{"id":63778,"date":"2021-09-04T16:21:42","date_gmt":"2021-09-04T10:51:42","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63778"},"modified":"2021-09-17T18:21:26","modified_gmt":"2021-09-17T12:51:26","slug":"rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 15 &#8211; Areas Of Parallelograms And Triangles (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124544\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-15-Areas-Of-Parallelograms-And-Triangles.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 15\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-15-Areas-Of-Parallelograms-And-Triangles.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-15-Areas-Of-Parallelograms-And-Triangles-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span style=\"font-weight: 400;\"><strong>RD Sharma Solutions Class 9 Maths Chapter 15 &#8211; Areas Of Parallelograms And Triangles:<\/strong> RD Sharma solution for chapter 15 CBSE Class 9 Maths <\/span><span style=\"font-weight: 400;\">is an important &amp; useful study resource for the students to prepare well for the exam.\u00a0<\/span><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 15 <span style=\"font-weight: 400;\">are created exercise-wise that assist the students to analyze their weaknesses. This also helps them to assess their exams preparation for scoring good marks.\u00a0<\/span><\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d4c49b36efd\" 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 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Solutions Class 9 Maths Chapter 15 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-15\" title=\"Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 15\u00a0\">Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 15\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#access-answers-of-rd-sharma-solutions-class-9-maths-chapter-15\" title=\"Access answers of RD Sharma Solutions Class 9 Maths Chapter 15\">Access answers of RD Sharma Solutions Class 9 Maths Chapter 15<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#rd-sharma-class-9-chapter-15-areas-of-parallelograms-and-triangles-ex-151\" title=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles Ex 15.1\">RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles Ex 15.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#rd-sharma-solutions-class-9-chapter-15-areas-of-parallelograms-and-triangles-ex-152\" title=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles Ex 15.2\">RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles Ex 15.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#class-9-maths-chapter-15-areas-of-parallelograms-and-triangles-rd-sharma-solutions-ex-153\" title=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Ex 15.3\">Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Ex 15.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#rd-sharma-mathematics-class-9-solutions-chapter-15-areas-of-parallelograms-and-triangles-ex-154\" title=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4\">RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 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-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#class-9-maths-chapter-15-areas-of-parallelograms-and-triangles-rd-sharma-solutions-ex-155\" title=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Ex 15.5\">Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Ex 15.5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#rd-sharma-class-9-book-chapter-15-areas-of-parallelograms-and-triangles-vsaqs\" title=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles VSAQS\">RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#rd-sharma-class-9-solution-chapter-15-areas-of-parallelograms-and-triangles-mcqs\" title=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles MCQS\">RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles MCQS<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#detailed-exercise-wise-explanation-with-listing-of-important-topics-in-the-exercise\" title=\"Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise\">Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#important-topics-from-rd-sharma-solutions-class-9-maths-chapter-15\" title=\"Important Topics from RD Sharma Solutions Class 9 Maths Chapter 15\">Important Topics from RD Sharma Solutions Class 9 Maths Chapter 15<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-15\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 15\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 15<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-15\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 15?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 15?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-15\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 15?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 15?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-15-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 15\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 15\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-chapter-15-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 15 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/15-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 15<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/15-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-15\"><\/span><strong>Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 15\u00a0<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-15-class-9-maths-exercise-15-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 15 Exercise 15.1<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-15-class-9-maths-exercise-15-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 15 Exercise 15.2<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-15-class-9-maths-exercise-15-3-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 15 Exercise 15.3<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-9-maths-chapter-15\"><\/span><strong>Access answers of <\/strong><strong>RD Sharma Solutions Class 9 Maths Chapter 15<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-15-areas-of-parallelograms-and-triangles-ex-151\"><\/span>RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles Ex 15.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Fill in the blanks: [NCERT]<br \/>(i) All points lying inside \/ outside a circle are called \u2026\u2026.. points \/ \u2026\u2026\u2026. points.<br \/>(ii) Circles having the same centre and different radii are called \u2026\u2026.. circles.<br \/>(iii) A point whose distance from the centre of a circle is greater than its radius lies in \u2026\u2026.. of the circle.<br \/>(iv) A continuous piece of a circle is \u2026\u2026.. of the circle.<br \/>(v) The longest chord of a circle is a \u2026\u2026\u2026 of the circle.<br \/>(vi) An arc is a \u2026\u2026.. when its ends are the ends of a diameter.<br \/>(vii) Segment of a circle is the region between an are and \u2026\u2026..of the circle.<br \/>(viii)A circle divides the plane, on which it lies, in \u2026\u2026.. parts.<br \/>Solution:<br \/>(i)\u00a0All points lying inside \/ outside a circle are called\u00a0interior\u00a0points \/\u00a0exterior\u00a0points.<br \/>(ii)\u00a0Circles having the same centre and different radii are called\u00a0concentric\u00a0circles.<br \/>(iii)\u00a0A point whose distance from the centre of a circle is greater than its radius lies in\u00a0exterior\u00a0of the circle.<br \/>(iv)\u00a0A continuous piece of a circle is\u00a0arc\u00a0of the circle.<br \/>(v)\u00a0The longest chord of a circle is a\u00a0diameter\u00a0of the circle.<br \/>(vi)\u00a0An arc is a\u00a0semi-circle\u00a0when its ends are the ends of a diameter.<br \/>(vii)\u00a0Segment of a circle is the region between an arc and\u00a0centre\u00a0of the circle.<br \/>(viii)\u00a0A circle divides the plane, on which it lies, in\u00a0three\u00a0parts.<\/p>\n<p>Question 2.<br \/>Write the truth value (T\/F) of the following with suitable reasons: [NCERT]<br \/>(i) A circle is a plane figure.<br \/>(ii) Line segment joining the centre to any point on the circle is a radius of the circle.<br \/>(iii) If a circle is divided into three equal arcs each is a major arc.<br \/>(iv) A circle has only finite number of equal chords.<br \/>(v) A chord of a cirlce, which is twice as long is its radius is a diameter of the circle.<br \/>(vi) Sector is the region between the chord and its corresponding arc.<br \/>(vii) The degree measure of an arc is the complement of the central angle containing the arc.<br \/>(viii)The degree measure of a semi-circle is 180\u00b0.<br \/>Solution:<br \/>(i)\u00a0True.<br \/>(ii)\u00a0True.<br \/>(iii)\u00a0True.<br \/>(iv)\u00a0False. As it has infinite number of equal chords.<br \/>(v)\u00a0True.<br \/>(vi)\u00a0False. It is a segment not sector.<br \/>(vii)\u00a0False. As total degree measure of a circle is 360\u00b0.<br \/>(viii)\u00a0True.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-15-areas-of-parallelograms-and-triangles-ex-152\"><\/span>RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles Ex 15.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.<br \/>Solution:<br \/>Radius of circle with centre O is OA = 8 cm<br \/>Length of chord AB = 12 cm<br \/>OC \u22a5 AB which bisects AB at C<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44920312954_433d96dc6a_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"153\" height=\"142\" \/><br \/>\u2234 AC = CB = 12 x\u00a0<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=\"mn\">1<\/span><span id=\"MathJax-Span-5\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 6 cm<br \/>In \u2206OAC,<br \/>OA2 = OC2 + AC2 (Pythagoras Theorem)<br \/>\u21d2 (8)2 = OC2 + (6)2<br \/>\u21d2 64 = OC2 + 36<br \/>OC2 = 64 \u2013 36 = 28<br \/>\u2234 OC =\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"msqrt\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mn\">28<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0=\u00a0<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"msqrt\"><span id=\"MathJax-Span-14\" class=\"mrow\"><span id=\"MathJax-Span-15\" class=\"mn\">4<\/span><span id=\"MathJax-Span-16\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-17\" class=\"mn\">7<\/span><\/span>\u2212\u2212\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0cm<br \/>= 2 x 2.6457 = 5.291 cm<\/p>\n<p>Question 2.<br \/>Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.<br \/>Solution:<br \/>Let AB be a chord of a circle with radius 10 cm. OC \u22a5 AB<br \/>\u2234 OA = 10 cm<br \/>OC = 5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/43827187230_b6340c7749_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.2 - 2\" width=\"179\" height=\"165\" \/><br \/>\u2235 OC divides AB into two equal parts<br \/>i.e. AC = CB<br \/>Now in right AOAC,<br \/>OA2 = OC2 + AC2 (Pythagoras Theorem)<br \/>\u21d2 (10)2 = (5)2 + AC2<br \/>\u21d2 100 = 25 + AC2<br \/>\u21d2 AC2 = 100 \u2013 25 = 75<br \/>\u2234 AC =\u00a0<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-18\" class=\"math\"><span id=\"MathJax-Span-19\" class=\"mrow\"><span id=\"MathJax-Span-20\" class=\"msqrt\"><span id=\"MathJax-Span-21\" class=\"mrow\"><span id=\"MathJax-Span-22\" class=\"mn\">75<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>=\u00a0<span id=\"MathJax-Element-5-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=\"msqrt\"><span id=\"MathJax-Span-26\" class=\"mrow\"><span id=\"MathJax-Span-27\" class=\"mn\">25<\/span><span id=\"MathJax-Span-28\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-29\" class=\"mn\">3<\/span><\/span>\u2212\u2212\u2212\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0= 5 x 1.732<br \/>\u2234 AB = 2 x AC = 2 x 5 x 1.732 = 10 x 1.732 = 17.32 cm<\/p>\n<p>Question 3.<br \/>Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.<br \/>Solution:<br \/>In a circle with centre O and radius 6 cm and a chord AB at a distance of 4 cm from the centre of the circle<br \/>i.e. OA = 6 cm and OL \u22a5 AB, OL = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1902\/44920312744_60f5001b41_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"186\" height=\"173\" \/><br \/>\u2235 Perpendicular OL bisects the chord AB at L 1<br \/>\u2234 AL = LB=<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-30\" class=\"math\"><span id=\"MathJax-Span-31\" class=\"mrow\"><span id=\"MathJax-Span-32\" class=\"mfrac\"><span id=\"MathJax-Span-33\" class=\"mn\">1<\/span><span id=\"MathJax-Span-34\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AB<br \/>Now in right \u2206OAL,<br \/>OA2 = OL2 + AL2 (Pythagoras Theorem)<br \/>(6)2 = (4)2 + AL2<br \/>\u21d2 36=16+AL2<br \/>\u21d2 AL2 = 36 \u2013 16 = 20<br \/>\u2234 AL =\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-35\" class=\"math\"><span id=\"MathJax-Span-36\" class=\"mrow\"><span id=\"MathJax-Span-37\" class=\"msqrt\"><span id=\"MathJax-Span-38\" class=\"mrow\"><span id=\"MathJax-Span-39\" class=\"mn\">20<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0=\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-40\" class=\"math\"><span id=\"MathJax-Span-41\" class=\"mrow\"><span id=\"MathJax-Span-42\" class=\"msqrt\"><span id=\"MathJax-Span-43\" class=\"mrow\"><span id=\"MathJax-Span-44\" class=\"mn\">4<\/span><span id=\"MathJax-Span-45\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-46\" class=\"mn\">5<\/span><\/span>\u2212\u2212\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0= 2 x 2.236 = 4.472 cm<br \/>\u2234 Chord AB = 4.472 x 2 = 8.944 = 8.94 cm<\/p>\n<p>Question 4.<br \/>Give a method to find the centre of a given circle.<br \/>Solution:<br \/>Steps of construction :<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44920312714_8ee3bab8bc_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"270\" height=\"223\" \/><br \/>(i) Take three distinct points on the circle say A, B and C.<br \/>(ii) Join AB and AC.<br \/>(iii) Draw the perpendicular bisectors of AB and AC which intersect each other at O.<br \/>O is the required centre of the given circle<\/p>\n<p>Question 5.<br \/>Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.<br \/>Solution:<br \/>Given : In circle with centre O<br \/>CD is the diameter and AB is the chord<br \/>which is bisected by diameter at E<br \/>OA and OB are joined<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44920312504_ceb1804c9f_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"226\" height=\"181\" \/><br \/>To prove : \u2220AOB = \u2220BOA<br \/>Proof : In \u2206OAE and \u2206OBE<br \/>OA = OB (Radii of the circle)<br \/>OE = OE (Common)<br \/>AE = EB (Given)<br \/>\u2234 \u2206OAE = \u2206OBE (SSS criterian)<br \/>\u2234 \u2220AOE = \u2220BOE (c.p.c.t.)<br \/>Hence diameter bisect the angle subtended by the chord AB.<\/p>\n<p>Question 6.<br \/>A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment AB = 5 cm.<br \/>(ii) Draw a perpendicular bisector of AB.<br \/>(iii) With centre A and radius 4 cm, draw an arc which intersects the perpendicular bisector at O.<br \/>(iv) With centre O and radius 4 cm, draw a circle which passes through A and B.<br \/>With radius 2 cm, we cannot draw the circle passing through A and B as diameter<br \/>i. e. 2 + 2 = 4 cm is shorder than 5 cm.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/44920312434_44eb9ac1d2_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"216\" height=\"289\" \/><\/p>\n<p>Question 7.<br \/>An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 9 cm.<br \/>(ii) With centres B and C, draw arcs of 9 cm radius which intersect each other at A.<br \/>(iii) Join AB and AC.<br \/>\u2206ABC is the required triangle.<br \/>(iv) Draw perpendicular bisectors of sides AB and BC which intersect each other at O.<br \/>(v) With centre O and radius OB, draw a circle which passes through A, B and C.<br \/>This is the require circle in which \u2206ABC is inscribed.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1923\/44920312334_ae53a67336_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"225\" height=\"220\" \/><br \/>On measuring its radius, it is 5.2 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1907\/44920312254_b75c2417b4_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"288\" height=\"150\" \/><\/p>\n<p>Question 8.<br \/>Given an arc of a circle, complete the circle.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Take three points A, B and C on the arc and join AB and BC.<br \/>(ii) Draw the perpendicular bisector of AB and BC which intersect each other at O.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1976\/44920312174_c69f5e47b7_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"260\" height=\"246\" \/><br \/>(iii) With centre O and radius OA or OB, complete the circle.<br \/>This is the required circle.<\/p>\n<p>Question 9.<br \/>Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?<br \/>Solution:<br \/>Below, three different pairs of circles are drawn:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44920312034_829981073d_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"286\" height=\"176\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920311994_81b081cf6c_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"273\" height=\"383\" \/><br \/>(i) In the first pair, two circles do not intersect each other. Therefore they have no point in common. .<br \/>(ii) In the second pair, two circles intersect (touch) each other at one point P. Therefore they have one point in common.<br \/>(iii) In the third pair, two circles intersect each other at two points. Therefore they have two points in common.<br \/>There is no other possibility of two circles intersecting each other.<br \/>Therefore, two circles have at the most two points in common.<\/p>\n<p>Question 10.<br \/>Suppose you are given a circle. Give a construction to find its centre.<br \/>Solution:<br \/>See Q. No. 4 of this exercise.<\/p>\n<p>Question 11.<br \/>The length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre? [NCERT]<br \/>Solution:<br \/>A circle with centre O and two parallel chords<br \/>AB and CD are AB = 6 cm, CD = 8 cm<br \/>Let OL \u22a5 AB and OM \u22a5 CD<br \/>\u2234 OL = 4 cm<br \/>Let OM = x cm<br \/>Let r be the radius of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44920311954_dc00545b3c_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"359\" height=\"601\" \/><\/p>\n<p>Question 12.<br \/>Two chords AB, CD of lengths 5 cm and 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.<br \/>Solution:<br \/>Let two chords AB and CD of length 5 cm and 11 cm are parallel to each other AB = 5 cm, CD = 11 cm<br \/>Distance between AB and LM = 3 cm<br \/>Join OB and OD<br \/>OL and OM are the perpendicular on CD and AB respectively. Which bisects AB and CD.<br \/>Let OL = x, then OM = (x + 3)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/44920311834_3e6a6e5810_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"228\" height=\"209\" \/><br \/>Now in right \u2206OLD,<br \/>OD2 = OL2 + LD2<br \/>= x2 + (5.5)2<br \/>Similarly in right \u2206OMB,<br \/>OB2 = OM2 + MB2 = (x + 3)2 + (2.5)2<br \/>But OD = OB (Radii of the circle)<br \/>\u2234 (x + 3)2 + (2.5)2 = x2 + (5.5)2<br \/>x2 + 6x + 9 + 6.25 = x2 + 30.25<br \/>6x = 30.25 \u2013 6.25 \u2013 9 = 15<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920311774_7feb40e6a7_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"354\" height=\"248\" \/><\/p>\n<p>Question 13.<br \/>Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.<br \/>Solution:<br \/>Given : A circle with centre O and a chord AB<br \/>Let M be the mid point of AB and OM is joined and produced to meet the minor arc AB at N<br \/>To prove : M is the mid point of arc AB<br \/>Construction : Join OA, OB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44920311664_c6d905ac62_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"220\" height=\"207\" \/><br \/>Proof: \u2235 M is mid point of AB<br \/>\u2234 OM \u22a5 AB<br \/>In AOAM and OBM,<br \/>OA = OB (Radii of the circle)<br \/>OM = OM (common)<br \/>AM = BM (M is mid point of AB)<br \/>\u2234 \u2206OAM = \u2206OBM (SSS criterian)<br \/>\u2234 \u2220AOM = \u2220BOM (c.p.c.t.)<br \/>\u21d2 \u2220AOM = \u2220BOM<br \/>But these are centre angles at the centre made by arcs AN and BN<br \/>\u2234 Arc AN = Arc BN<br \/>Hence N divides the arc in two equal parts<\/p>\n<p>Question 14.<br \/>Prove that two different circles cannot intersect each other at more than two points.<br \/>Solution:<br \/>Given : Two circles<br \/>To prove : They cannot intersect each other more than two points<br \/>Construction : Let two circles intersect each other at three points A, B and C<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44920311614_37175c683a_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"236\" height=\"162\" \/><br \/>Proof : Since two circles with centres O and O\u2019 intersect at A, B and C<br \/>\u2234 A, B and C are non-collinear points<br \/>\u2234 Circle with centre O passes through three points A, B and C<br \/>and circle with centre O\u2019 also passes through three points A, B and C<br \/>But one and only one circle can be drawn through three points<br \/>\u2234Our supposition is wrong<br \/>\u2234 Two circle cannot intersect each other not more than two points.<\/p>\n<p>Question 15.<br \/>Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. [NCERT]<br \/>Solution:<br \/>Let r be the radius of the circle with centre O.<br \/>Two parallel chords AB = 5 cm, CD = 11 cm<br \/>Let OL \u22a5 AB and OM \u22a5CD<br \/>\u2234 LM = 6 cm<br \/>Let OM = x, then<br \/>OL = 6 \u2013 x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/44920311564_0bb5877b7c_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"329\" height=\"463\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1977\/44920311374_fceb4821bc_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"305\" height=\"587\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-maths-chapter-15-areas-of-parallelograms-and-triangles-rd-sharma-solutions-ex-153\"><\/span>Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Ex 15.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha. [NCERT]<br \/>Solution:<br \/>\u2235 Distance between Isha and Ishita and Ishita and Nisha is same<br \/>\u2234 RS = SM = 24 m<br \/>\u2234They are equidistant from the centre<br \/>In right \u2206ORL,<br \/>OL\u00b2 = OR\u00b2 \u2013 RL\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44920311254_383604f433_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"325\" height=\"418\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44920311094_36d5b835be_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"333\" height=\"379\" \/><br \/>Hence distance between Ishita and Nisha = 38.4 m<\/p>\n<p>Question 2.<br \/>A circular park of radius 40 m is situated in a colony. Three beys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find thelength of the string of each phone. [NCERT]<br \/>Solution:<br \/>Radius of circular park = 40 m<br \/>Ankur, Amit and Anand are sitting at equal distance to each other By joining them, an equilateral triangle ABC is formed produce BO to L which is perpendicular bisector of AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44920311004_64b821c335_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"314\" height=\"238\" \/><br \/>\u2234 BL = 40 + 20 = 60 m (\u2235 O is centroid of \u2206ABC also)<br \/>Let a be the side of \u2206ABC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44920310744_f9a140b7b7_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"337\" height=\"125\" \/><\/p>\n<p>Hence the distance between each other = 40<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-47\" class=\"math\"><span id=\"MathJax-Span-48\" class=\"mrow\"><span id=\"MathJax-Span-49\" class=\"msqrt\"><span id=\"MathJax-Span-50\" class=\"mrow\"><span id=\"MathJax-Span-51\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0mRD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-mathematics-class-9-solutions-chapter-15-areas-of-parallelograms-and-triangles-ex-154\"><\/span>RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>In the figure, O is the centre of the circle. If \u2220APB = 50\u00b0, find \u2220AOB and \u2220OAB.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1925\/44920310614_ab5cb94069_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"178\" height=\"189\" \/><br \/>Solution:<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 src=\"https:\/\/farm2.staticflickr.com\/1952\/44920310484_18c15a1904_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"210\" height=\"193\" \/><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 =\u00a0<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-52\" class=\"math\"><span id=\"MathJax-Span-53\" class=\"mrow\"><span id=\"MathJax-Span-54\" class=\"mfrac\"><span id=\"MathJax-Span-55\" class=\"msubsup\"><span id=\"MathJax-Span-56\" class=\"texatom\"><span id=\"MathJax-Span-57\" class=\"mrow\"><span id=\"MathJax-Span-58\" class=\"mn\">80<\/span><\/span><\/span><span id=\"MathJax-Span-59\" class=\"texatom\"><span id=\"MathJax-Span-60\" class=\"mrow\"><span id=\"MathJax-Span-61\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-62\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 40\u00b0<\/p>\n<p>Question 2.<br \/>In the figure, O is the centre of the circle. Find \u2220BAC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/44920310424_f9a360aec3_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"219\" height=\"192\" \/><br \/>Solution:<br \/>In the circle with 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 src=\"https:\/\/farm2.staticflickr.com\/1932\/43827184490_c039d4da09_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"205\" height=\"198\" \/><br \/>\u2234 \u2220BOC = 2\u2220BAC<br \/>\u21d2 170\u00b0 = 2\u2220BAC<br \/>\u21d2 \u2220BAC =\u00a0<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-63\" class=\"math\"><span id=\"MathJax-Span-64\" class=\"mrow\"><span id=\"MathJax-Span-65\" class=\"mfrac\"><span id=\"MathJax-Span-66\" class=\"msubsup\"><span id=\"MathJax-Span-67\" class=\"texatom\"><span id=\"MathJax-Span-68\" class=\"mrow\"><span id=\"MathJax-Span-69\" class=\"mn\">170<\/span><\/span><\/span><span id=\"MathJax-Span-70\" class=\"texatom\"><span id=\"MathJax-Span-71\" class=\"mrow\"><span id=\"MathJax-Span-72\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-73\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 85\u00b0<br \/>\u2234 \u2220BAC = 85\u00b0<\/p>\n<p>Question 3.<br \/>If O is the centre of the circle, find the value of x in each of the following figures:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/44920310354_c5de485501_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"611\" height=\"499\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/43827184170_592a3f4a32_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"575\" height=\"455\" \/><br \/>Solution:<br \/>(i) 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 =\u00a0<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>\u2220BOC =\u00a0<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>\u00a0x 45\u00b0 =\u00a0<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=\"msubsup\"><span id=\"MathJax-Span-88\" class=\"texatom\"><span id=\"MathJax-Span-89\" class=\"mrow\"><span id=\"MathJax-Span-90\" class=\"mn\">45<\/span><\/span><\/span><span id=\"MathJax-Span-91\" class=\"texatom\"><span id=\"MathJax-Span-92\" class=\"mrow\"><span id=\"MathJax-Span-93\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-94\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br \/>\u2234 \u2220BPC = 22\u00a0<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-95\" class=\"math\"><span id=\"MathJax-Span-96\" class=\"mrow\"><span id=\"MathJax-Span-97\" class=\"mfrac\"><span id=\"MathJax-Span-98\" class=\"mn\">1<\/span><span id=\"MathJax-Span-99\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0 or x = 22\u00a0<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-100\" class=\"math\"><span id=\"MathJax-Span-101\" class=\"mrow\"><span id=\"MathJax-Span-102\" class=\"mfrac\"><span id=\"MathJax-Span-103\" class=\"mn\">1<\/span><span id=\"MathJax-Span-104\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0<br \/>(ii) \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 \/>(iii) 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 =\u00a0<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-105\" class=\"math\"><span id=\"MathJax-Span-106\" class=\"mrow\"><span id=\"MathJax-Span-107\" class=\"mfrac\"><span id=\"MathJax-Span-108\" class=\"mn\">1<\/span><span id=\"MathJax-Span-109\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOC =\u00a0<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-110\" class=\"math\"><span id=\"MathJax-Span-111\" class=\"mrow\"><span id=\"MathJax-Span-112\" class=\"mfrac\"><span id=\"MathJax-Span-113\" class=\"mn\">1<\/span><span id=\"MathJax-Span-114\" 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 \/>(iv) 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 \/>(v) 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 =\u00a0<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-115\" class=\"math\"><span id=\"MathJax-Span-116\" class=\"mrow\"><span id=\"MathJax-Span-117\" class=\"mfrac\"><span id=\"MathJax-Span-118\" class=\"mn\">1<\/span><span id=\"MathJax-Span-119\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB<br \/>\u21d2 x =\u00a0<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>\u00a0x 250\u00b0 = 125\u00b0<br \/>\u2234 x= 125\u00b0<br \/>(vi) 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 =\u00a0<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>\u00a0\u2220AOB<br \/>=\u00a0<span id=\"MathJax-Element-22-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-130\" class=\"math\"><span id=\"MathJax-Span-131\" class=\"mrow\"><span id=\"MathJax-Span-132\" class=\"mfrac\"><span id=\"MathJax-Span-133\" class=\"mn\">1<\/span><span id=\"MathJax-Span-134\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 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 \/>(vii) 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 \/>(viii)\u00a0In circle with centre O,<br \/>\u2220OBD = 40\u00b0<br \/>AB and CD are 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 \/>(ix) 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 \/>(x) 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 \/>(xi) 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 \/>(xii) 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>Question 4.<br \/>O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that \u2220BOD = \u2220A.<br \/>Solution:<br \/>Given : O is the circumcentre of \u2206ABC.<br \/>OD \u22a5 BC<br \/>OB is joined<br \/>To prove : \u2220BOD = \u2220A<br \/>Construction : Join OC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/43827183600_d73978edb1_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"243\" height=\"209\" \/><br \/>Proof : 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 =\u00a0<span id=\"MathJax-Element-23-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-135\" class=\"math\"><span id=\"MathJax-Span-136\" class=\"mrow\"><span id=\"MathJax-Span-137\" class=\"mfrac\"><span id=\"MathJax-Span-138\" class=\"mn\">1<\/span><span id=\"MathJax-Span-139\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\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>Question 5.<br \/>In the figure, O is the centre of the circle, BO is the bisector of \u2220ABC. Show that AB = BC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44920309554_47ec27db45_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"219\" height=\"208\" \/><br \/>Solution:<br \/>Given : In the figure, a circle with centre O OB is the bisector of \u2220ABC<br \/>To prove : AB = BC<br \/>Construction : Draw OL \u22a5 AB and OM \u22a5 BC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1935\/43827183350_b6d9d0ce39_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"215\" height=\"214\" \/><br \/>Proof: 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>Question 6.<br \/>In the figure, O and O\u2019 are centres of two circles intersecting at B and C. ACD is a straight line, find x.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1954\/44920309334_117b591e7a_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"249\" height=\"172\" \/><br \/>Solution:<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 src=\"https:\/\/farm2.staticflickr.com\/1933\/43827183230_bac744a5b1_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"283\" height=\"172\" \/><br \/>\u2234 \u2220ACB =<span id=\"MathJax-Element-24-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-140\" class=\"math\"><span id=\"MathJax-Span-141\" class=\"mrow\"><span id=\"MathJax-Span-142\" class=\"mfrac\"><span id=\"MathJax-Span-143\" class=\"mn\">1<\/span><span id=\"MathJax-Span-144\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB<br \/>=\u00a0<span id=\"MathJax-Element-25-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-145\" class=\"math\"><span id=\"MathJax-Span-146\" class=\"mrow\"><span id=\"MathJax-Span-147\" class=\"mfrac\"><span id=\"MathJax-Span-148\" class=\"mn\">1<\/span><span id=\"MathJax-Span-149\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 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<p>Question 7.<br \/>In the figure, if \u2220ACB = 40\u00b0, \u2220DPB = 120\u00b0, find \u2220CBD.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/44920309214_985ae43ea9_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"223\" height=\"195\" \/><br \/>Solution:<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>Question 8.<br \/>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.<br \/>Solution:<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 src=\"https:\/\/farm2.staticflickr.com\/1925\/43827182800_f36032df05_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"202\" height=\"218\" \/><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 =\u00a0<span id=\"MathJax-Element-26-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-150\" class=\"math\"><span id=\"MathJax-Span-151\" class=\"mrow\"><span id=\"MathJax-Span-152\" class=\"mfrac\"><span id=\"MathJax-Span-153\" class=\"mn\">1<\/span><span id=\"MathJax-Span-154\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220AOB =\u00a0<span id=\"MathJax-Element-27-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-155\" class=\"math\"><span id=\"MathJax-Span-156\" class=\"mrow\"><span id=\"MathJax-Span-157\" class=\"mfrac\"><span id=\"MathJax-Span-158\" class=\"mn\">1<\/span><span id=\"MathJax-Span-159\" 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 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>Question 9.<br \/>In the figure, it is given that O is the centre of the circle and \u2220AOC = 150\u00b0. Find \u2220ABC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1936\/43827182510_2cb07cb167_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"224\" height=\"183\" \/><br \/>Solution:<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 src=\"https:\/\/farm2.staticflickr.com\/1949\/45594768142_cc6e92043c_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles Ex 15.4\" width=\"222\" height=\"202\" \/><br \/>Reflex \u2220AOC = 2\u2220ABC<br \/>\u21d2 210\u00b0 = 2\u2220ABC<br \/>\u2234 \u2220ABC =\u00a0<span id=\"MathJax-Element-28-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-160\" class=\"math\"><span id=\"MathJax-Span-161\" class=\"mrow\"><span id=\"MathJax-Span-162\" class=\"mfrac\"><span id=\"MathJax-Span-163\" class=\"msubsup\"><span id=\"MathJax-Span-164\" class=\"texatom\"><span id=\"MathJax-Span-165\" class=\"mrow\"><span id=\"MathJax-Span-166\" class=\"mn\">210<\/span><\/span><\/span><span id=\"MathJax-Span-167\" class=\"texatom\"><span id=\"MathJax-Span-168\" class=\"mrow\"><span id=\"MathJax-Span-169\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-170\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 105\u00b0<\/p>\n<p>Question 10.<br \/>In the figure, O is the centre of the circle, prove that \u2220x = \u2220y + \u2220z.<br \/>Solution:<br \/>Given : In circle, O is centre<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/44920308634_33c32eb7a6_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"200\" height=\"238\" \/><br \/>To prove : \u2220x = \u2220y + \u2220z<br \/>Proof : \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>Question 11.<br \/>In the figure, O is the centre of a circle and PQ is a diameter. If \u2220ROS = 40\u00b0, find \u2220RTS.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1970\/45594767982_c7034c9ec6_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"222\" height=\"165\" \/><br \/>Solution:<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 src=\"https:\/\/farm2.staticflickr.com\/1955\/43827182050_44ff5b16bc_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"248\" height=\"185\" \/><br \/>\u2234 \u2220RQS =\u00a0<span id=\"MathJax-Element-29-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-171\" class=\"math\"><span id=\"MathJax-Span-172\" class=\"mrow\"><span id=\"MathJax-Span-173\" class=\"mfrac\"><span id=\"MathJax-Span-174\" class=\"mn\">1<\/span><span id=\"MathJax-Span-175\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220ROS<br \/>=\u00a0<span id=\"MathJax-Element-30-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-176\" class=\"math\"><span id=\"MathJax-Span-177\" class=\"mrow\"><span id=\"MathJax-Span-178\" class=\"mfrac\"><span id=\"MathJax-Span-179\" class=\"mn\">1<\/span><span id=\"MathJax-Span-180\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 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<h3><span class=\"ez-toc-section\" id=\"class-9-maths-chapter-15-areas-of-parallelograms-and-triangles-rd-sharma-solutions-ex-155\"><\/span>Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Ex 15.5<span class=\"ez-toc-section-end\"><\/span><\/h3>\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<br \/>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<br \/>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 centre 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 centre of the circle \u2220BOD =160\u00b0<br \/>ABCD is the cyclic quadrilateral<br \/>\u2235 Arc BAD subtends \u2220BOD is the angle at the centre 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\" \/><br \/>\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\" \/><br \/>Solution:<br \/>Given : ABCD is a cyclic quadrilateral in which AD || BC<br \/><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 cointerior 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 centre 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 centre 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 a.cyclic quadrilateral,<br \/>\u2234 \u2220APC + \u2220ABC = 180\u00b0<br \/>\u21d2 50\u00b0 + \u2220ABC = 180\u00b0 \u21d2 \u2220ABC =180\u00b0- 50\u00b0<br \/>\u2234 \u2220ABC =130\u00b0<br \/>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 diameiers of a circle with centre 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\" \/><br \/>\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)<br \/>\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 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\" \/><br \/>\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\" \/><br \/>\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<br \/>\u2234 \u2220A + \u2220D = 180\u00b0 (Sum of cointerior 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\" \/><br \/>\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 centre 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:<br \/>In the figure, O is the centre 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<br \/>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:<br \/>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\" \/><br \/>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<br \/><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\" \/><br \/>\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 centre 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<br \/>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<br \/>Arc AC subtends \u2220AOB at the centre 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:<br \/>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 = its interior opposite \u2220BAD<br \/>\u21d2 x = 78\u00b0<br \/>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)<br \/><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\" \/><br \/>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 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.<br \/>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\" \/><br \/>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<br \/>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<br \/>But \u2220DAC = \u2220BCA (Alternate angles)<br \/>\u2234 \u2220DAC = 60\u00b0<br \/>(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\" \/><br \/>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<br \/>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<br \/>(iii) In the figure, ABCD is a cyclic quadrilateral BD 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\" \/><br \/>\u2220A + \u2220C = 180\u00b0 (Sum of opposite angles of cyclic quad.)<br \/>\u2220A+ 100\u00b0= 180\u00b0<br \/>\u21d2 \u2220A= 180\u00b0- 100\u00b0<br \/>\u2234 \u2220A = 80\u00b0<br \/>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<br \/><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\" \/><br \/>To prove : The circles pass through the point of intersection of the diagonals of the rhombus ABCD<br \/>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<br \/>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.<br \/>Solution:<br \/>Given : In cyclic quadrilateral ABCD, AB = CD<br \/>AC and BD are the diagonals<br \/><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\" \/><br \/>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 are 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:<br \/>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\" \/><br \/>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 drawn on two sides of a triangle 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 other three angles of the trapezium.<br \/>Solution:<br \/>In the figure, ABCD is a trapezium in which AD || BC and \u2220B = 70\u00b0<br \/><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 cointerior angles)<br \/>\u21d2 \u2220A + 70\u00b0 = 180\u00b0<br \/>\u21d2 \u2220A= 180\u00b0- 70\u00b0 = 110\u00b0<br \/>\u2234 \u2220A = 110\u00b0<br \/>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:<br \/>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<br \/>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\" \/><br \/>To prove : The perpendicular bisectors of the sides are concurrent<br \/>Proof : \u2235 Each side of the cyclic quadrilateral is a chord of the circle and perpendicular of a chord passes through the centre of the circle<br \/>Hence the perpendicular bisectors of each side will pass through the centre O<br \/>Hence the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent<\/p>\n<p>Question 25.<br \/>Prove that the centre 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 centre of the circle.<br \/>Proof : Let O be the centre 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<br \/>\u2234 The diameters of the circle pass through the centre<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 : \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 centre and \u2220ACB at the remaining part of the circle \u2234 \u2220ACB =\u00a0<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 centre O and \u2220ACB is described in it<br \/>To prove : \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 centre 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 rijght 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 mid point 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 centre D and AC as 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 mid point 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<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-book-chapter-15-areas-of-parallelograms-and-triangles-vsaqs\"><\/span>RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If \u2220APB = 70\u00b0, find \u2220ACB.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/44731145945_c4583f6b09_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"266\" height=\"216\" \/><br \/>Solution:<br \/>Arc AB subtends \u2220AOB at the centre and \u2220APB at the remaining part of the circle<br \/>\u2234 \u2220AOB = 2\u2220APB = 2 x 70\u00b0 = 140\u00b0<br \/>Now in cyclic quadrilateral AOBC,<br \/>\u2220AOB + \u2220ACB = 180\u00b0 (Sum of the angles)<br \/>\u21d2 140\u00b0 +\u2220ACB = 180\u00b0<br \/>\u21d2 \u2220ACB = 180\u00b0 \u2013 140\u00b0 = 40\u00b0<br \/>\u2234 \u2220ACB = 40\u00b0<\/p>\n<p>Question 2.<br \/>In the figure, two congruent circles with centre O and O\u2019 intersect at A and B. If \u2220AO\u2019B = 50\u00b0, then find \u2220APB.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/44920315834_ef39646850_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"298\" height=\"164\" \/><br \/>Solution:<br \/>Two congruent circles with centres O and O\u2019 intersect at A and B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/43827190630_fcd1d9ac74_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"300\" height=\"163\" \/><br \/>\u2220AO\u2019B = 50\u00b0<br \/>\u2235 OA = OB = O\u2019A = 04B (Radii of the congruent circles)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44731145665_f0cbc35c3b_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"353\" height=\"208\" \/><\/p>\n<p>Question 3.<br \/>In the figure, ABCD is a cyclic quadrilateral in which \u2220BAD = 75\u00b0, \u2220ABD = 58\u00b0 and \u2220ADC = IT, AC and BD intersect at P. Then, find \u2220DPC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1913\/43827189940_e8b4cdcabb_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"183\" height=\"169\" \/><br \/>Solution:<br \/>\u2235 ABCD is a cyclic quadrilateral,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1941\/44731145335_7fb898c2ab_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"189\" height=\"165\" \/><br \/>\u2234 \u2220BAD + \u2220BCD = 180\u00b0<br \/>\u21d2 75\u00b0 + \u2220BCD \u2013 180\u00b0<br \/>\u21d2 \u2220BCD = 180\u00b0-75\u00b0= 105\u00b0 and \u2220ADC + \u2220ABC = 180\u00b0<br \/>\u21d2 77\u00b0 + \u2220ABC = 180\u00b0<br \/>\u21d2 \u2220ABC = 180\u00b0-77\u00b0= 103\u00b0<br \/>\u2234 \u2220DBC = \u2220ABC \u2013 \u2220ABD = 103\u00b0 \u2013 58\u00b0 = 45\u00b0<br \/>\u2235 Arc AD subtends \u2220ABD and \u2220ACD in the same segment of the circle 3<br \/>\u2234 \u2220ABD = \u2220ACD = 58\u00b0<br \/>\u2234 \u2220ACB = \u2220BCD \u2013 \u2220ACD = 105\u00b0 \u2013 58\u00b0 = 47\u00b0<br \/>Now in \u2206PBC,<br \/>Ext. \u2220DPC = \u2220PBC + \u2220PCB<br \/>=\u2220DBC + \u2220ACB = 45\u00b0 + 47\u00b0 = 92\u00b0<br \/>Hence \u2220DPC = 92\u00b0<\/p>\n<p>Question 4.<br \/>In the figure, if \u2220AOB = 80\u00b0 and \u2220ABC = 30\u00b0, then find \u2220CAO.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1976\/43827189720_6c20a2d831_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"182\" height=\"153\" \/><br \/>Solution:<br \/>In the figure, \u2220AOB = 80\u00b0, \u2220ABC = 30\u00b0<br \/>\u2235 Arc AB subtends \u2220AOB at the centre and<br \/>\u2220ACB at the remaining part of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1909\/44731145115_8fec7e5c05_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"192\" height=\"165\" \/><br \/>\u2234 \u2220ACB =\u00a0<span id=\"MathJax-Element-44-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-264\" class=\"math\"><span id=\"MathJax-Span-265\" class=\"mrow\"><span id=\"MathJax-Span-266\" class=\"mfrac\"><span id=\"MathJax-Span-267\" class=\"mn\">1<\/span><span id=\"MathJax-Span-268\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB =\u00a0<span id=\"MathJax-Element-45-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-269\" class=\"math\"><span id=\"MathJax-Span-270\" class=\"mrow\"><span id=\"MathJax-Span-271\" class=\"mfrac\"><span id=\"MathJax-Span-272\" class=\"mn\">1<\/span><span id=\"MathJax-Span-273\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 80\u00b0 = 40\u00b0<br \/>In \u2206OAB, OA = OB<br \/>\u2234 \u2220OAB = \u2220OBA<br \/>But \u2220OAB + \u2220OBA + \u2220AOB = 180\u00b0<br \/>\u2234 \u2220OAB + \u2220OBA + 80\u00b0 = 180\u00b0<br \/>\u21d2 \u2220OAB + \u2220OAB = 180\u00b0 \u2013 80\u00b0 = 100\u00b0<br \/>\u2234 2\u2220OAB = 100\u00b0<br \/>\u21d2 \u2220OAB =\u00a0<span id=\"MathJax-Element-46-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-274\" class=\"math\"><span id=\"MathJax-Span-275\" class=\"mrow\"><span id=\"MathJax-Span-276\" class=\"mfrac\"><span id=\"MathJax-Span-277\" class=\"msubsup\"><span id=\"MathJax-Span-278\" class=\"texatom\"><span id=\"MathJax-Span-279\" class=\"mrow\"><span id=\"MathJax-Span-280\" class=\"mn\">100<\/span><\/span><\/span><span id=\"MathJax-Span-281\" class=\"texatom\"><span id=\"MathJax-Span-282\" class=\"mrow\"><span id=\"MathJax-Span-283\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-284\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 50\u00b0<br \/>Similarly, in \u2206ABC,<br \/>\u2220BAC + \u2220ACB + \u2220ABC = 180\u00b0<br \/>\u2220BAC + 40\u00b0 + 30\u00b0 = 180\u00b0<br \/>\u21d2 \u2220BAC = 180\u00b0-30\u00b0-40\u00b0<br \/>= 180\u00b0-70\u00b0= 110\u00b0<br \/>\u2234 \u2220CAO = \u2220BAC \u2013 \u2220OAB<br \/>= 110\u00b0-50\u00b0 = 60\u00b0<\/p>\n<p>Question 5.<br \/>In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find \u2220BCD : \u2220ABE.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44731144935_20c251b023_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"218\" height=\"179\" \/><br \/>Solution:<br \/>In the figure, ABCD is a parallelogram and<br \/>CDE is a straight line<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/45594772772_2fddd815d3_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"221\" height=\"172\" \/><br \/>\u2235 ABCD is a ||gm<br \/>\u2234 \u2220A = \u2220C<br \/>and \u2220C = \u2220ADE (Corresponding angles)<br \/>\u21d2 \u2220BCD = \u2220ADE<br \/>Similarly, \u2220ABE = \u2220BED (Alternate angles)<br \/>\u2235 arc BD subtends \u2220BAD at the centre and<br \/>\u2220BED at the remaining part of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/43827188980_00124960a4_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"353\" height=\"235\" \/><\/p>\n<p>Question 6.<br \/>In the figure, AB is a diameter of the circle such that \u2220A = 35\u00b0 and \u2220Q = 25\u00b0, find \u2220PBR.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1977\/43827188550_98dc28c3d8_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"300\" height=\"166\" \/><br \/>Solution:<br \/>In the figure, AB is the diameter of the circle such that \u2220A = 35\u00b0 and \u2220Q = 25\u00b0, join OP.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/45594772232_62a6c945d6_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"312\" height=\"157\" \/><br \/>Arc PB subtends \u2220POB at the centre and<br \/>\u2220PAB at the remaining part of the circle<br \/>\u2234 \u2220POB = 2\u2220PAB = 2 x 35\u00b0 = 70\u00b0<br \/>Now in \u2206OP,<br \/>OP = OB radii of the circle<br \/>\u2234 \u2220OPB = \u2220OBP = 70\u00b0 (\u2235 \u2220OPB + \u2220OBP = 140\u00b0)<br \/>Now \u2220APB = 90\u00b0 (Angle in a semicircle)<br \/>\u2234 \u2220BPQ = 90\u00b0<br \/>and in \u2206PQB,<br \/>Ext. \u2220PBR = \u2220BPQ + \u2220PQB<br \/>= 90\u00b0 + 25\u00b0= 115\u00b0<br \/>\u2234 \u2220PBR = 115\u00b0<\/p>\n<p>Question 7.<br \/>In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, \u2220BQD =<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1975\/45594772052_3ace7d1fef_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"344\" height=\"226\" \/><br \/>Solution:<br \/>In the figure, P and Q are the centres of two circles which intersect each other at C and B<br \/>ACD is a straight line \u2220APB = 150\u00b0<br \/>Arc AB subtends \u2220APB at the centre and<br \/>\u2220ACB at the remaining part of the circle<br \/>\u2234 \u2220ACB =\u00a0<span id=\"MathJax-Element-47-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-285\" class=\"math\"><span id=\"MathJax-Span-286\" class=\"mrow\"><span id=\"MathJax-Span-287\" class=\"mfrac\"><span id=\"MathJax-Span-288\" class=\"mn\">1<\/span><span id=\"MathJax-Span-289\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220APB =\u00a0<span id=\"MathJax-Element-48-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-290\" class=\"math\"><span id=\"MathJax-Span-291\" class=\"mrow\"><span id=\"MathJax-Span-292\" class=\"mfrac\"><span id=\"MathJax-Span-293\" class=\"mn\">1<\/span><span id=\"MathJax-Span-294\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 150\u00b0 = 75\u00b0<br \/>But \u2220ACB + \u2220BCD = 180\u00b0 (Linear pair)<br \/>\u21d2 75\u00b0 + \u2220BCD = 180\u00b0<br \/>\u2220BCD = 180\u00b0-75\u00b0= 105\u00b0<br \/>Now arc BD subtends reflex \u2220BQD at the centre and \u2220BCD at the remaining part of the circle<br \/>Reflex \u2220BQD = 2\u2220BCD = 2 x 105\u00b0 = 210\u00b0<br \/>But \u2220BQD + reflex \u2220BQD = 360\u00b0<br \/>\u2234 \u2220BQD+ 210\u00b0 = 360\u00b0<br \/>\u2234 \u2220BQD = 360\u00b0 \u2013 210\u00b0 = 150\u00b0<\/p>\n<p>Question 8.<br \/>In the figure, if O is circumcentre of \u2206ABC then find the value of \u2220OBC + \u2220BAC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/43827188150_2b30d7e0e6_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"176\" height=\"167\" \/><br \/>Solution:<br \/>In the figure, join OC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/43827187910_2657bb0a1b_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"198\" height=\"154\" \/><br \/>\u2235 O is the circumcentre of \u2206ABC<br \/>\u2234 OA = OB = OC<br \/>\u2235 \u2220CAO = 60\u00b0 (Proved)<br \/>\u2234 \u2206OAC is an equilateral triangle<br \/>\u2234 \u2220AOC = 60\u00b0<br \/>Now, \u2220BOC = \u2220BOA + \u2220AOC<br \/>= 80\u00b0 + 60\u00b0 = 140\u00b0<br \/>and in \u2206OBC, OB = OC<br \/>\u2220OCB = \u2220OBC<br \/>But \u2220OCB + \u2220OBC = 180\u00b0 \u2013 \u2220BOC<br \/>= 180\u00b0- 140\u00b0 = 40\u00b0<br \/>\u21d2 \u2220OBC + \u2220OBC = 40\u00b0<br \/>\u2234 \u2220OBC =\u00a0<span id=\"MathJax-Element-49-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-295\" class=\"math\"><span id=\"MathJax-Span-296\" class=\"mrow\"><span id=\"MathJax-Span-297\" class=\"mfrac\"><span id=\"MathJax-Span-298\" class=\"msubsup\"><span id=\"MathJax-Span-299\" class=\"texatom\"><span id=\"MathJax-Span-300\" class=\"mrow\"><span id=\"MathJax-Span-301\" class=\"mn\">40<\/span><\/span><\/span><span id=\"MathJax-Span-302\" class=\"texatom\"><span id=\"MathJax-Span-303\" class=\"mrow\"><span id=\"MathJax-Span-304\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-305\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 20\u00b0<br \/>\u2220BAC = OAB + \u2220OAC = 50\u00b0 + 60\u00b0 = 110\u00b0<br \/>\u2234 \u2220OBC + \u2220BAC = 20\u00b0 + 110\u00b0 = 130\u00b0<\/p>\n<p>Question 9.<br \/>In the AOC is a diameter of the circle and arc AXB = 1\/2 arc BYC. Find \u2220BOC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/45594771402_cd68142a4d_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"204\" height=\"156\" \/><br \/>Solution:<br \/>In the figure, AOC is diameter arc AxB =\u00a0<span id=\"MathJax-Element-50-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-306\" class=\"math\"><span id=\"MathJax-Span-307\" class=\"mrow\"><span id=\"MathJax-Span-308\" class=\"mfrac\"><span id=\"MathJax-Span-309\" class=\"mn\">1<\/span><span id=\"MathJax-Span-310\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0arc BYC 1<br \/>\u2220AOB =\u00a0<span id=\"MathJax-Element-51-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-311\" class=\"math\"><span id=\"MathJax-Span-312\" class=\"mrow\"><span id=\"MathJax-Span-313\" class=\"mfrac\"><span id=\"MathJax-Span-314\" class=\"mn\">1<\/span><span id=\"MathJax-Span-315\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220BOC<br \/>\u21d2 \u2220BOC = 2\u2220AOB<br \/>But \u2220AOB + \u2220BOC = 180\u00b0<br \/>\u21d2 \u2220AOB + 2\u2220AOB = 180\u00b0<br \/>\u21d2 3 \u2220AOB = 180\u00b0<br \/>\u2234 \u2220AOB =\u00a0<span id=\"MathJax-Element-52-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-316\" class=\"math\"><span id=\"MathJax-Span-317\" class=\"mrow\"><span id=\"MathJax-Span-318\" class=\"mfrac\"><span id=\"MathJax-Span-319\" class=\"msubsup\"><span id=\"MathJax-Span-320\" class=\"texatom\"><span id=\"MathJax-Span-321\" class=\"mrow\"><span id=\"MathJax-Span-322\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-323\" class=\"texatom\"><span id=\"MathJax-Span-324\" class=\"mrow\"><span id=\"MathJax-Span-325\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-326\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0 = 60\u00b0<br \/>\u2234 \u2220BOC = 2 x 60\u00b0 = 120\u00b0<\/p>\n<p>Question 10.<br \/>In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that \u2220AED = 95\u00b0 and \u2220OBA = 30\u00b0. Find \u2220OAC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/45594771302_201d48c7d8_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"258\" height=\"198\" \/><br \/>Solution:<br \/>In the figure, ABCD is a cyclic quadrilateral<br \/>CD is produced to E such that \u2220ADE = 95\u00b0<br \/>O is the centre of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/44920313144_d991650cc8_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"230\" height=\"197\" \/><br \/>\u2235 \u2220ADC + \u2220ADE = 180\u00b0<br \/>\u21d2 \u2220ADC + 95\u00b0 = 180\u00b0<br \/>\u21d2 \u2220ADC = 180\u00b0-95\u00b0 = 85\u00b0<br \/>Now arc ABC subtends \u2220AOC at the centre and \u2220ADC at the remaining part of the circle<br \/>\u2235 \u2220AOC = 2\u2220ADC = 2 x 85\u00b0 = 170\u00b0<br \/>Now in \u2206OAC,<br \/>\u2220OAC + \u2220OCA + \u2220AOC = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 \u2220OAC = \u2220OCA (\u2235 OA = OC radii of circle)<br \/>\u2234 \u2220OAC + \u2220OAC + 170\u00b0 = 180\u00b0<br \/>2\u2220OAC = 180\u00b0- 170\u00b0= 10\u00b0<br \/>\u2234 \u2220OAC =\u00a0<span id=\"MathJax-Element-53-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-327\" class=\"math\"><span id=\"MathJax-Span-328\" class=\"mrow\"><span id=\"MathJax-Span-329\" class=\"mfrac\"><span id=\"MathJax-Span-330\" class=\"msubsup\"><span id=\"MathJax-Span-331\" class=\"texatom\"><span id=\"MathJax-Span-332\" class=\"mrow\"><span id=\"MathJax-Span-333\" class=\"mn\">10<\/span><\/span><\/span><span id=\"MathJax-Span-334\" class=\"texatom\"><span id=\"MathJax-Span-335\" class=\"mrow\"><span id=\"MathJax-Span-336\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-337\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 5\u00b0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-15-areas-of-parallelograms-and-triangles-mcqs\"><\/span>RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is<br \/>(a) 15 cm<br \/>(b) 16 cm<br \/>(c) 17 cm<br \/>(d) 34 cm<br \/>Solution:<br \/>Length of chord AB of circle = 16 cm<br \/>Distance from the centre OL = 15 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/44731149125_be6bdbcfca_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"181\" height=\"162\" \/><br \/>Let OA be the radius, then in right \u2206OAL,<br \/>OA<sup>2<\/sup>\u00a0= OL<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/>16<br \/>= (15)<sup>2<\/sup>\u00a0+\u00a0<span id=\"MathJax-Element-54-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-338\" class=\"math\"><span id=\"MathJax-Span-339\" class=\"mrow\"><span id=\"MathJax-Span-340\" class=\"mfrac\"><span id=\"MathJax-Span-341\" class=\"mn\">16<\/span><span id=\"MathJax-Span-342\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 15<sup>2<\/sup>\u00a0+ 8<sup>2<\/sup><br \/>= 225 + 64 = 289 = (17)<sup>2<\/sup><br \/>\u2234 OA = 17 cm<br \/>Hence radius of the circle = 17 cm (c)<\/p>\n<p>Question 2.<br \/>The radius of a circle Js 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45645337881_53f2daaa61_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"284\" height=\"71\" \/><br \/>Solution:<br \/>Radius of the cirlce (r) = 6 cm<br \/>Perpendicular distance from centre = ?<br \/>Length of chord = 8 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/30704328887_efee965913_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"173\" height=\"162\" \/><br \/>Let AB be chord, OL is the distance<br \/>In right \u2206OAL<br \/>OA<sup>2<\/sup>\u00a0= AL<sup>2<\/sup>\u00a0+ OL<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45645337671_a39943a457_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"349\" height=\"122\" \/><\/p>\n<p>Question 3.<br \/>If O is the centre of a circle of radius r and AB is a chord of the circle at a distance\u00a0<span id=\"MathJax-Element-55-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-343\" class=\"math\"><span id=\"MathJax-Span-344\" class=\"mrow\"><span id=\"MathJax-Span-345\" class=\"mfrac\"><span id=\"MathJax-Span-346\" class=\"mi\">r<\/span><span id=\"MathJax-Span-347\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0from O, then \u2220BAO =<br \/>(a) 60\u00b0<br \/>(b) 45\u00b0<br \/>(c) 30\u00b0<br \/>(d) 15\u00b0<br \/>Solution:<br \/>r is the radius of the circle with centre O<br \/>AB is the chord, at a distance of\u00a0<span id=\"MathJax-Element-56-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-348\" class=\"math\"><span id=\"MathJax-Span-349\" class=\"mrow\"><span id=\"MathJax-Span-350\" class=\"mfrac\"><span id=\"MathJax-Span-351\" class=\"mi\">r<\/span><span id=\"MathJax-Span-352\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0from the centre<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/30704328817_c70c969d7c_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"236\" height=\"278\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44731148765_f705629cf8_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"350\" height=\"127\" \/><\/p>\n<p>Question 4.<br \/>ABCD is a cyclic quadrilateral such that \u2220ADB = 30\u00b0 and \u2220DCA = 80\u00b0, then \u2220DAB=<br \/>(a) 70\u00b0<br \/>(b) 100\u00b0<br \/>(c) 125\u00b0<br \/>(d) 150\u00b0<br \/>Solution:<br \/>ABCD is a cyclic quadrilateral \u2220DCA = 80\u00b0 and \u2220ADB = 30\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/45645337371_7b04fb0f93_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"217\" height=\"199\" \/><br \/>\u2235\u2220ADB = \u2220ACB (Angles in the same segment)<br \/>\u2234 \u2220ACB = 30\u00b0<br \/>\u2234 \u2220BCD = 80\u00b0 + 30\u00b0 = 110\u00b0<br \/>\u2235 ABCD is a cyclic quadrilateral<br \/>\u2234\u2220BAD + \u2220BCD = 180\u00b0<br \/>\u21d2 \u2220BAD + 110\u00b0= 180\u00b0<br \/>\u21d2 \u2220BAD = 180\u00b0- 110\u00b0 = 70\u00b0<br \/>or \u2220DAB = 70\u00b0 (a)<\/p>\n<p>Question 5.<br \/>A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is<br \/>(a) 12 cm<br \/>(b) 14 cm<br \/>(c) 16 cm<br \/>(d) 18 cm<br \/>Solution:<br \/>In a circle AB chord = 14 cm<br \/>and distance from centre OL = 6 cm<br \/>Let r be the radius of the circle, then OA<sup>2<\/sup>\u00a0= AL<sup>2<\/sup>\u00a0+ OL<sup>2<\/sup><br \/>\u21d2 r<sup>2<\/sup>\u00a0= (7)<sup>2<\/sup>\u00a0+ (6)<sup>2<\/sup>\u00a0= 49 + 36 = 85<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1962\/44731148635_9c25055e7b_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"218\" height=\"207\" \/><br \/>In the same circle length of another chord CD = ?<br \/>Distance from centre = 2 cm<br \/>\u2234 r<sup>2<\/sup>\u00a0= OM<sup>2<\/sup>\u00a0+ MD<sup>2<\/sup><br \/>\u21d2 85 = (2)<sup>2<\/sup>\u00a0+ DM<sup>2<\/sup><br \/>\u21d2 85 = 4 + DM<sup>2<\/sup><br \/>\u21d2 DM<sup>2<\/sup>\u00a0= 85-4 = 81 = (9)<sup>2<\/sup><br \/>\u2234 DM = 9<br \/>\u2234 CD = 2 x DM = 2 x 9 = 18 cm<br \/>\u2234Length of another chord = 18 cm (d)<\/p>\n<p>Question 6.<br \/>One chord of a circle is known to be 10 cm. The radius of this circle must be<br \/>(a) 5 cm<br \/>(b) greater than 5 cm<br \/>(c) greater than or equal to 5 cm<br \/>(d) less than 5 cm<br \/>Solution:<br \/>Length of chord of a circle = 10 cm<br \/>Length of radius of the circle greater than half of the chord<br \/>More than\u00a0<span id=\"MathJax-Element-57-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-353\" class=\"math\"><span id=\"MathJax-Span-354\" class=\"mrow\"><span id=\"MathJax-Span-355\" class=\"mfrac\"><span id=\"MathJax-Span-356\" class=\"mn\">10<\/span><span id=\"MathJax-Span-357\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 5 cm (b)<\/p>\n<p>Question 7.<br \/>ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with O as centre and OC as radius. The length of the chord of this circle passing through C and B is<br \/>(a) 3 cm<br \/>(b) 4 cm<br \/>(c) 5 cm<br \/>(d) 6 cm<br \/>Solution:<br \/>In right \u2206ABC, \u2220B = 90\u00b0<br \/>AC = 5 cm, AB = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1948\/44731148615_67b97fd8af_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"200\" height=\"188\" \/><br \/>\u2234 BC<sup>2<\/sup>\u00a0= AC<sup>2\u00a0<\/sup>-AB<sup>2<\/sup><br \/>= 5<sup>2<\/sup>\u00a0\u2013 4<sup>2<\/sup>\u00a0= 25 \u2013 16<br \/>= 9 = (3)<sup>2<\/sup><br \/>\u2234 BC = 3 cm<br \/>\u2234 Length of chord BC = 3 cm (a)<\/p>\n<p>Question 8.<br \/>If AB, BC and CD are equal chords of a circle with O as centre, and AD diameter then \u2220AOB =<br \/>(a) 60\u00b0<br \/>(b) 90\u00b0<br \/>(c) 120\u00b0<br \/>(d) none of these<br \/>Solution:<br \/>In a circle chords AB = BC = CD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/44920319434_87e60f18ba_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"198\" height=\"198\" \/><br \/>O is the centre of the circle<br \/>\u2234 \u2220AOB = cannot be found (d)<\/p>\n<p>Question 9.<br \/>Let C be the mid-point of an arc AB of a circle such that m\u00a0<span id=\"MathJax-Element-58-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-358\" class=\"math\"><span id=\"MathJax-Span-359\" class=\"mrow\"><span id=\"MathJax-Span-360\" class=\"texatom\"><span id=\"MathJax-Span-361\" class=\"mrow\"><span id=\"MathJax-Span-362\" class=\"munderover\"><span id=\"MathJax-Span-363\" class=\"mrow\"><span id=\"MathJax-Span-364\" class=\"mi\">A<\/span><span id=\"MathJax-Span-365\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-366\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= 183\u00b0. If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle lies<br \/>(a) in the interior of S<br \/>(b) in the exterior of S<br \/>(c) on the segment AB<br \/>(d) on AB and bisects AB<br \/>Solution:<br \/><span id=\"MathJax-Element-59-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-367\" class=\"math\"><span id=\"MathJax-Span-368\" class=\"mrow\"><span id=\"MathJax-Span-369\" class=\"texatom\"><span id=\"MathJax-Span-370\" class=\"mrow\"><span id=\"MathJax-Span-371\" class=\"munderover\"><span id=\"MathJax-Span-372\" class=\"mrow\"><span id=\"MathJax-Span-373\" class=\"mi\">A<\/span><span id=\"MathJax-Span-374\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-375\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= 183\u00b0<br \/>\u2234 AB is the diameter of the circle with centre O and C is the mid point of arc AB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44731148335_e3f024b9b2_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"226\" height=\"210\" \/><br \/>Line segment AB = S<br \/>\u2234 Centre will lie on AB (c)<\/p>\n<p>Question 10.<br \/>In a circle, the major arc is 3 ti.nes the minor arc. The corresponding central angles and the degree measures of two arcs are<br \/>(a) 90\u00b0 and 270\u00b0<br \/>(b) 90\u00b0 and 90\u00b0<br \/>(c) 270\u00b0 adn 90\u00b0<br \/>(d) 60\u00b0 and 210\u00b0<br \/>Solution:<br \/>In a circle, major arc is 3 times the minor arc i.e. arc ACB = 3 arc ADB<br \/>\u2234 Reflex \u2220AOB = 3\u2220AOB<br \/>But angle at O = 360\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/44731148235_4ee99f12d7_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"222\" height=\"213\" \/><br \/>and let \u2220AOB = x<br \/>Then reflex \u2220ADB = x<br \/>x + 3x \u2013 360\u00b0<br \/>\u21d2 4x = 360\u00b0<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-60-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-376\" class=\"math\"><span id=\"MathJax-Span-377\" class=\"mrow\"><span id=\"MathJax-Span-378\" class=\"mfrac\"><span id=\"MathJax-Span-379\" class=\"msubsup\"><span id=\"MathJax-Span-380\" class=\"texatom\"><span id=\"MathJax-Span-381\" class=\"mrow\"><span id=\"MathJax-Span-382\" class=\"mn\">62<\/span><\/span><\/span><span id=\"MathJax-Span-383\" class=\"texatom\"><span id=\"MathJax-Span-384\" class=\"mrow\"><span id=\"MathJax-Span-385\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-386\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 90\u00b0<br \/>\u2234 3x = 90\u00b0 x 3 = 270\u00b0<br \/>Here angles are 270\u00b0 and 90\u00b0 (c)<\/p>\n<p>Question 11.<br \/>If A and B are two points on a circle such that m(<span id=\"MathJax-Element-61-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-387\" class=\"math\"><span id=\"MathJax-Span-388\" class=\"mrow\"><span id=\"MathJax-Span-389\" class=\"texatom\"><span id=\"MathJax-Span-390\" class=\"mrow\"><span id=\"MathJax-Span-391\" class=\"munderover\"><span id=\"MathJax-Span-392\" class=\"mrow\"><span id=\"MathJax-Span-393\" class=\"mi\">A<\/span><span id=\"MathJax-Span-394\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-395\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>) = 260\u00b0. A possible value for the angle subtended by arc BA at a point on the circle is<br \/>(a) 100\u00b0<br \/>(b) 75\u00b0<br \/>(c) 50\u00b0<br \/>(d) 25\u00b0<br \/>Solution:<br \/>A and B are two points on the circle such that reflex \u2220AOB = 260\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1908\/44731148115_d75e56e3eb_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"215\" height=\"204\" \/><br \/>\u2234 \u2220AOB = 360\u00b0 \u2013 260\u00b0 = 100\u00b0<br \/>C is a point on the circle<br \/>\u2234 By joining AC and BC,<br \/>\u2220ACB =\u00a0<span id=\"MathJax-Element-62-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-396\" class=\"math\"><span id=\"MathJax-Span-397\" class=\"mrow\"><span id=\"MathJax-Span-398\" class=\"mfrac\"><span id=\"MathJax-Span-399\" class=\"mn\">1<\/span><span id=\"MathJax-Span-400\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB =\u00a0<span id=\"MathJax-Element-63-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-401\" class=\"math\"><span id=\"MathJax-Span-402\" class=\"mrow\"><span id=\"MathJax-Span-403\" class=\"mfrac\"><span id=\"MathJax-Span-404\" class=\"mn\">1<\/span><span id=\"MathJax-Span-405\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 100\u00b0 = 50\u00b0 (c)<\/p>\n<p>Question 12.<br \/>An equilateral triangle ABC is inscribed in a circle with centre O. The measures of \u2220BOC is<br \/>(a) 30\u00b0<br \/>(b) 60\u00b0<br \/>(c) 90\u00b0<br \/>(d) 120\u00b0<br \/>Solution:<br \/>\u2206ABC is an equilateral triangle inscribed in a circle with centre O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1921\/44731148025_3646cccb2a_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"222\" height=\"213\" \/><br \/>\u2234 Measure of \u2220BOC = 2\u2220BAC<br \/>= 2 x 60\u00b0 = 120\u00b0 (d)<\/p>\n<p>Question 13.<br \/>If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a<br \/>(a) rhombus<br \/>(b) rectangle<br \/>(c) parallelogram<br \/>(d) square<br \/>Solution:<br \/>Two diameter of a circle AB and CD intersect each other at right angles<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44731147945_f878cd7255_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"254\" height=\"228\" \/><br \/>AD, DB, BC and CA are joined forming a quad. ABCD.<br \/>\u2235 The diagonals are equal and bisect each other at right angles<br \/>\u2234 ACBD is a square (d)<\/p>\n<p>Question 14.<br \/>In ABC is an arc of a circle and \u2220ABC = 135\u00b0, then the ratio of arc\u00a0<span id=\"MathJax-Element-64-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-406\" class=\"math\"><span id=\"MathJax-Span-407\" class=\"mrow\"><span id=\"MathJax-Span-408\" class=\"texatom\"><span id=\"MathJax-Span-409\" class=\"mrow\"><span id=\"MathJax-Span-410\" class=\"munderover\"><span id=\"MathJax-Span-411\" class=\"mrow\"><span id=\"MathJax-Span-412\" class=\"mi\">A<\/span><span id=\"MathJax-Span-413\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-414\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0to the circumference is<br \/>(a) 1 : 4<br \/>(b) 3 : 4<br \/>(c) 3 : 8<br \/>(d) 1 : 2<br \/>Solution:<br \/>Arc ABC of a circle and \u2220ABC = 135\u00b0<br \/>Join OA and OC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/44920318664_10fdf94989_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"207\" height=\"206\" \/><br \/>\u2234 Angle subtended by arc ABC at the centre = 2 x \u2220ABC = 2 x 135\u00b0 = 270\u00b0<br \/>Angle at the centre of the circle = 360\u00b0<br \/>\u2234 Ratio with circumference = 270\u00b0 : 360\u00b0 = 3:4 (b)<\/p>\n<p>Question 15.<br \/>The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is<br \/>(a) 60\u00b0<br \/>(b) 75\u00b0<br \/>(c) 120\u00b0<br \/>(d) 150\u00b0<br \/>Solution:<br \/>The chord of a circle = radius of the circle In the figure OA = OB = AB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/44731147735_bfaca94baa_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"214\" height=\"189\" \/><br \/>\u2234 \u2220AOB = 60\u00b0<br \/>(Each angle of an equilateral = 60\u00b0) (a)<\/p>\n<p>Question 16.<br \/>PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If \u2220QPR = 67\u00b0 and \u2220SPR = 72\u00b0, then \u2220QRS =<br \/>(a) 41\u00b0<br \/>(b) 23\u00b0<br \/>(c) 67\u00b0<br \/>(d) 18\u00b0<br \/>Solution:<br \/>PQRS is a cyclic quadrilateral with centre O and \u2220QPR = 67\u00b0<br \/>\u2220SPR = 72\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44920318454_e3202dce06_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"211\" height=\"200\" \/><br \/>\u2234 \u2220QPS = 67\u00b0 + 72\u00b0 = 139\u00b0<br \/>\u2235 \u2220QPS + \u2220QRS = 180\u00b0 (Sum of opposite angles of a cyclic quad.)<br \/>\u21d2 139\u00b0 + \u2220QRS = 180\u00b0<br \/>\u21d2 \u2220QRS = 180\u00b0 \u2013 139\u00b0 = 41\u00b0 (a)<\/p>\n<p>Question 17.<br \/>If A, B, C are three points on a circle with centre O such that \u2220AOB = 90\u00b0 and \u2220BOC = 120\u00b0, then \u2220ABC =<br \/>(a) 60\u00b0<br \/>(b) 75\u00b0<br \/>(c) 90\u00b0<br \/>(d) 135\u00b0<br \/>Solution:<br \/>A, B and C are three points on a circle with centre O<br \/>\u2220AOB = 90\u00b0 and \u2220BOC = 120\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44731147665_2d11e20e1c_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"214\" height=\"212\" \/><br \/>\u2234 \u2220AOC = 360\u00b0 \u2013 (120\u00b0 + 90\u00b0)<br \/>= 360\u00b0 -210\u00b0= 150\u00b0<br \/>But \u2220AOC is at the centre made by arc AC and \u2220ABC at the remaining part of the circle<br \/>\u2234 \u2220ABC =\u00a0<span id=\"MathJax-Element-65-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-415\" class=\"math\"><span id=\"MathJax-Span-416\" class=\"mrow\"><span id=\"MathJax-Span-417\" class=\"mfrac\"><span id=\"MathJax-Span-418\" class=\"mn\">1<\/span><span id=\"MathJax-Span-419\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220AOC<br \/>=\u00a0<span id=\"MathJax-Element-66-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-420\" class=\"math\"><span id=\"MathJax-Span-421\" class=\"mrow\"><span id=\"MathJax-Span-422\" class=\"mfrac\"><span id=\"MathJax-Span-423\" class=\"mn\">1<\/span><span id=\"MathJax-Span-424\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 150\u00b0 = 75\u00b0 (b)<\/p>\n<p>Question 18.<br \/>The greatest chord of a circle is called its<br \/>(a) radius<br \/>(b) secant<br \/>(c) diameter<br \/>(d) none of these<br \/>Solution:<br \/>The greatest chord of a circle is called its diameter. (c)<\/p>\n<p>Question 19.<br \/>Angle formed in minor segment of a circle is<br \/>(a) acute<br \/>(b) obtuse<br \/>(c) right angle<br \/>(d) none of these<br \/>Solution:<br \/>The angle formed in minor segment of a circle is obtuse angle. (b)<\/p>\n<p>Question 20.<br \/>Number of circles that can be drawn through three non-collinear points is<br \/>(a) 1<br \/>(b) 0<br \/>(c) 2<br \/>(d) 3<br \/>Solution:<br \/>The number of circles that can pass through three non-collinear points is only one. (a)<\/p>\n<p>Question 21.<br \/>In the figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =<br \/>(a) 45\u00b0<br \/>(b) 60\u00b0<br \/>(c) 75\u00b0<br \/>(d) 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44920318264_e06d621960_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"231\" height=\"204\" \/><br \/>Solution:<br \/>In the circle, AB and CD are two chords which intersect each other at P at right angle i.e. \u2220CPB = 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1901\/44731147545_d6d55e4bbb_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"220\" height=\"208\" \/><br \/>\u2220CAB and \u2220CDB are in the same segment<br \/>\u2234 \u2220CDB = \u2220CAB = x<br \/>Now in \u2206PDB,<br \/>Ext. \u2220CPB = \u2220D + \u2220DBP<br \/>\u21d2 90\u00b0 = x + y (\u2235 CD \u22a5 AB)<br \/>Hence x + y = 90\u00b0 (d)<\/p>\n<p>Question 22.<br \/>In the figure, if \u2220ABC = 45\u00b0, then \u2220AOC=<br \/>(a) 45\u00b0<br \/>(b) 60\u00b0<br \/>(c) 75\u00b0<br \/>(d) 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/44920317984_3af8533993_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"184\" height=\"157\" \/><br \/>Solution:<br \/>\u2235 arc AC subtends<br \/>\u2220AOC at the centre of the circle and \u2220ABC<br \/>at the remaining part of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/44731147405_0c352c52a4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"196\" height=\"163\" \/><br \/>\u2234 \u2220AOC = 2\u2220ABC<br \/>= 2 x 45\u00b0 = 90\u00b0<br \/>Hence \u2220AOC = 90\u00b0 (d)<\/p>\n<p>Question 23.<br \/>In the figure, chords AD and BC intersect each other at right angles at a point P. If \u2220D AB = 35\u00b0, then \u2220ADC =<br \/>(a) 35\u00b0<br \/>(b) 45\u00b0<br \/>(c) 55\u00b0<br \/>(d) 65\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44920317854_428a2229d1_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"186\" height=\"171\" \/><br \/>Solution:<br \/>Two chords AD and BC intersect each other at right angles at P, \u2220DAB = 35\u00b0<br \/>AB and CD are joined<br \/>In \u2206ABP,<br \/>Ext. \u2220APC = \u2220B + \u2220A<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/44731147225_36664d5533_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"171\" height=\"163\" \/><br \/>\u21d2 90\u00b0 = \u2220B + 35\u00b0<br \/>\u2220B = 90\u00b0 \u2013 35\u00b0 = 55\u00b0<br \/>\u2235 \u2220ABC and \u2220ADC are in the same segment<br \/>\u2234 \u2220ADC = \u2220ABC = 55\u00b0 (c)<\/p>\n<p>Question 24.<br \/>In the figure, O is the centre of the circle and \u2220BDC = 42\u00b0. The measure of \u2220ACB is<br \/>(a) 42\u00b0<br \/>(b) 48\u00b0<br \/>(c) 58\u00b0<br \/>(d) 52\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1931\/44731147105_8ac72fc40f_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"203\" height=\"216\" \/><br \/>Solution:<br \/>In the figure, O is the centre of the circle<br \/>\u2220BDC = 42\u00b0<br \/>\u2220ABC = 90\u00b0 (Angle in a semicircle)<br \/>and \u2220BAC and \u2220BDC are in the same segment of the circle.<br \/>\u2234 \u2220BAC = \u2220BDC = 42\u00b0<br \/>Now in \u2206ABC,<br \/>\u2220A + \u2220ABC + \u2220ACB = 180\u00b0 (Sum of angles of a triangle)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1962\/44920317574_5f9956dba3_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"217\" height=\"216\" \/><br \/>\u21d2 42\u00b0 + 90\u00b0 + \u2220ACB = 180\u00b0<br \/>\u21d2 132\u00b0 + \u2220ACB \u2013 180\u00b0<br \/>\u21d2 \u2220ACB = 180\u00b0 \u2013 132\u00b0 = 48\u00b0 (b)<\/p>\n<p>Question 25.<br \/>In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/44731146945_0f87563cc1_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"292\" height=\"94\" \/><br \/>Solution:<br \/>AB and CD are two diameters of a circle with centre O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920317444_d97b1ca7aa_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"325\" height=\"391\" \/><\/p>\n<p>Question 26.<br \/>Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44731146725_9e52e4aa3d_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"291\" height=\"92\" \/><br \/>Solution:<br \/>Two equal circles pass through the centre of the other and intersect each other at A and B<br \/>Let r be the radius of each circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44920317274_3a8c6aa131_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"345\" height=\"534\" \/><\/p>\n<p>Question 27.<br \/>If AB is a chord of a circle, P and Q are the two points on the circle different from A and B,then<br \/>(a) \u2220APB = \u2220AQB<br \/>(b) \u2220APB + \u2220AQB = 180\u00b0 or \u2220APB = \u2220AQB<br \/>(c) \u2220APB + \u2220AQB = 90\u00b0<br \/>(d) \u2220APB + \u2220AQB = 180\u00b0<br \/>Solution:<br \/>AB is chord of a circle,<br \/>P and Q are two points other than from points A and B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/44920317064_988501c844_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"226\" height=\"198\" \/><br \/>\u2235 \u2220APB and \u2220AQB are in the same segment of the circle<br \/>\u2234 \u2220APB = \u2220AQB (a)<\/p>\n<p>Question 28.<br \/>AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44731146465_d3b4af560a_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"302\" height=\"74\" \/><br \/>Solution:<br \/>AB and CD are two parallel chords of a circle with centre O<br \/>Let r be the radius of the circle AB = 6 cm, CD = 12 cm<br \/>and distance between them = 3 cm<br \/>Join OC and OA, LM = 3 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44920316944_3f8b6e5d15_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"227\" height=\"199\" \/><br \/>Let OM = x, then OL = x + 3<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44731146435_cde5a8713b_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"354\" height=\"550\" \/><\/p>\n<p>Question 29.<br \/>In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. This distance between the chords is 23 cm. If the length of one chord is 16 cm then the length of the other is<br \/>(a) 34 cm<br \/>(b) 15 cm<br \/>(c) 23 cm<br \/>(d) 30 cm<br \/>Solution:<br \/>Radius of a circle = 17 cm<br \/>The distance between two parallel chords = 23 cm<br \/>AB || CD and LM = 23 cm<br \/>Join OA and OC,<br \/>\u2234 OA = OC = 17 cm<br \/>Let OL = x, then OM = (23 \u2013 x) cm<br \/>AB = 16 cm<br \/>Now in right \u2206OAL,<br \/>OA<sup>2<\/sup>\u00a0= OL2<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/>\u21d2 (17)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/>\u21d2 289 = x<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/44920316624_d8daf0c782_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"353\" height=\"494\" \/><\/p>\n<p>Question 30.<br \/>In the figure, O is the centre of the circle such that \u2220AOC = 130\u00b0, then \u2220ABC =<br \/>(a) 130\u00b0<br \/>(b) 115\u00b0<br \/>(c) 65\u00b0<br \/>(d) 165\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920316304_eddfca0079_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"215\" height=\"191\" \/><br \/>Solution:<br \/>O is the centre of the circle and \u2220AOC = 130\u00b0<br \/>Reflex \u2220AOC = 360\u00b0 \u2013 130\u00b0 = 230\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1901\/44731146025_fa114dc527_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"226\" height=\"210\" \/><br \/>Now arc ADB subtends \u2220AOC at the centre and \u2220ABC at the remaining part of the circle<br \/>\u2234 \u2220ABC =\u00a0<span id=\"MathJax-Element-67-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-425\" class=\"math\"><span id=\"MathJax-Span-426\" class=\"mrow\"><span id=\"MathJax-Span-427\" class=\"mfrac\"><span id=\"MathJax-Span-428\" class=\"mn\">1<\/span><span id=\"MathJax-Span-429\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>reflex \u2220AOC<br \/>=\u00a0<span id=\"MathJax-Element-68-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-430\" class=\"math\"><span id=\"MathJax-Span-431\" class=\"mrow\"><span id=\"MathJax-Span-432\" class=\"mfrac\"><span id=\"MathJax-Span-433\" class=\"mn\">1<\/span><span id=\"MathJax-Span-434\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 230\u00b0= 115\u00b0 (b)<\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-listing-of-important-topics-in-the-exercise\"><\/span><strong>Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 15 exercise 15a:<\/strong> This exercise includes problems based on the same base and between the same parallels. These solutions assist the students to discover easy ways to solve difficult problems. The students can definitely score good marks in the exams by practising these study material. These solutions are correct &amp; reliable and framed as per the CBSE guidelines.\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 15 exercise 15b:<\/strong> This exercise contains problems based on area axioms &amp; polygon region. The students will study how to calculate Areas of Parallelograms and Triangles with the help of given RD Sharma Solutions. This exercise includes various topics and subtopics such as polygon regions, rectangular region, area axioms, &amp; triangular region. These<\/span> <span style=\"font-weight: 400;\">solutions enable the students to evaluate the areas of weaknesses so that they can improve to secure better marks in the final exam<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 15 exercise 15c<\/strong>: This exercise includes problem-based on mainly parallelograms &amp; triangles. The step by step solution provides valuable help to the students to prepare well for the exams. The students will study how to compute the area of any quadrilateral that includes triangles.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-from-rd-sharma-solutions-class-9-maths-chapter-15\"><\/span><span style=\"font-weight: 400;\">Important Topics from RD Sharma Solutions Class 9 Maths Chapter 15<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">RD Sharma solution for class 9 mathematics chapter 15 <\/span><span style=\"font-weight: 400;\">includes some important topics as listed below:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Introduction of Parallelograms &amp; Triangles<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Problems on the same base &amp; between the same parallels<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Figures on Geometric regions<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Area Axioms<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">A parallelogram on the same base and between the same parallels<\/span><\/li>\n<\/ul>\n<p>This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 15. If you have any doubts regarding the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Clas 9 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-15\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 15<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630753948294\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-15\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 15?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630754025569\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-15\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 15?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630754059697\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-15-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 15\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 9 Maths Chapter 15 &#8211; Areas Of Parallelograms And Triangles: RD Sharma solution for chapter 15 CBSE Class 9 Maths is an important &amp; useful study resource for the students to prepare well for the exam.\u00a0RD Sharma Solutions Class 9 Maths Chapter 15 are created exercise-wise that assist the students to &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 15 &#8211; Areas Of Parallelograms And Triangles (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 15 &#8211; Areas Of Parallelograms And Triangles (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124544,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3081,73335],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63778"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=63778"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63778\/revisions"}],"predecessor-version":[{"id":128878,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63778\/revisions\/128878"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124544"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63778"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63778"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63778"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}