{"id":72587,"date":"2021-02-08T13:31:54","date_gmt":"2021-02-08T08:01:54","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=72587"},"modified":"2021-02-08T13:31:58","modified_gmt":"2021-02-08T08:01:58","slug":"rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions\/","title":{"rendered":"RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions"},"content":{"rendered":"\n<p>RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions is based on the Volume of a Right Circular Cone, which is concerned under Surface Area And Volume of A Right Circular Cone. The solutions provided here can be useful for the students in preparing their own comfort zone strategically. Along with providing the answers, it can be useful in terms of a better understanding of the concepts.<\/p>\n<p>We have attached the PDF with this article, which helps students practice different types of questions related to the exercise. The PDF is free to access, prepared by our experts with the help of Previous Year\u2019s Question Paper, Text Book, and RD Sharma of Class 9.<\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/\" target=\"_blank\" rel=\"noopener noreferrer\"><strong>Learn about RD Sharma Class 9 Chapter 20 Surface Area And Volume of A Right Circular Cone<\/strong><\/a><\/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-69d16de661b99\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d16de661b99\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions\/#download-rd-sharma-chapter-20-class-9-maths-exercise-202-solutions-pdf\" title=\"Download RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions PDF\">Download RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions 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-chapter-20-class-9-maths-exercise-20-2-solutions\/#important-definitions-rd-sharma-chapter-20-class-9-maths-exercise-202-solutions\" title=\"Important Definitions RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions\">Important Definitions RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions\/#the-volume-of-a-right-circular-cone\" title=\"The volume of a Right Circular Cone\">The volume of a Right Circular Cone<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions\/#examples-of-volume-of-a-circular-cone-of-rd-sharma-chapter-20-class-9-maths-exercise-202-solutions\" title=\"Examples of Volume of a Circular Cone of RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions\">Examples of Volume of a Circular Cone of RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-chapter-20-class-9-maths-exercise-202-solutions-pdf\"><\/span>Download RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/Solutions-for-Class-9-Maths-Chapter-20-Surface-Area-and-Volume-of-A-Right-Circular-Cone-Exercise-20.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/Solutions-for-Class-9-Maths-Chapter-20-Surface-Area-and-Volume-of-A-Right-Circular-Cone-Exercise-20.2.pdf\">Solutions for Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone Exercise 20.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-definitions-rd-sharma-chapter-20-class-9-maths-exercise-202-solutions\"><\/span>Important Definitions RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>In this article, we will discuss the Volume of a Right Circular Cone with proof and examples. Go down for more information on the RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"the-volume-of-a-right-circular-cone\"><\/span><strong>The volume of a Right Circular Cone<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>The volume of a cone is one-third (\u2153) of the outcome of the area of the base and the height of the cone. The volume is measured in the names of cubic units.<\/p>\n<p>The volume of a right circular cone can be determined (calculated) by the following formula-<\/p>\n<p>The volume of a right circular cone = \u2153 (Base area \u00d7 Height)<\/p>\n<p>Where,<\/p>\n<p>Base Area = \u03c0 r2<\/p>\n<p>Hence, Volume = \u2153 \u03c0 r2h<\/p>\n<h3><span class=\"ez-toc-section\" id=\"examples-of-volume-of-a-circular-cone-of-rd-sharma-chapter-20-class-9-maths-exercise-202-solutions\"><\/span><strong>Examples of Volume of a Circular Cone of RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ques- Determine the volume of the right round cone with:<\/strong><\/p>\n<p><strong>(i) Radius 6cm, height 7cm<\/strong><\/p>\n<p><strong>(ii)Radius 3.5cm, height 12cm<\/strong><\/p>\n<p><strong>(iii) Height is 21cm and slant height 28cm<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p><strong>(i)<\/strong> Radius of cone(a)=6cm<\/p>\n<p>Height of cone(b)=7cm<\/p>\n<p>We understand, Volume of a right circular cone = 1\/3 \u03c0a2b<\/p>\n<p>By changing the values, we get<\/p>\n<p>= 1\/3 x 3.14 x 62 x 7<\/p>\n<p>= 264<\/p>\n<p>The volume of the right circular cone is 264 cm3<\/p>\n<p><strong>(ii)<\/strong> Radius of cone(a)=3.5 cm<\/p>\n<p>Height of cone(b)=12cm<\/p>\n<p>The volume of the right circular cone = 1\/3 \u03c0a2b<\/p>\n<p>By changing the values, we get<\/p>\n<p>= 1\/3 x 3.14 x 3.52 x 12<\/p>\n<p>= 154<\/p>\n<p>The volume of the right circular cone is 154 cm3<\/p>\n<p><strong>(iii)<\/strong> Height of cone(a)=21 cm<\/p>\n<p>Slant height of cone(b) = 28 cm<\/p>\n<p>Find the measure of c:<\/p>\n<p>We know, b2 = c2 + a2<\/p>\n<p>282 = c2 + 212<\/p>\n<p>or c = 7\u221a7<\/p>\n<p>Now,<\/p>\n<p>The volume of the right circular cone = 1\/3 \u03c0c2a<\/p>\n<p>By changing the values, we get<\/p>\n<p>= 1\/3 x 3.14 x (7\u221a7)2 x 21<\/p>\n<p>=7546<\/p>\n<p>The volume of the right circular cone is 7546 cm3<\/p>\n<p><strong>Ques- Determine the capacity in liters of a conical vessel with:<\/strong><\/p>\n<p><strong>(i) radius 7 cm, slant height 25 cm<\/strong><\/p>\n<p><strong>(ii) height 12 cm, slant height 13 cm.<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p><strong>(i)<\/strong> Radius of the cone(a) =7 cm<\/p>\n<p>Slant height of the cone (b) =25 cm<\/p>\n<p>As we know that, b2 = a2 + c2<\/p>\n<p>252 = 72 + c2<\/p>\n<p>or c = 24<\/p>\n<p>Now, Volume of the right circular cone = = 1\/3 \u03c0a2c<\/p>\n<p>By changing the values, we get<\/p>\n<p>= 1\/3 x 3.14 x (7)2 x 24<\/p>\n<p>= 1232<\/p>\n<p>The volume of a right circular cone is 1232 cm3 or 1.232 liters.<\/p>\n<p>[1 cm3 = 0.01 liter]<\/p>\n<p><strong>(ii)<\/strong> Height of cone(a)=12 cm<\/p>\n<p>Slant height of cone(b)=13 cm<\/p>\n<p>As we know that, b2 = c2 + a2<\/p>\n<p>132 = c2 + 122<\/p>\n<p>or c = 5<\/p>\n<p>Now, Volume of the right circular cone = 1\/3 \u03c0c2a<\/p>\n<p>By changing the values, we get<\/p>\n<p>= 1\/3 x 3.14 x (5)2 x 12<\/p>\n<p>= 314.28<\/p>\n<p>The volume of a right circular cone is 314.28 cm3 or 0.314 liters.<\/p>\n<p>[1 cm3 = 0.01 liters]<\/p>\n<p><strong>Ques- Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Determine the ratio of their volumes.<\/strong><\/p>\n<p><strong>Sol-<\/strong><\/p>\n<p>Let the heights of the cones are a and 3a and, the radii of their bases are 3b and b, respectively. Then, their volumes are<\/p>\n<p>The volume of the first cone (V1) = 1\/3 \u03c0(3b)2a<\/p>\n<p>Volume of second cone (V2) = 1\/3 \u03c0b2(3a)<\/p>\n<p>Now, V1\/V2 = 3\/1<\/p>\n<p>The ratio of the two volumes is 3:1.<\/p>\n<p><a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener noreferrer\"><strong>Know about CBSE Board<\/strong><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions is based on the Volume of a Right Circular Cone, which is concerned under Surface Area And Volume of A Right Circular Cone. The solutions provided here can be useful for the students in preparing their own comfort zone strategically. Along with providing the answers, &#8230; <a title=\"RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions\/\" aria-label=\"More on RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions\">Read more<\/a><\/p>\n","protected":false},"author":241,"featured_media":72588,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,2917,73719,73411,73410],"tags":[3081,3086,3085],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72587"}],"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\/241"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=72587"}],"version-history":[{"count":3,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72587\/revisions"}],"predecessor-version":[{"id":73966,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72587\/revisions\/73966"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/72588"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=72587"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=72587"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=72587"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}