{"id":124763,"date":"2021-09-06T14:41:48","date_gmt":"2021-09-06T09:11:48","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=124763"},"modified":"2021-09-08T18:26:36","modified_gmt":"2021-09-08T12:56:36","slug":"rd-sharma-class-9-solutions-chapter-20-exercise-20-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-20-exercise-20-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125597\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-20-Exercise-20.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-20-Exercise-20.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-20-Exercise-20.2-768x432.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 20.2&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 20.2 s<\/span><\/strong>olutions are provided here. These solutions are prepared by our experts and provide accurate answers to all the practice questions that students can refer to and prepare to solve during the exam. The chapter of <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<\/a> exercise 20.2 shows students many problems related to the surface area of \u200b\u200bstraight cones.<\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-6a33a9479585d\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-20-exercise-20-2\/#can-i-open-rd-sharma-class-9-solutions-chapter-20-exercise-202-pdf-on-my-smartphone\" title=\"Can I open RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 PDF on my smartphone?\">Can I open RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 PDF on my smartphone?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-20-exercise-202-pdf\"><\/span>Download\u00a0<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 20.2 PDF\u00a0<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>\u00a0<\/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\/20.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/20.2.pdf\">RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-20-exercise-202\"><\/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 20.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: Find the volume of the right circular 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>\u00a0Radius of cone(r)=6cm<\/p>\n<p>Height of cone(h)=7cm<\/p>\n<p>We know, Volume of a right circular cone = 1\/3 \u03c0r<sup>2<\/sup>h<\/p>\n<p>By substituting the values, we get<\/p>\n<p>= 1\/3 x 3.14 x 6<sup>2<\/sup>\u00a0x 7<\/p>\n<p>= 264<\/p>\n<p>Volume of a right circular cone is 264 cm<sup>3<\/sup><\/p>\n<p><strong>(ii)<\/strong>\u00a0Radius of cone(r)=3.5 cm<\/p>\n<p>Height of cone(h)=12cm<\/p>\n<p>Volume of a right circular cone = 1\/3 \u03c0r<sup>2<\/sup>h<\/p>\n<p>By substituting the values, we get<\/p>\n<p>= 1\/3 x 3.14 x 3.5<sup>2<\/sup>\u00a0x 12<\/p>\n<p>=154<\/p>\n<p>Volume of a right circular cone is 154 cm<sup>3<\/sup><\/p>\n<p><strong>(iii)<\/strong>\u00a0Height of cone(h)=21 cm<\/p>\n<p>Slant height of cone(l) = 28 cm<\/p>\n<p>Find the measure of r:<\/p>\n<p>We know, l<sup>2\u00a0<\/sup>= r<sup>2\u00a0<\/sup>+ h<sup>2<\/sup><\/p>\n<p>28<sup>2<\/sup>\u00a0= r<sup>2\u00a0<\/sup>+ 21<sup>2<\/sup><\/p>\n<p>or r = 7\u221a7<\/p>\n<p>Now,<\/p>\n<p>Volume of a right circular cone = 1\/3 \u03c0r<sup>2<\/sup>h<\/p>\n<p>By substituting the values, we get<\/p>\n<p>= 1\/3 x 3.14 x (7\u221a7)<sup>2<\/sup>\u00a0x 21<\/p>\n<p>=7546<\/p>\n<p>Volume of a right circular cone is 7546 cm<sup>3<\/sup><\/p>\n<p><strong>Question 2: Find the capacity in litres 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>\u00a0Radius of the cone(r) =7 cm<\/p>\n<p>Slant height of the cone (l) =25 cm<\/p>\n<p>As we know that, l<sup>2\u00a0<\/sup>= r<sup>2\u00a0<\/sup>+ h<sup>2<\/sup><\/p>\n<p>25<sup>2<\/sup>\u00a0= 7<sup>2\u00a0<\/sup>+ h<sup>2<\/sup><\/p>\n<p>or h = 24<\/p>\n<p>Now, Volume of a right circular cone = = 1\/3 \u03c0r<sup>2<\/sup>h<\/p>\n<p>By substituting the values, we get<\/p>\n<p>= 1\/3 x 3.14 x (7)<sup>2<\/sup>\u00a0x 24<\/p>\n<p>= 1232<\/p>\n<p>Volume of a right circular cone is 1232 cm<sup>3<\/sup>\u00a0or 1.232 litres<\/p>\n<p>[1 cm<sup>3<\/sup>\u00a0= 0.01 liter]<\/p>\n<p><strong>(ii)<\/strong>\u00a0Height of cone(h)=12 cm<\/p>\n<p>Slant height of cone(l)=13 cm<\/p>\n<p>As we know that, l<sup>2\u00a0<\/sup>= r<sup>2\u00a0<\/sup>+ h<sup>2<\/sup><\/p>\n<p>13<sup>2<\/sup>\u00a0= r<sup>2\u00a0<\/sup>+ 12<sup>2<\/sup><\/p>\n<p>or r = 5<\/p>\n<p>Now, Volume of a right circular cone = 1\/3 \u03c0r<sup>2<\/sup>h<\/p>\n<p>By substituting the values, we get<\/p>\n<p>= 1\/3 x 3.14 x (5)<sup>2<\/sup>\u00a0x 12<\/p>\n<p>= 314.28<\/p>\n<p>Volume of a right circular cone is 314.28 cm<sup>3<\/sup>\u00a0or 0.314 litres.<\/p>\n<p>[1 cm<sup>3\u00a0<\/sup>= 0.01 liters]<\/p>\n<p><strong>Question 3: Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the heights of the cones be h and 3h and radii of their bases be 3r and r respectively. Then, their volumes are<\/p>\n<p>Volume of first cone (V1) = 1\/3 \u03c0(3r)<sup>2<\/sup>h<\/p>\n<p>Volume of second cone (V2) = 1\/3 \u03c0r<sup>2<\/sup>(3h)<\/p>\n<p>Now, V1\/V2 = 3\/1<\/p>\n<p>Ratio of two volumes is 3:1.<\/p>\n<p><strong>Question 4: The radius and the height of a right circular cone are in the ratio 5:12. If its volume is 314 cubic meter, find the slant height and the radius. (Use \u03c0=3.14).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us assume the ratio of radius and the height of a right circular cone to be x.<\/p>\n<p>Then, radius be 5x and height be 12x<\/p>\n<p>We know, l<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>= (5x)<sup>\u00a02<\/sup>\u00a0+ (12x)<sup>2<\/sup><\/p>\n<p>= 25 x<sup>2<\/sup>\u00a0+ 144 x<sup>2<\/sup><\/p>\n<p>or l = 13x<\/p>\n<p>Therefore, slant height is 13 m.<\/p>\n<p>Now it is given that volume of cone = 314 m<sup>3<\/sup><\/p>\n<p>\u21d2 1\/3\u03c0r<sup>2<\/sup>h = 314<\/p>\n<p>\u21d2 1\/3 x 3.14 x (25x<sup>2<\/sup>\u00a0) x (12x) = 314<\/p>\n<p>\u21d2 x<sup>3<\/sup>=1<\/p>\n<p>or x = 1<\/p>\n<p>So, radius = 5x 1 = 5 m<\/p>\n<p>Therefore ,<\/p>\n<p>Answer: Slant height = 13m<\/p>\n<p>Radius = 5m<\/p>\n<p><strong>Question 5: The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use \u03c0=3.14).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the ratio of radius and height of a right circular cone be y.<\/p>\n<p>Radius of cone(r) = 5y<\/p>\n<p>Height of cone (h) =12y<\/p>\n<p>Now we know, l<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>= (5y)<sup>\u00a02<\/sup>\u00a0+ (12y)<sup>2<\/sup><\/p>\n<p>= 25 y<sup>2<\/sup>\u00a0+ 144 y<sup>2<\/sup><\/p>\n<p>or l = 13y<\/p>\n<p>Now, volume of the cone is given 2512cm<sup>3<\/sup><\/p>\n<p>\u21d2 1\/3\u03c0r<sup>2<\/sup>h=2512<\/p>\n<p>\u21d2 1\/3 x 3.14 x (5y)<sup>2<\/sup>\u00a0x 12y = 2512<\/p>\n<p>\u21d2 y<sup>3<\/sup>\u00a0= (2512 x 3)\/(3.14 x 25 x 12) = 8<\/p>\n<p>or y = 2<\/p>\n<p>Therefore,<\/p>\n<p>Radius of cone = 5y = 5\u00d72 = 10cm<\/p>\n<p>Slant height (l) =13y = 13\u00d72 = 26cm<\/p>\n<p><strong>Question 6: The ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2 : 3. Find the ratio of their vertical heights.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the ratio of the radius be x and ratio of the volume be y.<\/p>\n<p>Then, Radius of 1st cone (r<sub>1<\/sub>) =2x<\/p>\n<p>Radius of 2nd cone (r<sub>2<\/sub>) =3x<\/p>\n<p>Volume of 1st cone (V<sub>1<\/sub>)= 4y<\/p>\n<p>Volume of 2nd cone (V<sub>2<\/sub>)= 5y<\/p>\n<p>We know formula for volume of a cone = 1\/3\u03c0r<sup>2<\/sup>h<\/p>\n<p>Let h<sub>1<\/sub>\u00a0and h<sub>2<\/sub>\u00a0be the heights of respective cones.<\/p>\n<p><img class=\"alignnone size-full wp-image-124781\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/rd-sharma-class-9-maths-chapter-20-ex-20-2-solutio-1st-image.png\" alt=\"\" width=\"349\" height=\"67\"><\/p>\n<p>Therefore, heights are in the ratio of 9 : 5.<\/p>\n<p><strong>Question 7: A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We are given, a cylinder and a cone are having equal radii of their bases and heights.<\/p>\n<p>Let, radius of the cone = radius of the cylinder = r and<\/p>\n<p>Height of the cone = height of the cylinder = h<\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-124780\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/rd-sharma-class-9-maths-chapter-20-ex-20-2-q7-2nd-image.png\" alt=\"\" width=\"284\" height=\"66\"><\/p>\n<p>Therefore, ratio of their volumes is 3:1.<\/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 20.2. If you have any query feel free to ask in the comment section.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-20-exercise-202\"><\/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 20.2<\/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-1630918934429\" 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 id=\"faq-question-1630918965351\" 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-202-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 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 20.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630918966276\" 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-1630918967155\" 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-202-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 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 20.2 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 solutions are provided here. These solutions are prepared by our experts and provide accurate answers to all the practice questions that students can refer to and prepare to solve during the exam. The chapter of RD Sharma Class 9 Solutions Chapter 20 exercise 20.2 shows students &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-20-exercise-20-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":125597,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[3086,76786,76788],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124763"}],"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=124763"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124763\/revisions"}],"predecessor-version":[{"id":125654,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124763\/revisions\/125654"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125597"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=124763"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=124763"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=124763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}