{"id":63883,"date":"2023-04-05T06:00:00","date_gmt":"2023-04-05T00:30:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63883"},"modified":"2023-12-22T10:35:38","modified_gmt":"2023-12-22T05:05:38","slug":"rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 20 &#8211; Surface Area And Volume of A Right Circular Cone (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124634\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-20-Surface-Area-And-Volume-of-A-Right-Circular-Cone.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 20\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-20-Surface-Area-And-Volume-of-A-Right-Circular-Cone.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-20-Surface-Area-And-Volume-of-A-Right-Circular-Cone-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 9 Maths Chapter 20:\u00a0<\/strong>It<span style=\"font-weight: 400;\"> is the best resource for students to prepare confidently for the final exam. The exercise problems included in this chapter are prepared with in-depth knowledge. These solutions are created according to the weightage assigned in the exam. <\/span><span style=\"font-weight: 400;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions class 9 Maths<\/a> chapter 20 <\/span><span style=\"font-weight: 400;\">\u00a0assist the students with their problem-solving abilities &amp; examine their understanding of this chapter.\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-69d0fe9f0f664\" 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-69d0fe9f0f664\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/#download-rd-sharma-solutions-class-9-maths-chapter-20-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 20 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 20 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-20-surface-area-and-volume-of-a-right-circular-cone\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-20\" title=\"Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 20\">Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 20<\/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-20-surface-area-and-volume-of-a-right-circular-cone\/#access-solutions-of-rd-sharma-solutions-class-9-maths-chapter-20\" title=\"Access solutions of RD Sharma Solutions Class 9 Maths Chapter 20\">Access solutions of RD Sharma Solutions Class 9 Maths Chapter 20<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/#detailed-exercise-wise-explanation-with-listing-of-important-topics\" title=\"Detailed Exercise-wise Explanation with Listing of Important Topics\">Detailed Exercise-wise Explanation with Listing of Important Topics<\/a><ul class='ez-toc-list-level-3'><li class='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-20-surface-area-and-volume-of-a-right-circular-cone\/#rd-sharma-class-9-chapter-20-exercise-20a\" title=\"RD Sharma class 9 chapter 20 exercise 20a:\">RD Sharma class 9 chapter 20 exercise 20a:<\/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-20-surface-area-and-volume-of-a-right-circular-cone\/#rd-sharma-class-9-chapter-20-exercise-20b\" title=\"RD Sharma class 9 chapter 20 exercise 20b:\">RD Sharma class 9 chapter 20 exercise 20b:<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/#important-topics-from-class-9-maths-chapter-20\" title=\"Important Topics from Class 9 Maths Chapter 20\">Important Topics from Class 9 Maths Chapter 20<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-20\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 20\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 20<\/a><ul class='ez-toc-list-level-3'><li class='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-20-surface-area-and-volume-of-a-right-circular-cone\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-20\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 20?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 20?<\/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-20-surface-area-and-volume-of-a-right-circular-cone\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-20\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 20?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 20?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-20-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 20\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 20\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-20-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 20 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/20-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 20<\/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\/20-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-20\"><\/span><strong>Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 20<\/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-20-class-9-maths-exercise-20-1-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 20 exercise 20a<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-20-class-9-maths-exercise-20-2-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 20 exercise 20b<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-solutions-of-rd-sharma-solutions-class-9-maths-chapter-20\"><\/span><strong>Access solutions of RD Sharma Solutions Class 9 Maths Chapter 20<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1:\u00a0<\/strong>Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.<\/p>\n<p>Solution:<\/p>\n<p>The slant height of cone (l) = 60 cm<\/p>\n<p>The radius of the base of the cone (r) = 21 cm<\/p>\n<p>Now,<\/p>\n<p>Curved surface area of the right circular cone = \u03c0rl = 22\/7 x 21 x 60 = 3960 cm<sup>2<\/sup><\/p>\n<p>Therefore the curved surface area of the right circular cone is 3960 cm<sup>2<\/sup><\/p>\n<p><strong>Question 2: The radius of a cone is 5cm and the vertical height is 12cm. Find the area of the curved surface.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The radius of cone (r) = 5 cm<\/p>\n<p>Height of cone (h) = 12 cm<\/p>\n<p>Find the Slant Height of cone (l):<\/p>\n<p>We know, l<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>l<sup>2\u00a0<\/sup>= 5<sup>2\u00a0<\/sup>+12<sup>2<\/sup><\/p>\n<p>l<sup>2\u00a0<\/sup>= 25 + 144 = 169<\/p>\n<p>Or l = 13 cm<\/p>\n<p>Now,<\/p>\n<p>C.S.A = \u03c0rl =3.14 x 5 x 13 = 204.28<\/p>\n<p>Therefore, the curved surface area of the cone is 204.28 cm<sup>2<\/sup><\/p>\n<p><strong>Question 3: The radius of a cone is 7 cm and the area of the curved surface is 176 cm<sup>2<\/sup>\u00a0.Find the slant height.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of cone(r) = 7 cm<\/p>\n<p>Curved surface area(C.S.A)= 176cm<sup>2<\/sup><\/p>\n<p>We know, C.S.A. = \u03c0rl<\/p>\n<p>\u21d2\u03c0rl = 176<\/p>\n<p>\u21d2 22\/7 x 7 x l = 176<\/p>\n<p>or l = 8<\/p>\n<p>Therefore, the slant height of the cone is 8 cm.<\/p>\n<p><strong>Question 4: The height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Height of cone(h) = 21 cm<\/p>\n<p>The slant height of cone (l) = 28 cm<\/p>\n<p>We know that, l<sup>2\u00a0<\/sup>= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>28<sup>2<\/sup>=r<sup>2<\/sup>+21<sup>2<\/sup><\/p>\n<p>r<sup>2<\/sup>=28<sup>2<\/sup>\u221221<sup>2<\/sup><\/p>\n<p>or r= 7\u221a7 cm<\/p>\n<p>Now,<\/p>\n<p>Area of the circular base = \u03c0r<sup>2<\/sup><\/p>\n<p>= 22\/7 x (7\u221a7 )<sup>2<\/sup><\/p>\n<p>=1078<\/p>\n<p>Therefore, the area of the base is 1078 cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 5: Find the total surface area of a right circular cone with a radius 6 cm and a height of 8 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The radius of cone (r) = 6 cm<\/p>\n<p>Height of cone (h) = 8 cm<\/p>\n<p>The total Surface area of the cone (T.S.A)=?<\/p>\n<p>Find the slant height of the cone:<\/p>\n<p>We know, l<sup>2\u00a0<\/sup>= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>=6<sup>2<\/sup>+8<sup>2<\/sup><\/p>\n<p>= 36 + 64<\/p>\n<p>= 100<\/p>\n<p>or l = 10 cm<\/p>\n<p>Now,<\/p>\n<p>Total Surface area of the cone (T.S.A) = Curved surface area of cone + Area of circular base<\/p>\n<p>= \u03c0rl + \u03c0r<sup>2<\/sup><\/p>\n<p>= (22\/7 x 6 x 10) + (22\/7 x 6 x 6)<\/p>\n<p>= 1320\/7 + 792\/7<\/p>\n<p>= 301.71<\/p>\n<p>Therefore, the area of the base is 301.71cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 6: Find the curved surface area of a cone with a base radius of 5.25 cm and a slant height of 10 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Base radius of the cone(r) = 5.25 cm<\/p>\n<p>Slant height of the cone(l) = 10 cm<\/p>\n<p>Curved surface area (C.S.A) = \u03c0rl<\/p>\n<p>=22\/7 x 5.25 x 10<\/p>\n<p>= 165<\/p>\n<p>Therefore, the curved surface area of the cone is 165cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 7: Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Diameter of the cone(d)=24 m<\/p>\n<p>So, radius of the cone(r)= diameter\/ 2 = 24\/2 m = 12m<\/p>\n<p>Slant height of the cone(l) = 21 m<\/p>\n<p>T.S.A = Curved surface area of cone + Area of the circular base<\/p>\n<p>= \u03c0rl+ \u03c0r<sup>2<\/sup><\/p>\n<p>= (22\/7 x 12 x 21) + (22\/7 x 12 x 12)<\/p>\n<p>= 1244.57<\/p>\n<p>Therefore, the total surface area of the cone is 1244.57 m<sup>2<\/sup>.<\/p>\n<p><strong>Question 8: The area of the curved surface of a cone is 60 \u03c0 cm<sup>2<\/sup>. If the slant height of the cone is 8 cm, find the radius of the base.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Curved surface area(C.S.A)= 60 \u03c0 cm<sup>2<\/sup><\/p>\n<p>Slant height of the cone(l) = 8 cm<\/p>\n<p>We know, Curved surface area(C.S.A )=\u03c0rl<\/p>\n<p>\u21d2 \u03c0rl = 60 \u03c0<\/p>\n<p>\u21d2 r x 8 = 60<\/p>\n<p>or r = 60\/8 = 7.5<\/p>\n<p>Therefore, the radius of the base of the cone is 7.5 cm.<\/p>\n<p><strong>Question 9: The curved surface area of a cone is 4070 cm<sup>2<\/sup> and the diameter is 70 cm . What is its slant height? (Use \u03c0 =22\/7)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Diameter of the cone(d) = 70 cm<\/p>\n<p>So, radius of the cone(r)= diameter\/2 = 70\/2 cm = 35 cm<\/p>\n<p>Curved surface area = 4070 cm<sup>2<\/sup><\/p>\n<p>Now,<\/p>\n<p>We know that curved surface area = \u03c0rl<\/p>\n<p>So, \u03c0rl = 4070<\/p>\n<p>By substituting the values, we get<\/p>\n<p>22\/7 x 35 x l = 4070<\/p>\n<p>or l = 37<\/p>\n<p>Therefore, the slant height of the cone is 37 cm.<\/p>\n<p><strong>Question 10: The radius and slant height of a cone are in the ratio 4:7. If its curved surface area is 792 cm<sup>2<\/sup>, find its radius. (Use \u03c0 =22\/7)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Curved surface area = 792 cm<sup>2<\/sup><\/p>\n<p>The radius and slant height of a cone are in the ratio 4:7 (Given)<\/p>\n<p>Let 4x be the radius and 7x be the height of the cone.<\/p>\n<p>Now,<\/p>\n<p>Curved surface area (C.S.A.) = \u03c0rl<\/p>\n<p>So, 22\/7 x (4x) x (7x) = 792<\/p>\n<p>or x<sup>2\u00a0<\/sup>= 9<\/p>\n<p>or x = 3<\/p>\n<p>Therefore, Radius = 4x = 4(3) cm = 12 cm<\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-listing-of-important-topics\"><\/span><strong>Detailed Exercise-wise Explanation with Listing of Important Topics<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-20-exercise-20a\"><\/span>RD Sharma class 9 chapter 20 exercise 20a:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">This exercise includes topics related to the right circular cone &amp; surface area of a right circular cone. RD Sharma class 9 chapter 20 exercise 20a assists the students in understanding each topic in a simplified way. These solutions enable them to learn online as well as offline in an efficient &amp; simplified way.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-20-exercise-20b\"><\/span>RD Sharma class 9 chapter 20 exercise 20b:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">This exercise includes a topic based on the volume of a right circular cone. These solutions are the best study material to practice the concepts of this chapter to get excellent marks in the exams. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 20 exercise 20b enables the students to improve their question-solving efficiency by considering these solutions. The students can develop problem-solving abilities by practicing exercise-wise problems regularly.\u00a0<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-from-class-9-maths-chapter-20\"><\/span><span style=\"font-weight: 400;\">Important Topics from Class 9 Maths Chapter 20<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 9 Maths Chapter 20 <span style=\"font-weight: 400;\">covers some important concepts that are listed below:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Introduction of a right circular cone<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">About vertex, axis, radius, base, &amp; slant height.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Calculate the surface area of a right circular cone<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Find the volume of a right circular cone<\/span><\/li>\n<\/ul>\n<p>This is a complete blog on RD Sharma Solutions Class 9 Maths Chapter 20. For more doubts regarding the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 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-20\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 20<\/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-1630759038770\" 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-20\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 20?<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-1630759063169\" 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-20\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 20?<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-1630759078035\" 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-20-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 20\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 20:\u00a0It is the best resource for students to prepare confidently for the final exam. The exercise problems included in this chapter are prepared with in-depth knowledge. These solutions are created according to the weightage assigned in the exam. RD Sharma Solutions class 9 Maths chapter 20 \u00a0assist the &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 20 &#8211; Surface Area And Volume of A Right Circular Cone (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-20-surface-area-and-volume-of-a-right-circular-cone\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 20 &#8211; Surface Area And Volume of A Right Circular Cone (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124634,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63883"}],"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=63883"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63883\/revisions"}],"predecessor-version":[{"id":525560,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63883\/revisions\/525560"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124634"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63883"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63883"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}