{"id":125928,"date":"2021-09-13T20:00:59","date_gmt":"2021-09-13T14:30:59","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125928"},"modified":"2021-09-13T20:04:36","modified_gmt":"2021-09-13T14:34:36","slug":"rd-sharma-class-9-solutions-chapter-18-exercise-18-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-18-exercise-18-1\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 (Updated 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125943\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-18-Exercise-18.1.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-18-Exercise-18.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-18-Exercise-18.1-768x432.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 18 Exercise 18.1&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 18 Exercise 18.1<\/strong>: Our team of experts have created <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-18-surface-area-and-volume-of-a-cuboid-and-cube\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 18<\/a> Exercise 18.1. You can easily get the PDF for the same from the link provided below.\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-69d91c83c1155\" 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-69d91c83c1155\"><\/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-18-exercise-18-1\/#download-rd-sharma-class-9-solutions-chapter-18-exercise-181\" title=\"Download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1\">Download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1<\/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-18-exercise-18-1\/#access-rd-sharma-class-9-solutions-chapter-18-exercise-181\" title=\"Access RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1\">Access RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1<\/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-18-exercise-18-1\/#faq-rd-sharma-class-9-solutions-chapter-18-exercise-181\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1\">FAQ: RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1<\/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-18-exercise-18-1\/#can-i-download-rd-sharma-class-9-solutions-chapter-18-exercise-181-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 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-18-exercise-18-1\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions?\">What are the benefits of studying RD Sharma Class 9 Solutions?<\/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-18-exercise-18-1\/#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-18-exercise-181\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1&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 18 Exercise 18.1<\/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\/18.1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/18.1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-18-exercise-181\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1&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 18 Exercise 18.1<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.<br \/>Solution:<br \/>Length of cuboid (l) = 80 cm<br \/>Breadth (b) = 40 cm<br \/>Height (h) = 20 cm<br \/>(i) \u2234 Lateral surface area = 2h(l + b)<br \/>= 2 x 20(80 + 40) cm\u00b2<br \/>= 40 x 120 = 4800 cm\u00b2<br \/>(ii) Total surface area = 2(lb + bh + hl)<br \/>= 2(80 x 40 + 40 x 20 + 20 x 80) cm\u00b2<br \/>= 2(3200 + 800 + 1600) cm\u00b2<br \/>= 5600 x 2 = 11200 cm\u00b2<\/p>\n<p>Question 2.<br \/>Find the lateral surface area and total surface area of a cube of edge 10 cm.<br \/>Solution:<br \/>Edge of cube (a) = 10 cm<br \/>(i) \u2234 Lateral surface area = 4a\u00b2<br \/>= 4 x (10)\u00b2 = 4 x 100 cm\u00b2= 400 cm\u00b2<br \/>(ii) Total surface area = 6a\u00b2 = 6 x(10)\u00b2 cm\u00b2<br \/>= 6 x 100 = 600 cm\u00b2<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_2_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_2_anchor\"><iframe id=\"aswift_2\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=1057982765&amp;pi=t.aa~a.1381849204~i.8~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1630454644&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-class-9-solutions-chapter-18-surface-areas-and-volume-of-a-cuboid-and-cube%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8OHmiQYQkMCJsMKosN5ZEjkAU_Bnjh1_NT_U3wQZgN6FV7ofeBRZEi5nnKe0JcwZUzuvYzSPPo91Ps4csv2C9G0NhdaFWtt-pg8&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631176112709&amp;bpp=3&amp;bdt=3551&amp;idt=-M&amp;shv=r20210901&amp;mjsv=m202109080101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280&amp;nras=2&amp;correlator=3626722227617&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631176111&amp;ga_hid=297232170&amp;ga_fc=0&amp;u_tz=330&amp;u_his=6&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=1692&amp;biw=1349&amp;bih=600&amp;scr_x=0&amp;scr_y=414&amp;eid=44748448%2C44747620%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;pvsid=31424176682072&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C600&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;fu=128&amp;bc=31&amp;ifi=3&amp;uci=a!3&amp;btvi=1&amp;fsb=1&amp;xpc=tEvqxVywho&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=164\" name=\"aswift_2\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!3\" data-google-query-id=\"CMWp28W88fICFe_MKAUdOZsG4A\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 3.<br \/>Find the ratio of the total surface area and lateral surface area of a cube.<br \/>Solution:<br \/>Let a be the edge of the cube, then Total surface area = 6a2\u00b2<br \/>and lateral surface area = 4a\u00b2<br \/>Now ratio between total surface area and lateral surface area = 6a\u00b2 : 4a\u00b2 = 3 : 2<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_3_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_3_anchor\"><iframe id=\"aswift_3\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=1215781487&amp;pi=t.aa~a.1381849204~i.10~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1630454644&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-class-9-solutions-chapter-18-surface-areas-and-volume-of-a-cuboid-and-cube%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8OHmiQYQkMCJsMKosN5ZEjkAU_Bnjh1_NT_U3wQZgN6FV7ofeBRZEi5nnKe0JcwZUzuvYzSPPo91Ps4csv2C9G0NhdaFWtt-pg8&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631176112709&amp;bpp=5&amp;bdt=3551&amp;idt=5&amp;shv=r20210901&amp;mjsv=m202109080101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280&amp;nras=3&amp;correlator=3626722227617&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631176111&amp;ga_hid=297232170&amp;ga_fc=0&amp;u_tz=330&amp;u_his=6&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=2154&amp;biw=1349&amp;bih=600&amp;scr_x=0&amp;scr_y=414&amp;eid=44748448%2C44747620%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;pvsid=31424176682072&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C600&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;fu=128&amp;bc=31&amp;ifi=4&amp;uci=a!4&amp;btvi=2&amp;fsb=1&amp;xpc=80XFADdIqD&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=296\" name=\"aswift_3\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!4\" data-google-query-id=\"CJDi48W88fICFTbMKAUdHi8C0w\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 4.<br \/>Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?\u00a0 \u00a0[NCERT]<br \/>Solution:<br \/>Length of box (l) = 80 cm<br \/>Breadth (b) = 40 cm<br \/>and height (h) = 20 cm<br \/>\u2234 Total surface area = 2(lb + bh + hl)<br \/>= 2[80 x 40 + 40 x 20 + 20 x 80] cm\u00b2<br \/>= 2[3200 + 800 + 1600] cm\u00b2 = 2 x 5600 = 11200 cm\u00b2<br \/>Size of paper sheet = 40 cm<br \/>\u2234 Area of one sheet = (40 cm)\u00b2 = 1600 cm\u00b2<br \/>\u2234 No. of sheets required for the box = 11200 = 1600 = 7 sheets<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_4_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_4_anchor\"><iframe id=\"aswift_4\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=152193507&amp;pi=t.aa~a.1381849204~i.12~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1630454644&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-class-9-solutions-chapter-18-surface-areas-and-volume-of-a-cuboid-and-cube%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8OHmiQYQkMCJsMKosN5ZEjkAU_Bnjh1_NT_U3wQZgN6FV7ofeBRZEi5nnKe0JcwZUzuvYzSPPo91Ps4csv2C9G0NhdaFWtt-pg8&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631176112729&amp;bpp=5&amp;bdt=3571&amp;idt=5&amp;shv=r20210901&amp;mjsv=m202109080101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280&amp;nras=4&amp;correlator=3626722227617&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631176111&amp;ga_hid=297232170&amp;ga_fc=0&amp;u_tz=330&amp;u_his=6&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=2850&amp;biw=1349&amp;bih=600&amp;scr_x=0&amp;scr_y=1036&amp;eid=44748448%2C44747620%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;psts=AGkb-H-iPR1ZfgpJGOXipKqSuqpEURW8tAmQi4ZOBi5UNpKYfgNZeNaCUkTU-jMpijhYMT5a1PbiLC7u0A&amp;pvsid=31424176682072&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C600&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;fu=128&amp;bc=31&amp;ifi=5&amp;uci=a!5&amp;btvi=3&amp;fsb=1&amp;xpc=kiKItqJzI4&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=577\" name=\"aswift_4\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!5\" data-google-query-id=\"CJuN5MW88fICFR_IKAUdvQkNYA\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 5.<br \/>The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of \u20b97.50 m\u00b2.<br \/>Solution:<br \/>Length of a room (l) = 5m<br \/>Breadth (b) = 4 m<br \/>and height (h) = 3 m<br \/>\u2234 Area of 4 walls = 2(l + b) x h<br \/>= 2(5 + 4) x 3 = 6 x 9 = 54 m\u00b2<br \/>and area of ceiling = l x b = 5 x 4 = 20 m\u00b2<br \/>\u2234 Total area = 54 + 20 = 74 m2<br \/>Rate of white washing = 7.50 per m\u00b2<br \/>\u2234 Total cost = \u20b974 x 7.50 = \u20b9555<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_5_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_5_anchor\"><iframe id=\"aswift_5\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=1295003436&amp;pi=t.aa~a.1381849204~i.14~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1630454644&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-class-9-solutions-chapter-18-surface-areas-and-volume-of-a-cuboid-and-cube%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8OHmiQYQkMCJsMKosN5ZEjkAU_Bnjh1_NT_U3wQZgN6FV7ofeBRZEi5nnKe0JcwZUzuvYzSPPo91Ps4csv2C9G0NhdaFWtt-pg8&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631176112749&amp;bpp=5&amp;bdt=3591&amp;idt=5&amp;shv=r20210901&amp;mjsv=m202109080101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280%2C750x280&amp;nras=5&amp;correlator=3626722227617&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631176111&amp;ga_hid=297232170&amp;ga_fc=0&amp;u_tz=330&amp;u_his=6&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=3494&amp;biw=1349&amp;bih=600&amp;scr_x=0&amp;scr_y=1109&amp;eid=44748448%2C44747620%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;psts=AGkb-H-iPR1ZfgpJGOXipKqSuqpEURW8tAmQi4ZOBi5UNpKYfgNZeNaCUkTU-jMpijhYMT5a1PbiLC7u0A&amp;pvsid=31424176682072&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C600&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;fu=128&amp;bc=31&amp;ifi=6&amp;uci=a!6&amp;btvi=4&amp;fsb=1&amp;xpc=fXHxTphlzC&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=697\" name=\"aswift_5\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!6\" data-google-query-id=\"CK3c58W88fICFeTIKAUdvK8Oyw\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 6.<br \/>Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.<br \/>Solution:<br \/>Let each side of a cube = a cm<br \/>Then surface area = 6a\u00b2 cm\u00b2<br \/>and surface area of 3 such cubes = 3 x 6a\u00b2 = 18a\u00b2 cm\u00b2<br \/>By placing three cubes side by side we get a cuboid whose,<br \/>Length (l) = a x 3 = 3a<br \/>Breadth (b) = a<br \/>Height (h) = a<br \/>\u2234 Total surface area = 2(lb + bh + hf)<br \/>= 2[3a x a+a x a+a x 3a] cm\u00b2<br \/>= 2[3a\u00b2 + a\u00b2 + 3a\u00b2] = 14 a\u00b2<br \/>\u2234 Ratio between their surface areas = 14a\u00b2 : 18a\u00b2 = 7 : 9<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_6_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_6_anchor\"><iframe id=\"aswift_6\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=2534421946&amp;pi=t.aa~a.1381849204~i.16~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1630454644&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-class-9-solutions-chapter-18-surface-areas-and-volume-of-a-cuboid-and-cube%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8OHmiQYQkMCJsMKosN5ZEjkAU_Bnjh1_NT_U3wQZgN6FV7ofeBRZEi5nnKe0JcwZUzuvYzSPPo91Ps4csv2C9G0NhdaFWtt-pg8&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631176112768&amp;bpp=4&amp;bdt=3610&amp;idt=4&amp;shv=r20210901&amp;mjsv=m202109080101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280%2C750x280%2C750x280%2C1349x600&amp;nras=7&amp;correlator=3626722227617&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631176111&amp;ga_hid=297232170&amp;ga_fc=0&amp;u_tz=330&amp;u_his=6&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=4190&amp;biw=1349&amp;bih=600&amp;scr_x=0&amp;scr_y=1836&amp;eid=44748448%2C44747620%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;psts=AGkb-H-iPR1ZfgpJGOXipKqSuqpEURW8tAmQi4ZOBi5UNpKYfgNZeNaCUkTU-jMpijhYMT5a1PbiLC7u0A&amp;pvsid=31424176682072&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C600&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;cms=2&amp;fu=128&amp;bc=31&amp;ifi=7&amp;uci=a!7&amp;btvi=5&amp;fsb=1&amp;xpc=1TIQPXBWhj&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=798\" name=\"aswift_6\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!7\" data-google-query-id=\"CIjo7cW88fICFZvnKAUd-JQL1g\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 7.<br \/>A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.<br \/>Solution:<br \/>Side of cube = 4 cm<br \/>But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64<br \/>Now surface area of one cube = 6 x (1)\u00b2<br \/>= 6 x 1=6 cm\u00b2<br \/>and surface area of 64 cubes = 6 x 64 cm\u00b2 = 384 cm\u00b2<\/p>\n<p>Question 8.<br \/>The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.<br \/>Solution:<br \/>Let h be the height of the room<br \/>Length (l) = 18 m<br \/>and width (b) = 12 m<br \/>Now surface area of floor and roof = 2 x lb = 2 x 18 x 12 m\u00b2<br \/>= 432 m\u00b2<br \/>and surface area of 4-walls = 2h (l + b)<br \/>= 2h(18 + 12) = 2 x 30h m\u00b2 = 60h m\u00b2<br \/>\u2235 The surface are of 4-walls and area of floor and roof are equal<br \/>\u2234 60h = 432<br \/>\u21d2 h =\u00a0<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"mn\">432<\/span><span id=\"MathJax-Span-5\" class=\"mn\">60<\/span><\/span><\/span><\/span><\/span>\u00a0=\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"mfrac\"><span id=\"MathJax-Span-9\" class=\"mn\">72<\/span><span id=\"MathJax-Span-10\" class=\"mn\">10<\/span><\/span><\/span><\/span><\/span>\u00a0m<br \/>\u2234 Height = 7.2m<\/p>\n<p>Question 9.<br \/>Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if the cost of tiles is \u20b9360 per dozen. [NCERT]<br \/>Solution:<br \/>Edge of cubical tank = 1.5 m<br \/>\u2234 Area of 4 walls = 4 (side)\u00b2 = 4(1.5)\u00b2 m\u00b2 = 4 x 225 = 9 m\u00b2<br \/>Area of floor = (1.5)\u00b2 = 2.25 m\u00b2<br \/>\u2234 Total surface area = 9 + 2.25 = 11.25 m\u00b2<br \/>Edge of square tile = 25 m = 0.25 m\u00b2<br \/>\u2234 Area of 1 tile = (0.25)2 = .0625 m\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/31774381718_62a8c3055e_o.png\" alt=\"RD Sharma Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"310\" height=\"113\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1971\/44732595715_4a29abd118_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"327\" height=\"82\" \/><\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_8_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_8_anchor\"><iframe id=\"aswift_8\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=3490167035&amp;pi=t.aa~a.1381849204~i.22~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1630454644&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-class-9-solutions-chapter-18-surface-areas-and-volume-of-a-cuboid-and-cube%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8OHmiQYQkMCJsMKosN5ZEiYAU_BnjhcTcEioiTZzdi3wsUFiesdkO_C1Y96_EjYuOy5HNLW21A&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631176112805&amp;bpp=5&amp;bdt=3628&amp;idt=5&amp;shv=r20210901&amp;mjsv=m202109080101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280%2C750x280%2C750x280%2C1349x600%2C750x280%2C750x280&amp;nras=9&amp;correlator=3626722227617&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631176111&amp;ga_hid=297232170&amp;ga_fc=0&amp;u_tz=330&amp;u_his=6&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=4524&amp;biw=1349&amp;bih=600&amp;scr_x=0&amp;scr_y=2137&amp;eid=44748448%2C44747620%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;psts=AGkb-H-iPR1ZfgpJGOXipKqSuqpEURW8tAmQi4ZOBi5UNpKYfgNZeNaCUkTU-jMpijhYMT5a1PbiLC7u0A&amp;pvsid=31424176682072&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C600&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;fu=128&amp;bc=31&amp;jar=2021-09-09-08&amp;ifi=9&amp;uci=a!9&amp;btvi=7&amp;fsb=1&amp;xpc=bYRkQgvqHY&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=M\" name=\"aswift_8\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!9\" data-google-query-id=\"COzyopPB8fICFajIKAUdaVIC4A\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 10.<br \/>Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.<br \/>Solution:<br \/>Let edge of a cube = a<br \/>Total surface area = 6a2<br \/>By increasing edge at 50%,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1956\/31774381358_f0450c6df3_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"350\" height=\"425\" \/><\/p>\n<p>Question 11.<br \/>A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of \u20b95 per metre sheet, sheet being 2 m wide.<br \/>Solution:<br \/>Length of iron tank (l) = 12 m<br \/>Breadth (b) = 9 m<br \/>Depth (h) = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/44732595475_45ef3243d2_o.png\" alt=\"Surface Areas and Volume of a Cuboid and Cube Class 9 RD Sharma Solutions\" width=\"192\" height=\"142\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/31774381078_539c415504_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" height=\"244\" \/><\/p>\n<p>Question 12.<br \/>Ravish wanted to make a temporary shelter for his car by making a box-like structure with tarpaulin that covers all the four sides and the top of tire car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make (he shelter of height 2.5 m with base dimensions 4 m x 3 m? [NCERT]<br \/>Solution:<br \/>Length of base (l) = 4m<br \/>Breadth (b) = 3 m<br \/>Height (h) = 2.5 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/44921829294_f047c64157_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"317\" height=\"267\" \/><\/p>\n<p>Question 13.<br \/>An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of \u20b950 per sq. metre.<br \/>Solution:<br \/>Length of open wood box (L) = 1.48 m = 148 cm<br \/>Breadth (B) = 1.16 m = 116 cm<br \/>and height (H) = 8.3 dm = 83 cm<br \/>Thickness of wood = 3 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1975\/31774380768_c4e0a112cb_o.png\" alt=\"Class 9 Maths Chapter 18 Surface Areas and Volume of a Cuboid and Cube RD Sharma Solutions\" width=\"343\" height=\"607\" \/><\/p>\n<p>Question 14.<br \/>The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at \u20b93.50 per square metre.<br \/>Solution:<br \/>Length of room (l) = 12.5 m<br \/>Breadth (b) = 9 m<br \/>and height (h) = 7 m<br \/>\u2234 Total area of walls = 2h(l + b)<br \/>= 2 x 7[12.5 + 9] = 14 x 21.5 m\u00b2\u00a0= 301 m\u00b2<br \/>Area of 2 doors of 2.5 m x 1.2 m<br \/>= 2 x 2.5 x 1.2 m\u00b2 = 6 m\u00b2<br \/>and area of 4 window of 1.5 m x 1 m<br \/>= 4 x 1.5 x 1 = 6 m\u00b2<br \/>\u2234 Remaining area of walls = 301 \u2013 (6 + 6)<br \/>= 301 \u2013 12 = 289 m\u00b2<br \/>Rate of painting the walls = \u20b93.50 per m\u00b2<br \/>\u2234 Total cost = 289 x 3.50 = \u20b91011.50<\/p>\n<p>Question 15.<br \/>The paint in a certain container is sufficient to paint on area equal to 9.375 m2. How many bricks of dimension 22.5 cm x 10 cm x 7.5 cm can be painted out of this container? [NCERT]<br \/>Solution:<br \/>Area of place for painting = 9.375 m\u00b2<br \/>Dimension of one brick = 22.5 cm x 10 cm x 7.5 cm<br \/>\u2234 Surface area of one bricks = 2 (lb + bh + hl)<br \/>= 2[22.5 x 10 + 10 x 7.5 + 7.5 x 22.5] cm2<br \/>= 2[225 + 75 + 168.75]<br \/>= 2 x 468.75 cm\u00b2 = 937.5 cm\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1941\/44732594985_36266c5c99_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"325\" height=\"175\" \/><\/p>\n<p>Question 16.<br \/>The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of \u20b98 and \u20b99.50 per m2 is \u20b91248. Find the dimensions of the box.<br \/>Solution:<br \/>Ratio in the dimensions of a cuboidal box = 2 : 3 : 4<br \/>Let length (l) = 4x<br \/>Breadth (b) = 3.v<br \/>and height (h) = 2x<br \/>\u2234 Total surface area = 2 [lb + bh + hl]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/31774380118_24ff67fdf4_o.png\" alt=\"RD Sharma Class 9 Book Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"319\" height=\"24\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/44732594825_177cd8c9bf_o.png\" alt=\"Surface Areas and Volume of a Cuboid and Cube With Solutions PDF RD Sharma Class 9 Solutions\" width=\"305\" height=\"370\" \/><\/p>\n<p>Question 17.<br \/>The cost of preparing the walls of a room 12 m long at the rate of \u20b91.35 per square metre is \u20b9340.20 and the cost of matting the floor at 85 paise per square metre is \u20b991.80. Find the height of the room.<br \/>Solution:<br \/>Cost of preparing walls of a room = \u20b9340.20<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/31774379878_d286d264db_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"357\" height=\"437\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1927\/44732594455_8269184053_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"335\" height=\"135\" \/><\/p>\n<p>Question 18.<br \/>The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 metres. The cost of decorating its walls (including doors and windows) at \u20b96.60 per square metre is \u20b95082. Find the length and breadth of the room<br \/>Solution:<br \/>Ratio in length and breadth = 4:3<br \/>and height (h) = 5.5 m<br \/>Cost of decorating the walls of a room including doors and windows = \u20b95082<br \/>Rate = \u20b96.60 per m\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/31774379558_8c26a020b7_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"347\" height=\"338\" \/><\/p>\n<p>Question 19.<br \/>A wooden bookshelf has external dimensions as follows: Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf.\u00a0 [NCERT]<br \/>Solution:<br \/>Length (l) = 85 cm<br \/>Breadth (b) = 25 cm<br \/>and height (h) = 110 cm<br \/>Thickness of plank = 5 cm<br \/>Surface area to be polished<br \/>= [(100 x 85) + 2 (110 x 25) + 2 (85 x 25) + 2 (110 x 5) + 4 (75 x 5)]<br \/>= (9350 + 5500 + 4250 + 1100 + 1500) cm\u00b2 = 21700 cm\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/44732594145_e239944722_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"270\" height=\"311\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/31774379128_43b35e11ee_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"348\" height=\"401\" \/><\/p>\n<p>We have included all the information regarding <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1<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 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-18-exercise-181\"><\/span>FAQ: RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1<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-1631178082299\" 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-18-exercise-181-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 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 18 Exercise 18.1 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178198626\" 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\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions?<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-1631178200153\" 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 18 Exercise 18.1: Our team of experts have created RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1. You can easily get the PDF for the same from the link provided below.\u00a0 RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 \u00a0 &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 (Updated 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-18-exercise-18-1\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 18 Exercise 18.1 (Updated 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":125943,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76832],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125928"}],"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=125928"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125928\/revisions"}],"predecessor-version":[{"id":127207,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125928\/revisions\/127207"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125943"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125928"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125928"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125928"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}