{"id":125931,"date":"2021-09-13T20:00:32","date_gmt":"2021-09-13T14:30:32","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125931"},"modified":"2021-09-13T20:04:29","modified_gmt":"2021-09-13T14:34:29","slug":"rd-sharma-class-9-solutions-chapter-18-exercise-18-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-18-exercise-18-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125944\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-18-Exercise-18.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-18-Exercise-18.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-18-Exercise-18.2-768x432.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 18.2&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;}\"><strong>RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2<\/strong> are available here. You can get the detailed <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>\u00a0Exercise 18.2. Our experts have created these solutions in the detailed mannner which will be very beneficial for your class 9 exams.\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-69da072c93946\" 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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Solutions Chapter 18 Exercise 18.2 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2 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-18-2\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-18-exercise-182\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2?\">What are the benefits of studying RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2?<\/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-18-2\/#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-182\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2&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 18 Exercise 18.2<\/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\/18.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/18.2.pdf\">RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-18-exercise-182\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2&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 18 Exercise 18.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? [NCERT]<br \/>Solution:<br \/>Length of water tank (l) = 6 m<br \/>Breadth (b) = 5 m<br \/>and depth (h) = 4.5 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1902\/44732593765_f567f0cd3a_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"217\" height=\"158\" \/><br \/>\u2234 Volume of water in it = lbh<br \/>= 6 x 5 x 4.5 m<sup>3<\/sup>\u00a0= 135 m<sup>3<\/sup><br \/>Capacity of water in litres = 135 x 1000 litres (1 m<sup>3<\/sup>\u00a0= 1000 l)<br \/>= 135000 litres<\/p>\n<p>Question 2.<br \/>A cubical vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? [NCERT]<br \/>Solution:<br \/>Length of vessal (l) = 10 m<br \/>Breadth (b) = 8 m<br \/>Volume = 380 m<sup>3<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1931\/31774378678_7795587040_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"354\" height=\"52\" \/><\/p>\n<p>Question 3.<br \/>Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of \u20b930 per m<sup>3<\/sup>. [NCERT]<br \/>Solution:<br \/>Length of pit (l) = 8m<br \/>Width (b) = 6 m<br \/>and depth (h) = 3 m<br \/>\u2234 Volume of earth digout = lbh<br \/>= 8 x 6 x 3 = 144 m<sup>3<\/sup><br \/>Cost of digging the pit at the rate of \u20b930 per m<sup>3<\/sup><br \/>= 144 x 30 = \u20b94320<\/p>\n<p>Question 4.<br \/>If the areas of three adjacent faces of a cuboid are 8 cm<sup>2<\/sup>, 18 cm<sup>2<\/sup>\u00a0and 25 cm<sup>2<\/sup>. Find the volume of the cuboid.<br \/>Solution:<br \/>Let x, y, z be the three adjacent faces of the cuboid, then<br \/>x = 8 cm<sup>2<\/sup>, y = 18 cm<sup>2<\/sup>, z = 25 cm<sup>2<\/sup><br \/>and let l, b, h are the dimensions of the cuboid, then<br \/>x = lb = 8 cm<sup>2<\/sup><br \/>y = bh = 18 cm<sup>2<\/sup><br \/>z = hl = 25 cm<sup>2<\/sup><br \/>\u2234 Volume = lbh<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1963\/44732593695_31d5c7b0f0_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"324\" height=\"71\" \/><\/p>\n<p>Question 5.<br \/>The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.<br \/>Solution:<br \/>Let breadth of a room (b) = x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44921828384_3bfe282e17_o.png\" alt=\"Surface Areas and Volume of a Cuboid and Cube Class 9 RD Sharma Solutions\" width=\"316\" height=\"279\" \/><\/p>\n<p>Question 6.<br \/>Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.<\/p>\n<p>Solution:<\/p>\n<p>Edge of first cube = 6 cm<br \/>Edge of second cube = 8 cm<br \/>and edge of third cube = 10 cm<br \/>\u2234 Volume of 3 cubes = (6)<sup>3<\/sup>\u00a0+ (8)<sup>3<\/sup>\u00a0+ (10)3 cm<sup>3<\/sup><br \/>= 216 + 512 + 1000 cm<sup>3<\/sup><br \/>= 1728 cm<sup>3<\/sup><br \/>\u2234 Edge of so formed cube<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44921828154_595d56c857_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"271\" height=\"41\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1952\/44921828094_20a6fa867c_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"309\" height=\"124\" \/><\/p>\n<p>Question 7.<br \/>Two cubes, each of volume 512 cm<sup>3<\/sup>\u00a0are joined end to end. Find the surface area of the resulting cuboid.<\/p>\n<p>Solution:<\/p>\n<p>Volume of each volume = 512 cm<sup>3<\/sup><br \/>\u2234 Side (edge) =\u00a0<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"mroot\"><span id=\"MathJax-Span-14\" class=\"mn\">512<\/span>\u2212\u2212\u2212\u221a<span id=\"MathJax-Span-15\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span><br \/>=<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-16\" class=\"math\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"mroot\"><span id=\"MathJax-Span-19\" class=\"msubsup\"><span id=\"MathJax-Span-20\" class=\"texatom\"><span id=\"MathJax-Span-21\" class=\"mrow\"><span id=\"MathJax-Span-22\" class=\"mn\">8<\/span><\/span><\/span><span id=\"MathJax-Span-23\" class=\"texatom\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"mn\">3<\/span><\/span><\/span><\/span>\u2212\u2212\u221a<span id=\"MathJax-Span-26\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0 = 8 cm<br \/>Now by joining two cubes, then Length of so formed cuboid (l)<br \/>= 8 + 8 = 16 cm<br \/>Breadth (b) = 8 cm<br \/>and height (h) = 8 cm<br \/>\u2234 Surface area = 2(lb + bh + hl)<br \/>= 2[16 x 8 + 8 x 8 + 8 x 16] cm<sup>2<\/sup><br \/>= 2[128 + 64 + 128] cm<sup>2<\/sup><br \/>= 2 x 320 = 640 cm<sup>2<\/sup><\/p>\n<p>Question 8.<br \/>A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.<\/p>\n<p>Solution:<\/p>\n<p>Edge of metal cube = 12 cm<br \/>\u2234 Its volume = (Edge)<sup>3<\/sup>\u00a0= (12)<sup>3<\/sup>\u00a0cm3<sup>3<\/sup><br \/>= 1728 cm<sup>3<\/sup><br \/>It is melted and form 3 cubes<br \/>Edge of one smaller cube = 6 cm<br \/>and edge of second smaller cube = 8 cm<br \/>\u2234 Volume of two smaller cubes = (6)<sup>3<\/sup>\u00a0+ (8)<sup>3<\/sup>\u00a0cm<sup>3<\/sup><br \/>= 216 + 512 cm<sup>3<\/sup>\u00a0= 728 cm<sup>3<\/sup><br \/>\u2234 Volume of third smaller cube = 1728 \u2013 728 = 1000 cm<sup>3<\/sup><br \/>\u2234 Edge of the third cube =\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-27\" class=\"math\"><span id=\"MathJax-Span-28\" class=\"mrow\"><span id=\"MathJax-Span-29\" class=\"mroot\"><span id=\"MathJax-Span-30\" class=\"mn\">1000<\/span>\u2212\u2212\u2212\u2212\u221a<span id=\"MathJax-Span-31\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span><br \/>=\u00a0<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-32\" class=\"math\"><span id=\"MathJax-Span-33\" class=\"mrow\"><span id=\"MathJax-Span-34\" class=\"mroot\"><span id=\"MathJax-Span-35\" class=\"mrow\"><span id=\"MathJax-Span-36\" class=\"mo\">(<\/span><span id=\"MathJax-Span-37\" class=\"mn\">10<\/span><span id=\"MathJax-Span-38\" class=\"mo\">)<\/span><span id=\"MathJax-Span-39\" class=\"mn\">3<\/span><\/span>\u2212\u2212\u2212\u2212\u221a<span id=\"MathJax-Span-40\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0 cm = 10 cm<\/p>\n<p>Question 9.<br \/>The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?<br \/>Solution:<\/p>\n<p>Length of cinema hall (l) = 100 m<br \/>Breadth (b) = 50 m<br \/>and height (h) = 18 m<br \/>\u2234 Volume of air in it = lbh<br \/>= 100 x 50 x 18 m<sup>3\u00a0<\/sup>= 90000 m<sup>3<\/sup><br \/>Air required for one person = 150 m<sup>3<\/sup><br \/>\u2234 Number of persons in the hall =\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-41\" class=\"math\"><span id=\"MathJax-Span-42\" class=\"mrow\"><span id=\"MathJax-Span-43\" class=\"mfrac\"><span id=\"MathJax-Span-44\" class=\"mn\">90000<\/span><span id=\"MathJax-Span-45\" class=\"mn\">150<\/span><\/span><\/span><\/span><\/span>\u00a0= 600 persons<\/p>\n<p>Question 10.<br \/>Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in width and 5 cm thick is 112 kg. Find the length of the block.<br \/>Solution:<\/p>\n<p>Weight of 1 cm3 = 0.25 kg<br \/>Breadth of the block (b) = 28 cm<br \/>Thickness (h) = 5 cm<br \/>and total weight of the block = 112 kg<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1962\/44732593225_d182cf85e7_o.png\" alt=\"Class 9 Maths Chapter 18 Surface Areas and Volume of a Cuboid and Cube RD Sharma Solutions\" width=\"332\" height=\"174\" \/><\/p>\n<p>Question 11.<br \/>A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a liquid can be placed in it? Also, find the volume of the wood used in it.<br \/>Solution:<\/p>\n<p>Outer length of the closed wooden box (l) = 25 cm<br \/>Breadth (b) = 18 cm<br \/>and height (h) = 15 cm<br \/>Width of wood = 2 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1971\/44732593115_b20d14da2f_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"239\" height=\"201\" \/><br \/>\u2234 Inner length = 25 \u2013 2\u00d72 = 25- 4 = 21cm<br \/>Breadth =18- 2\u00d72 = 18-4 = 14 cm<br \/>and height =15- 2\u00d72 = 15- 4=11 cm<br \/>Now outer volume = 25 x 18 x 15 cm<sup>3<\/sup>\u00a0= 6750 cm<sup>3<\/sup><br \/>and inner volume = 21 x 14 x 11 cm<sup>3<\/sup>\u00a0= 3234 cm<sup>3<\/sup><br \/>(i) Inner volume = 3234 cm<sup>3<\/sup><br \/>(ii) Volume of wood = 6750 \u2013 3234 = 3516 cm<sup>3<\/sup><\/p>\n<p>Question 12.<br \/>The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box?<br \/>Solution:<\/p>\n<p>External length of a closed wooden box (L) = 48 cm<br \/>Width (B) = 36 cm<br \/>and height (H) = 30 cm<br \/>Thickness of wood = 1.5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44921827854_e3b18afa9a_o.png\" alt=\"RD Sharma Class 9 Book Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"232\" height=\"182\" \/><br \/>\u2234 Internal length (l) = 48 \u2013 2 x 1.5 cm = 48 \u2013 3 = 45 cm<br \/>Width (b) = 36 \u2013 2 x 1.5 cm<br \/>= 36 \u2013 3 = 33 cm<br \/>and height (h) = 30 \u2013 2 x 1.5 cm<br \/>= 30 \u2013 3 = 27 cm<br \/>Now volume of internal box<br \/>= lbh = 45 x 33 x 27 cm<sup>3<\/sup><br \/>Volume of one bricks = 6 x 3 x 0.75 cm<sup>3<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/44732592955_653d4c2f11_o.png\" alt=\"Surface Areas and Volume of a Cuboid and Cube With Solutions PDF RD Sharma Class 9 Solutions\" width=\"314\" height=\"134\" \/><\/p>\n<p>Question 13.<br \/>A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.<br \/>Solution:<\/p>\n<p>Edge of a cube = 9 cm<br \/>Volume of cube = (9)<sup>3<\/sup>\u00a0cm<sup>3<\/sup><br \/>= 729 cm<sup>3<\/sup><br \/>Now length of vessel (l) = 15 cm<br \/>and breadth (b) = 12 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1963\/44921827764_02f713270a_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"324\" height=\"362\" \/><\/p>\n<p>Question 14.<br \/>A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7 m deep and the earth taken out is spread evenly on the field. By how many metres is the level of the field raised? Give the answer to the second place of decimal.<br \/>Solution:<\/p>\n<p>Length of a field (l) = 200 m<br \/>Breadth (b) = 150 m<br \/>Length of plot = 50 m<br \/>and breadth = 40 m<br \/>Depth of plot = 7 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44732592815_03e7f48154_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"361\" height=\"463\" \/><\/p>\n<p>Question 15.<br \/>A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a comer of the field and the earth taken out is spread- over the remaining area of the field. Find out the extent to which the level of the field has been raised.<\/p>\n<p>Solution:<\/p>\n<p>Length of a field (L) = 18m<br \/>and width (B) = 15 m<br \/>Length of pit (l) = 7.5 m<br \/>Breadth (b) = 6 m<br \/>and depth (h) = 0.8 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44732592555_4fb2568853_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"237\" height=\"178\" \/><br \/>\u2234 Volume of earth dugout = lbh<br \/>= 7.5 x 6 x 0.8 m<sup>3<\/sup><br \/>= 45 x 0.8 =\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-46\" class=\"math\"><span id=\"MathJax-Span-47\" class=\"mrow\"><span id=\"MathJax-Span-48\" class=\"mfrac\"><span id=\"MathJax-Span-49\" class=\"mrow\"><span id=\"MathJax-Span-50\" class=\"mn\">45<\/span><span id=\"MathJax-Span-51\" class=\"mi\">x<\/span><span id=\"MathJax-Span-52\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-53\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>\u00a0= 36 m3<br \/>Total area of the field = L x B<br \/>= 18 x 15 = 270 m<sup>2<\/sup><br \/>and area of pit = lb = 7.5 x 6 = 45 m<sup>2<\/sup><br \/>\u2234 Remaining area of the field excluding pit<br \/>= 270 \u2013 45 = 225 m<sup>2<\/sup><br \/>Let by spreading the earth on the remaining part of the field, the height = h<br \/>= 225 x h = 36<br \/>\u21d2 h =\u00a0<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-54\" class=\"math\"><span id=\"MathJax-Span-55\" class=\"mrow\"><span id=\"MathJax-Span-56\" class=\"mfrac\"><span id=\"MathJax-Span-57\" class=\"mn\">36<\/span><span id=\"MathJax-Span-58\" class=\"mn\">225<\/span><\/span><\/span><\/span><\/span>\u00a0=\u00a0<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-59\" class=\"math\"><span id=\"MathJax-Span-60\" class=\"mrow\"><span id=\"MathJax-Span-61\" class=\"mfrac\"><span id=\"MathJax-Span-62\" class=\"mn\">4<\/span><span id=\"MathJax-Span-63\" class=\"mn\">25<\/span><\/span><\/span><\/span><\/span>= 0.16 m = 16 cm<br \/>\u2234 Level of field raised = 16 cm<\/p>\n<p>Question 16.<br \/>A village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m x 15m x 6 m. For how many days will the water of this tank last? [NCERT]<br \/>Solution:<br \/>Population of a village = 4000<br \/>Water required per head per day = 150 litres<br \/>\u2234 Total water required = 4000 x 150 litres = 600000 litres<br \/>Dimensions of a tank = 20mx 15mx6m<br \/>\u2234 Volume of tank = 20 x 15 x 6 m3 = 1800 m<sup>3<\/sup><br \/>Capacity of water in litres = 1800 x 1000 litres (1 m<sup>3<\/sup>\u00a0= 1000 litres)<br \/>= 1800000 litres<br \/>The water will last for =\u00a0<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-64\" class=\"math\"><span id=\"MathJax-Span-65\" class=\"mrow\"><span id=\"MathJax-Span-66\" class=\"mfrac\"><span id=\"MathJax-Span-67\" class=\"mn\">1800000<\/span><span id=\"MathJax-Span-68\" class=\"mn\">600000<\/span><\/span><\/span><\/span><\/span>\u00a0= 3 days<\/p>\n<p>Question 17.<br \/>A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the figure. If the edge of each cube is 3 cm, find the volume of the structure built by the child. [NCERT]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1902\/44732592365_f0b6b5e125_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"289\" height=\"256\" \/><br \/>Solution:<br \/>No. of cubes at the given structure = 1+ 2 + 3+ 4 + 5 = 15<br \/>Edge of one cube = 3 cm<br \/>\u2234 Volume of one cube = (3)<sup>3<\/sup>\u00a0= 3 x 3 x 3 cm<sup>3<\/sup>\u00a0= 27 cm<sup>3<\/sup><br \/>\u2234Volume of the structure = 27 x 15 cm<sup>3<\/sup>\u00a0= 405 cm<sup>3<\/sup><\/p>\n<p>Question 18.<br \/>A godown measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown. [NCERT]<br \/>Solution:<br \/>Length of godown (L) = 40 m<br \/>Breadth (B) = 25 m<br \/>and height (H) = 10 m<br \/>\u2234 Volume of godown = LBH<br \/>= 40 x 25 x 10 = 10000 m<sup>3<\/sup><br \/>Dimension of one wooden crates = 1.5 m x 1.25 m x 0.5 m<br \/>\u2234Volume of one crate = 1.5 x 1.25 x 0.5 m<sup>3<\/sup>\u00a0= 0.9375 m<sup>3<\/sup><br \/>\u2234 Number of crates to be stored in the<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/44921827164_73527bbc0b_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"331\" height=\"144\" \/><\/p>\n<p>Question 19.<br \/>A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required? [NCERT]<br \/>Solution:<br \/>Length of wall (L) = 10 m = 1000 cm<br \/>Height (H) = 4 m = 400 cm<br \/>Thickness (B) = 24 cm = 24 cm<br \/>\u2234 Volume of wall = LBH = 1000 x 24 x 400 cm3 = 9600000 cm<sup>3<\/sup><br \/>Dimensions of one brick = 24 cm x 12 cm x 8 cm = 2304 cm<sup>3<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44732591805_89324e4c08_o.png\" alt=\"RD Sharma Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"340\" height=\"115\" \/><\/p>\n<p>Question 20.<br \/>If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/45596329312_5d2f6a5598_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"178\" height=\"59\" \/><br \/>Solution:<br \/>a, b, c are the dimensions of a cuboid S is the surface area and V is the volume<br \/>\u2234 V = abc and S = 2(lb + bc + ca)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/44732591605_3230f506bc_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"235\" height=\"187\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/45596329112_5b8df1bcd4_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"274\" height=\"54\" \/><\/p>\n<p>Question 21.<br \/>The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V<sup>2\u00a0<\/sup>= xyz.<br \/>Solution:<br \/>Let a, b, c are the dimensions of a cuboid then,<br \/>x = ab, y = bc, z = ca<br \/>and V = abc<br \/>Now L.H.S. = V<sup>2<\/sup><br \/>= (abc)<sup>2\u00a0<\/sup>= a<sup>2<\/sup>b<sup>2<\/sup>c<sup>2<\/sup><br \/>= ab.bc.ca = xyz = R.H.S.<br \/>Hence V<sup>2<\/sup>\u00a0= xyz<\/p>\n<p>Question 22.<br \/>A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? [NCERT]<br \/>Solution:<br \/>Speed of water in a river = 2 km\/hr<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/44732591405_ee3f917e99_o.png\" alt=\"Surface Areas and Volume of a Cuboid and Cube Class 9 RD Sharma Solutions\" width=\"361\" height=\"280\" \/><\/p>\n<p>Question 23.<br \/>Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?<br \/>Solution:<br \/>Width of canal (b) = 30 dm = 3 m<br \/>Depth (h) = 12 dm = 1.2 m<br \/>Speed of water = 100 km\/hr<br \/>Length of water flow in 30 minutes =\u00a0<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-69\" class=\"math\"><span id=\"MathJax-Span-70\" class=\"mrow\"><span id=\"MathJax-Span-71\" class=\"mfrac\"><span id=\"MathJax-Span-72\" class=\"mn\">1<\/span><span id=\"MathJax-Span-73\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0hr<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/45596328762_bfb1789479_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"347\" height=\"366\" \/><\/p>\n<p>Question 24.<br \/>Half cubic metre of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold- sheet.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1925\/44732590925_39ba542d50_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"315\" height=\"273\" \/><\/p>\n<p>Question 25.<br \/>How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 g, find the weight of the empty box in kg.<br \/>Solution:<br \/>External length of open box (L) = 36 cm<br \/>Breadth (B) = 25 cm<br \/>and Height (H) = 16.5 cm<br \/>Width of iron sheet used = 1.5 cm<br \/>\u2234 Inner length (l) = 36 \u2013 1.5 x 2 = 36 \u2013 3 = 33 cm<br \/>Breadth (b) = 25 \u2013 2 x 1.5 = 25 \u2013 3 = 22 cm<br \/>and Height (h) = 16.5 \u2013 1.5 = 15 cm<br \/>\u2234 Volume of the iron used = Outer volume \u2013 Inner volume<br \/>= 36 x 25 x 16.5 \u2013 33 x 22 x 15<br \/>= 14850 \u2013 10890 = 3960 cm3<br \/>Weight of 1 cm3 = 15 g<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45596328362_fe510c9b80_o.png\" alt=\"Class 9 Maths Chapter 18 Surface Areas and Volume of a Cuboid and Cube RD Sharma Solutions Ex 18.2\" width=\"315\" height=\"76\" \/><\/p>\n<p>Question 26.<br \/>A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.<br \/>Solution:<br \/>Base of the container = 5 cm x 5 cm<br \/>Level of water upto 1 cm from the top After placing a cube in it, the water rises to the top and 2 cubic cm of water overflows,<br \/>(i) \u2234 Volume of water = 5 x 5 x 1 + 2 = 25 + 2 = 27 cm<sup>3<\/sup><br \/>\u2234 Volume of cube = 27 cm<sup>3<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1951\/30705855387_4672e48857_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"354\" height=\"315\" \/><\/p>\n<p>Question 27.<br \/>A rectangular tank is 80 m long and 25 m broad. Water-flows into it through a pipe whose cross-section is 25 cm<sup>2<\/sup>, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes.<br \/>Solution:<br \/>Length of tank (l) = 80 m<br \/>Breadth (b) = 25 m<br \/>Area of cross section of the month of pipe = 25 cm<sup>2<\/sup><br \/>and speed of water-flow =16 km\/h<br \/>\u2234 Volume of water is 45 minutes<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1918\/30705855007_89850cbb55_o.png\" alt=\"RD Sharma Class 9 Book Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"264\" height=\"311\" \/><\/p>\n<p>Question 28.<br \/>Water in a rectangular reservoir having base 80 m by 60 m is 6.5 m deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 15 km\/hr.<\/p>\n<p>Solution:<\/p>\n<p>Length of reservoir (l) = 80 m<br \/>Breadth (b) = 60 m<br \/>and depth (h) = 6.5 m<br \/>\u2234 Volume of water in it = lbh = 80 x 60 x 6.5 m<sup>3<\/sup>\u00a0= 31200 m<sup>3<\/sup><br \/>Area of cross-section of the month of pipe = 20 x 20 = 400 cm<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/30705854757_394bc284d1_o.png\" alt=\"Surface Areas and Volume of a Cuboid and Cube With Solutions PDF RD Sharma Class 9 Solutions\" width=\"317\" height=\"105\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/45596327492_38d0052de7_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 18 Surface Areas and Volume of a Cuboid and Cube\" width=\"339\" height=\"207\" \/><\/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 18.2<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-182\"><\/span>FAQ: RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2<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-1631178091182\" 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-182-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2 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 18 Exercise 18.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178194654\" 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-18-exercise-182\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2, students can earn higher academic grades. Our experts solve RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2 with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178196305\" 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 18.2 are available here. You can get the detailed RD Sharma Class 9 Solutions Chapter 18\u00a0Exercise 18.2. Our experts have created these solutions in the detailed mannner which will be very beneficial for your class 9 exams.\u00a0 RD Sharma Class 9 Maths Solutions Download RD Sharma Class &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-18-exercise-18-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 18 Exercise 18.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":125944,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76835],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125931"}],"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=125931"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125931\/revisions"}],"predecessor-version":[{"id":127209,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125931\/revisions\/127209"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125944"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}