{"id":126062,"date":"2023-03-23T08:11:00","date_gmt":"2023-03-23T02:41:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126062"},"modified":"2023-12-01T10:38:59","modified_gmt":"2023-12-01T05:08:59","slug":"rd-sharma-class-9-solutions-chapter-19-exercise-19-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-1\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126129\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-19-Exercise-19.1.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-19-Exercise-19.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-19-Exercise-19.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1&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;}\"><strong>RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1<\/strong>: This exercise deals with a surface area of a cylinder. Practice problems on surface areas of a circular cylinder through solved RD Sharma solutions\u00a0to have a better understanding of the topic. The solutions that are provided here have been prepared by subject experts to further help students boost their scoring potential in the examination.<\/span><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1&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;}\">These solutions are created by the expert&#8217;s team. You can easily download <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 19<\/a> Exercise 19.1 from the link provided in this article.\u00a0<\/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 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-69d5628bb16ff\" 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-69d5628bb16ff\"><\/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-class-9-solutions-chapter-19-exercise-19-1\/#download-rd-sharma-class-9-solutions-chapter-19-exercise-191\" title=\"Download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1\">Download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1<\/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-class-9-solutions-chapter-19-exercise-19-1\/#access-rd-sharma-class-9-solutions-chapter-19-exercise-191\" title=\"Access RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1\">Access RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1<\/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-class-9-solutions-chapter-19-exercise-19-1\/#faq-rd-sharma-class-9-solutions-chapter-19-exercise-191\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1\">FAQ: RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1<\/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-class-9-solutions-chapter-19-exercise-19-1\/#can-i-download-rd-sharma-class-9-solutions-chapter-19-exercise-191-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1 PDF free?<\/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-class-9-solutions-chapter-19-exercise-19-1\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-19-exercise-191\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1?\">What are the benefits of studying RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1?<\/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-class-9-solutions-chapter-19-exercise-19-1\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-19-exercise-191\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1&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 19 Exercise 19.1<\/span><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\/09\/19.1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/19.1.pdf\">RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-19-exercise-191\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1&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 19 Exercise 19.1<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: Curved surface area of a right circular cylinder is 4.4 m<sup>2<\/sup>. If the radius of the base of the cylinder is 0.7 m. Find its height.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of the base of the cylinder = r = 0.7 m (Given)<\/p>\n<p>Curved surface area of cylinder = C.S.A = 4.4m<strong><sup>2\u00a0<\/sup><\/strong>(Given)<\/p>\n<p>Let \u2018h\u2019 be the height of the cylinder.<\/p>\n<p>We know, curved surface area of a cylinder = 2\u03c0rh<\/p>\n<p>Therefore,<\/p>\n<p>2\u03c0rh = 4.4<\/p>\n<p>2 x 3.14 x 0.7 x h = 4.4<\/p>\n<p>[using \u03c0=3.14 ]<\/p>\n<p>or h = 1<\/p>\n<p>Therefore the height of the cylinder is 1 m.<\/p>\n<p><strong>Question 2: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Height of cylinder (h) = Length of cylindrical pipe = 28 m or 2800 cm (Given)<\/p>\n<p>[1 m = 100 cm]<\/p>\n<p>Diameter of circular end of pipe = 5 cm (given)<\/p>\n<p>Let \u2018r\u2019 be the radius of circular end, then r = diameter\/2 = 5\/2 cm<\/p>\n<p>We know, Curved surface area of cylindrical pipe = 2\u03c0rh<\/p>\n<p>= 2 x 3.14 x 5\/2 x 2800<\/p>\n<p>[using \u03c0 = 3.14]<\/p>\n<p>= 44000<\/p>\n<p>Therefore, the area of radiating surface is 44000 cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 3: A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m<sup>2<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Height of cylindrical pillar (h) = 3.5 m<\/p>\n<p>Radius of circular end of pillar ( r) = 50\/2 cm = 25 cm = 0.25 m<\/p>\n<p>[As radius = half of the diameter] and [1 m = 100 cm]<\/p>\n<p>Curved surface area of cylindrical pillar = 2\u03c0rh<\/p>\n<p>= 2 x 3.14 x 0.25 x 3.5<\/p>\n<p>= 5.5<\/p>\n<p>Curved surface area of cylindrical pillar is 5.5 m.<\/p>\n<p>Find the cost:<\/p>\n<p>Cost of whitewashing 1m<sup>2<\/sup>\u00a0is Rs 12.50 (Given)<\/p>\n<p>Cost of whitewashing 5.5 m<sup>2<\/sup>\u00a0area = Rs. 12.50 x 5.5 = Rs. 68.75<\/p>\n<p>Thus the cost of whitewashing the pillar is Rs 68.75.<\/p>\n<p><strong>Question 4: It is required to make a closed cylindrical tank of height 1 m and the base diameter of 140 cm from a metal sheet. How many square meters of the sheet are required for the same?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Height of cylindrical tank (h) = 1 m<\/p>\n<p>Base radius of cylindrical tank (r) = diameter\/2 = 140\/2 cm = 70 cm = 0.7 m<\/p>\n<p>[1 m = 100 cm]<\/p>\n<p>Now,<\/p>\n<p>Area of sheet required = Total surface area of tank (TSA) = 2\u03c0r(h + r)<\/p>\n<p>=2 x 3.14 x 0.7(1 + 0.7)<\/p>\n<p>= 7.48<\/p>\n<p>Therefore, 7.48 m<sup>2<\/sup>\u00a0metal sheet is required to make required closed cylindrical tank.<\/p>\n<p><strong>Question 5: A solid cylinder has a total surface area of 462 cm<sup>2<\/sup>. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Total surface area of a cylinder = 462 cm<sup>2\u00a0<\/sup>(Given)<\/p>\n<p>As per given statement:<\/p>\n<p>Curved or lateral surface area = 1\/3 (Total surface area)<\/p>\n<p>\u21d2 2\u03c0rh = 1\/3(462)<\/p>\n<p>\u21d2 2\u03c0rh = 154<\/p>\n<p>\u21d2 h = 49\/2r \u2026.(1)<\/p>\n<p>[Using \u03c0 = 22\/7]<\/p>\n<p>Again,<\/p>\n<p>Total surface area = 462 cm<sup>2<\/sup><\/p>\n<p>2\u03c0r(h + r) = 462<\/p>\n<p>2\u03c0r(49\/2r + r) = 462<\/p>\n<p>or 49 + 2r<sup>2<\/sup>\u00a0= 147<\/p>\n<p>or 2r<sup>2<\/sup>\u00a0= 98<\/p>\n<p>or r = 7<\/p>\n<p>Substitute the value of r in equation (1), and find the value of h.<\/p>\n<p>h = 49\/2(7) = 49\/14 = 7\/2<\/p>\n<p>Height (h) = 7\/2 cm<\/p>\n<p>Answer: Radius = 7 cm and height = 7\/2 cm of the cylinder<\/p>\n<p><strong>Question 6: The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.cm and the area of the base ring is 115.5 sq.cm and height is 7 cm. Find the thickness of the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total surface area of hollow cylinder = 4620 cm<sup>2<\/sup><\/p>\n<p>Height of cylinder (h) = 7 cm<\/p>\n<p>Area of base ring = 115.5 cm<sup>2<\/sup><\/p>\n<p>To find: Thickness of the cylinder<\/p>\n<p>Let \u2018r<sub>1<\/sub>\u2019 and \u2018r<sub>2<\/sub>\u2019 are the inner and outer radii of the hollow cylinder respectively.<\/p>\n<p>Then, \u03c0r<sub>2<\/sub><sup>2<\/sup>\u00a0\u2013 \u03c0r<sub>1<\/sub><sup>2<\/sup>\u00a0= 115.5 \u2026\u2026.(1)<\/p>\n<p>And,<\/p>\n<p>2\u03c0r<sub>1<\/sub>h +2\u03c0r<sub>2<\/sub>h+ 2(\u03c0r<sub>2<\/sub><sup>2<\/sup>\u00a0\u2013 \u03c0r<sub>1<\/sub><sup>2<\/sup>) = 4620<\/p>\n<p>Or 2\u03c0h (r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>\u00a0) + 2 x 115.5 = 4620<\/p>\n<p>(Using equation (1) and h = 7 cm)<\/p>\n<p>or 2\u03c07 (r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>\u00a0) = 4389<\/p>\n<p>or \u03c0 (r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>\u00a0) = 313.5 \u2026.(2)<\/p>\n<p>Again, from equation (1),<\/p>\n<p>\u03c0r<sub>2<\/sub><sup>2<\/sup>\u00a0\u2013 \u03c0r<sub>1<\/sub><sup>2<\/sup>\u00a0= 115.5<\/p>\n<p>or \u03c0(r<sub>2<\/sub>\u00a0+ r<sub>1<\/sub>) (r<sub>2<\/sub>\u00a0\u2013 r<sub>1<\/sub>) = 115.5<\/p>\n<p>[using identity: a^2 \u2013 b^2 = (a \u2013 b)(a + b)]<\/p>\n<p>Using result of equation (2),<\/p>\n<p>313.5 (r<sub>2<\/sub>\u00a0\u2013 r<sub>1<\/sub>) = 115.5<\/p>\n<p>or r<sub>2<\/sub>\u00a0\u2013 r<sub>1\u00a0<\/sub>= 7\/19 = 0.3684<\/p>\n<p>Therefore, thickness of the cylinder is 7\/19 cm or 0.3684 cm.<\/p>\n<p><strong>Question 7: Find the ratio between the total surface area of a cylinder to its curved surface area, given that height and radius of the tank are 7.5 m and 3.5 m.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Height of cylinder (h) = 7.5 m<\/p>\n<p>Radius of cylinder (r) = 3.5 m<\/p>\n<p>We know, Total Surface Area of cylinder (T.S.A) = 2\u03c0r(r+h)<\/p>\n<p>And, Curved surface area of a cylinder(C.S.A) = 2\u03c0rh<\/p>\n<p>Now, Ratio between the total surface area of a cylinder to its curved surface area is<\/p>\n<p>T.S.A\/C.S.A = 2\u03c0r(r+h)\/2\u03c0rh<\/p>\n<p>= (r + h)\/h<\/p>\n<p>= (3.5 + 7.5)\/7.5<\/p>\n<p>= 11\/7.5<\/p>\n<p>= 22\/15 or 22:15<\/p>\n<p>Therefore the required ratio is 22:15.<\/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 19 Exercise 19.1&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 19 Exercise 19.1<\/span><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;}\">. If you have any queries 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-19-exercise-191\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1&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 19 Exercise 19.1<\/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-1631191206079\" 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-19-exercise-191-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1 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 19 Exercise 19.1 PDF for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631191275033\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-19-exercise-191\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 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-1631191276127\" 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>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1: This exercise deals with a surface area of a cylinder. Practice problems on surface areas of a circular cylinder through solved RD Sharma solutions\u00a0to have a better understanding of the topic. The solutions that are provided here have been prepared by subject experts to further help &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-1\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 19 Exercise 19.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":126129,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76836],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126062"}],"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=126062"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126062\/revisions"}],"predecessor-version":[{"id":515141,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126062\/revisions\/515141"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126129"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126062"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126062"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}