{"id":124757,"date":"2021-09-06T14:32:22","date_gmt":"2021-09-06T09:02:22","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=124757"},"modified":"2023-11-29T10:56:47","modified_gmt":"2023-11-29T05:26:47","slug":"rd-sharma-class-9-solutions-chapter-20-exercise-20-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-20-exercise-20-1\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125585\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-20-Exercise-20.1.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-20-Exercise-20.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-20-Exercise-20.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1:&nbsp;<\/span><\/strong><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">You can download <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\">RD Sharma Class 9 Solutions Chapter 20 Maths<\/a> Exercise 20.1 here. These solutions are created by our experts. They have provided you the accurate answers which will be very beneficial for your class 9 exams. The questions are solved in an organised manner. These solutions are available in free PDF form which you can save easily for long time period.&nbsp;<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 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class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-20-exercise-20-1\/#what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\" title=\"What are the benefits of studying from RD Sharma Solutions Class 12?\">What are the benefits of studying from RD Sharma Solutions Class 12?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-20-exercise-201-pdf\"><\/span>Download&nbsp;<span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"\n{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 PDF&nbsp;<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>&nbsp;<\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/20.1-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/20.1-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-20-exercise-201\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1:&nbsp;<\/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>Slant height of cone (l) = 60 cm<\/p>\n<p>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 vertical height is 12cm. Find the area of the curved surface.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of cone (r) = 5 cm<\/p>\n<p>Height of cone (h) = 12 cm<\/p>\n<p>Find Slant Height of cone (l):<\/p>\n<p>We know, l<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>&nbsp;+ h<sup>2<\/sup><\/p>\n<p>l<sup>2&nbsp;<\/sup>= 5<sup>2&nbsp;<\/sup>+12<sup>2<\/sup><\/p>\n<p>l<sup>2&nbsp;<\/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 area of curved surface is 176 cm<sup>2<\/sup>&nbsp;.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, slant height of the cone is 8 cm.<\/p>\n<p><strong>Question 4: The height of a cone 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>Slant height of cone (l) = 28 cm<\/p>\n<p>We know that, l<sup>2&nbsp;<\/sup>= r<sup>2<\/sup>&nbsp;+ 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, 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 radius 6 cm and height 8 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of cone (r) = 6 cm<\/p>\n<p>Height of cone (h) = 8 cm<\/p>\n<p>Total Surface area of the cone (T.S.A)=?<\/p>\n<p>Find slant height of cone:<\/p>\n<p>We know, l<sup>2&nbsp;<\/sup>= r<sup>2<\/sup>&nbsp;+ 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, 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 base radius 5.25 cm and slant height 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, 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 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 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, 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 be 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, 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>&nbsp;and 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, 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, slant height of 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 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&nbsp;<\/sup>= 9<\/p>\n<p>or x = 3<\/p>\n<p>Therefore, Radius = 4x = 4(3) cm = 12 cm<\/p>\n<p>We have included all the information regarding <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1. If you have any query feel free to ask in the comment section.&nbsp;<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-20-exercise-201\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630918000137\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-class-9-solutions-chapter-20-exercise-201-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630918019271\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-class-9-solutions-chapter-20-exercise-201-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630918022584\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630918023728\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\"><\/span>What are the benefits of studying from RD Sharma Solutions Class 12?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1:&nbsp;You can download RD Sharma Class 9 Solutions Chapter 20 Maths Exercise 20.1 here. These solutions are created by our experts. They have provided you the accurate answers which will be very beneficial for your class 9 exams. The questions are solved in an organised manner. These &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-20-exercise-20-1\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":125585,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[3086,76789,76790],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124757"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=124757"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124757\/revisions"}],"predecessor-version":[{"id":513809,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124757\/revisions\/513809"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125585"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=124757"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=124757"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=124757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}