC = 60\u00b0 and \u2220BCA = 20\u00b0, find \u2220ADC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/43827179850_c55ba32191_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"181\" height=\"190\" \/><br \/>Solution:<\/p>\n<p>In \u2206ABC,<br \/>\u2220BAC + \u2220ABC + \u2220ACB = 180\u00b0 (Sum of angles of a triangle)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1956\/43827179740_3e961250e1_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"184\" height=\"189\" \/><\/p>\n<p>60\u00b0 + \u2220ABC + 20\u00b0 = 180\u00b0<br \/>\u2220ABC + 80\u00b0 = 180\u00b0<br \/>\u2234 \u2220ABC = 180\u00b0 -80\u00b0= 100\u00b0<br \/>\u2235 ABCD is a cyclic quadrilateral,<br \/>\u2234 \u2220ABC + \u2220ADC = 180\u00b0<br \/>100\u00b0 + \u2220ADC = 180\u00b0<br \/>\u2234 \u2220ADC = 180\u00b0- 100\u00b0 = 80\u00b0<\/p>\n<p>Question 13.<br \/>In the figure, if ABC is an equilateral triangle. Find \u2220BDC and \u2220BEC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/45594766462_e4a6305831_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"184\" height=\"197\" \/><br \/>Solution:<br \/>In a circle, \u2206ABC is an equilateral triangle<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1934\/30704316757_5c587e395c_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"184\" height=\"193\" \/><\/p>\n<p>\u2234 \u2220A = 60\u00b0<br \/>\u2235 \u2220BAC and \u2220BDC are in the same segment<br \/>\u2234 \u2220BAC = \u2220BDC = 60\u00b0<br \/>\u2235 BECD is a cyclic quadrilateral<br \/>\u2234 \u2220BDC + \u2220BEC = 180\u00b0<br \/>\u21d2 60\u00b0 + \u2220BEC = 180\u00b0<br \/>\u21d2 \u2220BEC = 180\u00b0-60\u00b0= 120\u00b0<br \/>Hence \u2220BDC = 60\u00b0 and \u2220BEC = 120\u00b0<\/p>\n<p>Question 14.<br \/>In the figure, O is the center of the circle. If \u2220CEA = 30\u00b0, find the values of x, y, and z.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/30704316687_88dbc85a2f_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"173\" height=\"194\" \/><br \/>Solution:<br \/>\u2220AEC and \u2220ADC are in the same segment<br \/>\u2234 \u2220AEC = \u2220ADC = 30\u00b0<br \/>\u2234 z = 30\u00b0<\/p>\n<p>ABCD is a cyclic quadrilateral<br \/>\u2234 \u2220B + \u2220D = 180\u00b0<br \/>\u21d2 x + z = 180\u00b0<br \/>\u21d2 x + 30\u00b0 = 180\u00b0<br \/>\u21d2 x = 180\u00b0 \u2013 30\u00b0 = 150\u00b0<\/p>\n<p>Arc AC subtends \u2220AOB at the center and \u2220ADC at the remaining part of the circle<br \/>\u2234 \u2220AOC = 2\u2220D = 2 x 30\u00b0 = 60\u00b0<br \/>\u2234 y = 60\u00b0<br \/>Hence x = 150\u00b0, y \u2013 60\u00b0 and z = 30\u00b0<\/p>\n<p>Question 15.<br \/>In the figure, \u2220BAD = 78\u00b0, \u2220DCF = x\u00b0 and \u2220DEF = y\u00b0. Find the values of x and y.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/30704316537_fe8e66d2d5_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"265\" height=\"156\" \/><br \/>Solution:<\/p>\n<p>In the figure, two circles intersect each other at C and D<br \/>\u2220BAD = 78\u00b0, \u2220DCF = x, \u2220DEF = y<br \/>ABCD is a cyclic quadrilateral<br \/>\u2234 Ext. \u2220DCF = it&#8217;s interior opposite \u2220BAD<br \/>\u21d2 x = 78\u00b0<\/p>\n<p>In cyclic quadrilateral CDEF,<br \/>\u2220DCF + \u2220DEF = 180\u00b0<br \/>\u21d2 78\u00b0 + y = 180\u00b0<br \/>\u21d2 y = 180\u00b0 \u2013 78\u00b0<br \/>y = 102\u00b0<br \/>Hence x = 78\u00b0, and y- 102\u00b0<\/p>\n<p>Question 16.<br \/>In a cyclic quadrilateral ABCD, if \u2220A \u2013 \u2220C = 60\u00b0, prove that the smaller of two is 60\u00b0.<br \/>Solution:<br \/>In cyclic quadrilateral ABCD,<br \/>\u2220A \u2013 \u2220C = 60\u00b0<br \/>But \u2220A + \u2220C = 180\u00b0 (Sum of opposite angles)<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45594766222_463f81ed9f_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"248\" height=\"198\" \/><\/p>\n<p>Adding, 2\u2220A = 240\u00b0 \u21d2 \u2220A =\u00a0<span id=\"MathJax-Element-36-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-212\" class=\"math\"><span id=\"MathJax-Span-213\" class=\"mrow\"><span id=\"MathJax-Span-214\" class=\"mfrac\"><span id=\"MathJax-Span-215\" class=\"msubsup\"><span id=\"MathJax-Span-216\" class=\"texatom\"><span id=\"MathJax-Span-217\" class=\"mrow\"><span id=\"MathJax-Span-218\" class=\"mn\">62<\/span><\/span><\/span><span id=\"MathJax-Span-219\" class=\"texatom\"><span id=\"MathJax-Span-220\" class=\"mrow\"><span id=\"MathJax-Span-221\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-222\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 120\u00b0 and subtracting<br \/>2\u2220C = 120\u00b0 \u21d2 \u2220C =\u00a0<span id=\"MathJax-Element-37-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-223\" class=\"math\"><span id=\"MathJax-Span-224\" class=\"mrow\"><span id=\"MathJax-Span-225\" class=\"mfrac\"><span id=\"MathJax-Span-226\" class=\"msubsup\"><span id=\"MathJax-Span-227\" class=\"texatom\"><span id=\"MathJax-Span-228\" class=\"mrow\"><span id=\"MathJax-Span-229\" class=\"mn\">120<\/span><\/span><\/span><span id=\"MathJax-Span-230\" class=\"texatom\"><span id=\"MathJax-Span-231\" class=\"mrow\"><span id=\"MathJax-Span-232\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-233\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 60\u00b0<br \/>\u2234 The smaller angle of the two is 60\u00b0.<\/p>\n<p>Question 17.<br \/>In the figure, ABCD is a cyclic quadrilateral. Find the value of x.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1969\/30704316387_255df795f4_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"210\" height=\"256\" \/><br \/>Solution:<br \/>\u2220CDE + \u2220CDA = 180\u00b0 (Linear pair)<br \/>\u21d2 80\u00b0 + \u2220CDA = 180\u00b0<br \/>\u21d2 \u2220CDA = 180\u00b0 \u2013 80\u00b0 = 100\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/45594766142_ca645e6e88_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"211\" height=\"279\" \/><br \/>In cyclic quadrilateral ABCD,<br \/>Ext. \u2220ABF = Its interior opposite angle \u2220CDA = 100\u00b0<br \/>\u2234 x = 100\u00b0<\/p>\n<p>Question 18.<br \/>ABCD is a cyclic quadrilateral in which:<br \/>(i) BC || AD, \u2220ADC =110\u00b0 and \u2220B AC = 50\u00b0. Find \u2220DAC.<br \/>(ii) \u2220DBC = 80\u00b0 and \u2220BAC = 40\u00b0. Find \u2220BCD.<br \/>(iii) \u2220BCD = 100\u00b0 and \u2220ABD = 70\u00b0, find \u2220ADB.<\/p>\n<p>Solution:<br \/>(i) In the figure,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1935\/31772954248_e8d67d7c37_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"221\" height=\"195\" \/><\/p>\n<p>ABCD is a cyclic quadrilateral and AD || BC, \u2220ADC = 110\u00b0<br \/>\u2220BAC = 50\u00b0<br \/>\u2235 \u2220B + \u2220D = 180\u00b0 (Sum of opposite angles)<br \/>\u21d2 \u2220B + 110\u00b0 = 180\u00b0<br \/>\u2234 \u2220B = 180\u00b0- 110\u00b0 = 70\u00b0<\/p>\n<p>Now in \u2206ABC,<br \/>\u2220CAB + \u2220ABC + \u2220BCA = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 50\u00b0 + 70\u00b0 + \u2220BCA = 180\u00b0<br \/>\u21d2 120\u00b0 + \u2220BCA = 180\u00b0<br \/>\u21d2 \u2220BCA = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<\/p>\n<p>But \u2220DAC = \u2220BCA (Alternate angles)<br \/>\u2234 \u2220DAC = 60\u00b0<\/p>\n<p>(ii) In cyclic quadrilateral ABCD,<br \/>Diagonals AC and BD are joined \u2220DBC = 80\u00b0, \u2220BAC = 40\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1913\/30704316247_40c156f069_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"219\" height=\"203\" \/><\/p>\n<p>Arc DC subtends \u2220DBC and \u2220DAC in the same segment<br \/>\u2234 \u2220DBC = \u2220DAC = 80\u00b0<br \/>\u2234 \u2220DAB = \u2220DAC + \u2220CAB = 80\u00b0 + 40\u00b0 = 120\u00b0<\/p>\n<p>But \u2220DAC + \u2220BCD = 180\u00b0 (Sum of opposite angles of a cyclic quad.)<br \/>\u21d2 120\u00b0 +\u2220BCD = 180\u00b0<br \/>\u21d2 \u2220BCD = 180\u00b0- 120\u00b0 = 60\u00b0<\/p>\n<p>(iii) In the figure, ABCD is a cyclic quadrilateral BD that is joined<br \/>\u2220BCD = 100\u00b0<br \/>and \u2220ABD = 70\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1951\/45594766062_76895cf80a_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"232\" height=\"218\" \/><\/p>\n<p>\u2220A + \u2220C = 180\u00b0 (Sum of opposite angles of the cyclic quad.)<br \/>\u2220A+ 100\u00b0= 180\u00b0<br \/>\u21d2 \u2220A= 180\u00b0- 100\u00b0<br \/>\u2234 \u2220A = 80\u00b0<\/p>\n<p>Now in \u2206ABD,<br \/>\u2220A + \u2220ABD + \u2220ADB = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 80\u00b0 + 70\u00b0 + \u2220ADB = 180\u00b0<br \/>\u21d2 150\u00b0 +\u2220ADB = 180\u00b0<br \/>\u21d2 \u2220ADB = 180\u00b0- 150\u00b0 = 30\u00b0<br \/>\u2234 \u2220ADB = 30\u00b0<\/p>\n<p>Question 19.<br \/>Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.<br \/>Solution:<br \/>Given: ABCD is a rhombus. Four circles are drawn on the sides AB, BC, CD, and DA respectively<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1961\/30704316127_a8c7f67932_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"308\" height=\"280\" \/><\/p>\n<p>To prove: The circles pass through the point of intersection of the diagonals of the rhombus ABCD<\/p>\n<p>Proof: ABCD is a rhombus whose diagonals AC and BD intersect each other at O<br \/>\u2235 The diagonals of a rhombus bisect each other at right angles<br \/>\u2234 \u2220AOB = \u2220BOC = \u2220COD = \u2220DOA = 90\u00b0<\/p>\n<p>Now when \u2220AOB = 90\u00b0<br \/>and a circle described on AB as diameter will pass through O<br \/>Similarly, the circles on BC, CD, and DA as diameter, will also pass through O<\/p>\n<p>Question 20.<br \/>If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that is diagonals are equal.<\/p>\n<p>Solution:<br \/>Given: In cyclic quadrilateral ABCD, AB = CD<br \/>AC and BD are the diagonals<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1945\/45594765622_de59d91420_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"227\" height=\"202\" \/><\/p>\n<p>To prove: AC = BC<br \/>Proof: \u2235 AB = CD<br \/>\u2234 arc AB = arc CD<br \/>Adding arc BC to both sides, then arc AB + arc BC = arc BC + arc CD<br \/>\u21d2 arc AC = arc BD<br \/>\u2234 AC = BD<br \/>Hence diagonal of the cyclic quadrilateral is equal.<\/p>\n<p>Question 21.<br \/>Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).<br \/>Solution:<\/p>\n<p>Given: In \u2206ABC, circles are drawn on sides AB and AC<br \/>To prove: Circles drawn on AB and AC intersect at D which lies on BC, the third side<br \/>Construction: Draw AD \u22a5 BC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1963\/30704315857_41465ccaf1_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"212\" height=\"194\" \/><\/p>\n<p>Proof: \u2235 AD \u22a5 BC<br \/>\u2234 \u2220ADB = \u2220ADC = 90\u00b0<br \/>So, the circles drawn on sides AB and AC as diameter will pass through D<br \/>Hence circles are drawn on two sides of a triangle and pass through D, which lies on the third side.<\/p>\n<p>Question 22.<br \/>ABCD is a cyclic trapezium with AD || BC. If \u2220B = 70\u00b0, determine the other three angles of the trapezium.<br \/>Solution:<br \/>In the figure, ABCD is a trapezium in which AD || BC and \u2220B = 70\u00b0<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1945\/45594765422_f93f468aa5_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"223\" height=\"196\" \/><br \/>\u2235 AD || BC<br \/>\u2234 \u2220A + \u2220B = 180\u00b0 (Sum of co-interior angles)<br \/>\u21d2 \u2220A + 70\u00b0 = 180\u00b0<br \/>\u21d2 \u2220A= 180\u00b0- 70\u00b0 = 110\u00b0<br \/>\u2234 \u2220A = 110\u00b0<\/p>\n<p>But \u2220A + \u2220C = 180\u00b0 and \u2220B + \u2220D = 180\u00b0 (Sum of opposite angles of a cyclic quadrilateral)<br \/>\u2234 110\u00b0 + \u2220C = 180\u00b0<br \/>\u21d2 \u2220C = 180\u00b0- 110\u00b0 = 70\u00b0<br \/>and 70\u00b0 + \u2220D = 180\u00b0<br \/>\u21d2 \u2220D = 180\u00b0 \u2013 70\u00b0 = 110\u00b0<br \/>\u2234 \u2220A = 110\u00b0, \u2220C = 70\u00b0 and \u2220D = 110\u00b0<\/p>\n<p>Question 23.<br \/>In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If \u2220DBC = 55\u00b0 and \u2220BAC = 45\u00b0, find \u2220BCD.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1948\/30704315747_a3a6d5dcb2_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"168\" height=\"156\" \/><br \/>Solution:<\/p>\n<p>In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are drawn \u2220DBC = 55\u00b0 and \u2220BAC = 45\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/30704315677_443411ebbc_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"168\" height=\"156\" \/><br \/>\u2235 \u2220BAC and \u2220BDC are in the same segment<br \/>\u2234 \u2220BAC = \u2220BDC = 45\u00b0<\/p>\n<p>Now in ABCD,<br \/>\u2220DBC + \u2220BDC + \u2220BCD = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 55\u00b0 + 45\u00b0 + \u2220BCD = 180\u00b0<br \/>\u21d2 100\u00b0 + \u2220BCD = 180\u00b0<br \/>\u21d2 \u2220BCD = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br \/>Hence \u2220BCD = 80\u00b0<\/p>\n<p>Question 24.<br \/>Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.<br \/>Solution:<br \/>Given: ABCD is a cyclic quadrilateral<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/45594765172_764fe6590f_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"280\" height=\"258\" \/><\/p>\n<p>To prove: The perpendicular bisectors of the sides are concurrent<\/p>\n<p>Proof: \u2235 Each side of the cyclic quadrilateral is a chord of the circle and the perpendicular of a chord passes through the centre of the circle<br \/>Hence the perpendicular bisectors of each side will pass through the center O<br \/>Hence the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent<\/p>\n<p>Question 25.<br \/>Prove that the center of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.<br \/>Solution:<br \/>Given: ABCD is a cyclic rectangle and diagonals AC and BD intersect each other at O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1924\/44731137375_4c25340400_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"210\" height=\"196\" \/><br \/>To prove: O is the point of intersection is the center of the circle.<\/p>\n<p>Proof: Let O be the center of the circle- circumscribing the rectangle ABCD<br \/>Since each angle of a rectangle is a right angle and AC is the chord of the circle<br \/>\u2234 AC will be the diameter of the circle Similarly, we can prove that diagonal BD is also the diameter of the circle<\/p>\n<p>\u2234 The diameters of the circle pass through the center<br \/>Hence the point of intersection of the diagonals of the rectangle is the centre of the circle.<\/p>\n<p>Question 26.<br \/>ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:<br \/>(i) AD || BC<br \/>(ii) EB = EC.<br \/>Solution:<br \/>Given: ABCD is a cyclic quadrilateral in which sides BA and CD are produced to meet at E and EA = ED<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45645326121_4ae98333cd_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"287\" height=\"188\" \/><br \/>To prove :<br \/>(i) AD || BC<br \/>(ii) EB = EC<br \/>Proof: \u2235 EA = ED<br \/>\u2234 In \u2206EAD<br \/>\u2220EAD = \u2220EDA (Angles opposite to equal sides)<br \/>In a cyclic quadrilateral ABCD,<br \/>Ext. \u2220EAD = \u2220C<br \/>Similarly Ext. \u2220EDA = \u2220B<br \/>\u2235 \u2220EAD = \u2220EDA<br \/>\u2234 \u2220B = \u2220C<br \/>Now in \u2206EBC,<br \/>\u2235 \u2220B = \u2220C<br \/>\u2234 EC = EB (Sides opposite to equal sides)<br \/>and \u2220EAD = \u2220B<br \/>But these are corresponding angles<br \/>\u2234 AD || BC<\/p>\n<p>Question 27.<br \/>Prove that the angle in a segment shorter than a semicircle is greater than a right angle.<br \/>Solution:<br \/>Given: A segment ACB shorter than a semicircle and an angle \u2220ACB inscribed in it<br \/>To prove\u00a0 \u2220ACB &lt; 90\u00b0<br \/>Construction: Join OA and OB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/30704315277_0ae47def5f_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"249\" height=\"232\" \/><br \/>Proof: Arc ADB subtends \u2220AOB at the center and \u2220ACB at the remaining part of the circle \u2234 \u2220ACB = <span id=\"MathJax-Element-38-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-234\" class=\"math\"><span id=\"MathJax-Span-235\" class=\"mrow\"><span id=\"MathJax-Span-236\" class=\"mfrac\"><span id=\"MathJax-Span-237\" class=\"mn\">1<\/span><span id=\"MathJax-Span-238\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220AOB But \u2220AOB &gt; 180\u00b0 (Reflex angle)<br \/>\u2234 \u2220ACB &gt;\u00a0<span id=\"MathJax-Element-39-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-239\" class=\"math\"><span id=\"MathJax-Span-240\" class=\"mrow\"><span id=\"MathJax-Span-241\" class=\"mfrac\"><span id=\"MathJax-Span-242\" class=\"mn\">1<\/span><span id=\"MathJax-Span-243\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x [80\u00b0<br \/>\u21d2 \u2220ACB &gt; 90\u00b0<\/p>\n<p>Question 28.<br \/>Prove that the angle in a segment greater than a semi-circle is less than a right angle<br \/>Solution:<br \/>Given: A segment ACB, greater than a semicircle with center O and \u2220ACB is described in it<br \/>To prove\u00a0 \u2220ACB &lt; 90\u00b0<br \/>Construction: Join OA and OB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1921\/45645325991_62c2597508_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"217\" height=\"235\" \/><br \/>Proof: Arc ADB subtends \u2220AOB at the center and \u2220ACB at the remaining part of the circle<br \/>\u2234 \u2220ACB =<span id=\"MathJax-Element-40-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-244\" class=\"math\"><span id=\"MathJax-Span-245\" class=\"mrow\"><span id=\"MathJax-Span-246\" class=\"mfrac\"><span id=\"MathJax-Span-247\" class=\"mn\">1<\/span><span id=\"MathJax-Span-248\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220AOB<br \/>But \u2220AOB &lt; 180\u00b0 (A straight angle) 1<br \/>\u2234 \u2220ACB &lt;\u00a0<span id=\"MathJax-Element-41-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-249\" class=\"math\"><span id=\"MathJax-Span-250\" class=\"mrow\"><span id=\"MathJax-Span-251\" class=\"mfrac\"><span id=\"MathJax-Span-252\" class=\"mn\">1<\/span><span id=\"MathJax-Span-253\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 180\u00b0<br \/>\u21d2 \u2220ACB &lt;90\u00b0<br \/>Hence \u2220ACB &lt; 90\u00b0<\/p>\n<p>Question 29.<br \/>Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.<br \/>Solution:<br \/>Given: In a right-angled \u2206ABC<br \/>\u2220B = 90\u00b0, D is the midpoint of hypotenuse AC. DB is joined.<br \/>To prove : BD =\u00a0<span id=\"MathJax-Element-42-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-254\" class=\"math\"><span id=\"MathJax-Span-255\" class=\"mrow\"><span id=\"MathJax-Span-256\" class=\"mfrac\"><span id=\"MathJax-Span-257\" class=\"mn\">1<\/span><span id=\"MathJax-Span-258\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC<br \/>Construction: Draw a circle with center D and AC as the diameter<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1909\/30704315147_6b66e4c936_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"221\" height=\"196\" \/><br \/>Proof: \u2235 \u2220ABC = 90\u00b0<br \/>\u2234 The circle drawn on AC as diameter will pass through B<br \/>\u2234 BD is the radius of the circle<br \/>But AC is the diameter of the circle and D is the midpoint of AC<br \/>\u2234 AD = DC = BD<br \/>\u2234 BD=\u00a0<span id=\"MathJax-Element-43-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-259\" class=\"math\"><span id=\"MathJax-Span-260\" class=\"mrow\"><span id=\"MathJax-Span-261\" class=\"mfrac\"><span id=\"MathJax-Span-262\" class=\"mn\">1<\/span><span id=\"MathJax-Span-263\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC<\/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 15.5&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 15.5<\/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.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-15-exercise-155\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5&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 15.5<\/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-1631088241649\" 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-155-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 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 15.5 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088353084\" 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-155-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 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 15.5 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088354292\" 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-155\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RRD Sharma Class 9 Solutions Chapter 15 Exercise 15.5, students can earn higher academic grades. Our experts solve RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088357429\" 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 15.5: PDF is available for free. You can easily download PDF Sharma RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5. These solutions will be very beneficial for your 9th-grade exams. RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-5\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 15 Exercise 15.5 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127679,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76824],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125333"}],"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=125333"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125333\/revisions"}],"predecessor-version":[{"id":504343,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125333\/revisions\/504343"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127679"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125333"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125333"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125333"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}