[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-297452-105":3,"detail-sidebar-cat-0-en-105":85,"doc-detail-297452-en":135},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":78,"head_meta":80,"extra_data":82,"updated_unix":84},105,"en","chapter-13-surface-area-and-volumes-exercise-131","Chapter-13: Surface Area and Volumes - Exercise 13.1","","Chapter 13 “Surface Area and Volumes” provides worked solutions for Grade 10 NCERT Exercise 13.1. The problems cover surface area of a cuboid formed by joining cubes, inner surface area of a vessel combining a hollow hemisphere and hollow cylinder, total surface area of a composite toy (cone on a hemisphere), and a solid formed by a cubical block surmounted by a hemisphere. Each solution states given data, outlines key formulas, and computes final areas with clear steps.",{"@graph":14,"@context":77},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/chapter-13-surface-area-and-volumes-exercise-131/297452/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/chapter-13-surface-area-and-volumes-exercise-131/297452.png","ImageObject",300,407,{"name":42,"@type":43},"นรินทร์","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-20","2026-09-18",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",15,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69,73],{"name":60,"@type":61,"acceptedAnswer":62},"How is the surface area of the cuboid in Q1 determined?","Question",{"text":63,"@type":64},"First compute the cube edge length from the given cube volume, then treat the joined shape as a cuboid. Use the cuboid surface area formula 2(lb + bh + lh).","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"What surfaces are included when finding the inner surface area in Q2?",{"text":68,"@type":64},"The inner surface consists of the curved surface area of the hemisphere plus the curved surface area of the cylinder. The cylinder height is the total vessel height minus the hemisphere height.",{"name":70,"@type":61,"acceptedAnswer":71},"In Q3, which areas contribute to the total surface area of the toy?",{"text":72,"@type":64},"The total surface area includes the curved surface areas of the hemisphere and the cone. The cone’s slant height is found using l = r^2 + h^2 as used in the solution.",{"name":74,"@type":61,"acceptedAnswer":75},"For Q4, why is the maximum hemisphere diameter limited by the cube?",{"text":76,"@type":64},"The hemisphere’s rim cannot extend beyond the top surface of the cube. The greatest diameter equals the cube’s edge, otherwise part of the hemisphere would lie outside the cube.","https://schema.org",{"og:url":32,"og:type":79,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":81,"canonical":32},"index,follow",{"doc_id":83,"site_id":7},297452,1789744708,{"code":4,"msg":86,"data":87},"success",[88,92,96,99,104,109,114,119,124,127,131],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":93,"show_sort_weight":94,"slug":95},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":97,"slug":98},70,"exam",{"id":100,"doc_module":4,"doc_module_name":25,"category_name":101,"show_sort_weight":102,"slug":103},5,"Comic",60,"comic",{"id":105,"doc_module":4,"doc_module_name":25,"category_name":106,"show_sort_weight":107,"slug":108},6,"Technology",50,"technology",{"id":110,"doc_module":4,"doc_module_name":25,"category_name":111,"show_sort_weight":112,"slug":113},7,"Healthcare",40,"healthcare",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},8,"Research & Report",30,"research-report",{"id":120,"doc_module":4,"doc_module_name":25,"category_name":121,"show_sort_weight":122,"slug":123},9,"Religion & Spirituality",20,"religion-spirituality",{"id":122,"doc_module":4,"doc_module_name":25,"category_name":125,"show_sort_weight":122,"slug":126},"World Cup","world-cup",{"id":128,"doc_module":4,"doc_module_name":25,"category_name":129,"show_sort_weight":128,"slug":130},10,"Lifestyle","lifestyle",{"id":132,"doc_module":4,"doc_module_name":25,"category_name":133,"show_sort_weight":100,"slug":134},19,"General","general",{"code":4,"msg":86,"data":136},{"doc_id":83,"user_id":137,"nickname":42,"user_avatar":138,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":139,"file_id":140,"file_url":141,"file_type":142,"file_size":143,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":55,"language":144,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":145,"faqs":146,"seo_title":147,"seo_description":12,"update_tm":84,"read_time":148},2336475104957,"https://ap-avatar.wpscdn.com/avatar/22000c4c6bd8a5076e1?x-image-process=image/resize,m_fixed,w_180,h_180&k=1787554080175789136","Thankyou for downloading our FREE NCERT Solutions PDF  \nTHE MATH EXPERT  \nCrack School ~~ ~~ & ~~  ~~Board Exams  \nWith Cuemath you will get  \n5x more attention than in school  \nCertified Math Experts  \nInteractive Simulations  \n9 out of 10 Cuemath Students score 90+ in math exams within 6 months of classes  \nBook a Demo  \nChapter-13: Surface Area and Volumes  \nExercise 13.1 (Page 244 of Grade 10 NCERT)  \nQ1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.  \nDifficulty Level:  \nMedium  \nKnown/given:  \nTwo cubes each of volume 64cm3 are joined end to end.  \nUnknown:  \nSurface area of the resulting cuboid when two cubes are joined end to end.  \nReasoning:  \nWe will find the length of the edge of each cube by using the formula for volume of a cube = a3 , where length ofthe edge is a .  \nAs the cubes are joined end to end, they will appear as follows  \nUsing the formula for Surface area of a cuboid = 2 (lb + bh + lh ) , where l, b and h are length, breadth and height respectively. We’ll be able to get the answer.  \nSolution:  \nLet the length of the edge of each cube is a Therefore, volume of the cube = a3  \n| Grade 10-NCERT – Math\u003Cbr>volume of the cube, a3 = 64cm3\u003Cbr>a3 = 64cm3\u003Cbr>a = 64cm \u003Cbr>a = 3(4cm )3\u003Cbr>a = 4cm | |\n| --- | --- |\n| Therefore,\u003Cbr>Length of the resulting cuboid, l = a = 4cm\u003Cbr>Breadth of the resulting cuboid, b = a = 4cm\u003Cbr>Height of the resulting cuboid, h = 2a = 2􀁵 4cm = 8cm\u003Cbr>Surface area of the resulting cuboid = 2 (lb + bh + lh )\u003Cbr>= 2 (4cm 􀁵 4cm + 4cm 􀁵8cm + 4cm 􀁵8cm )\u003Cbr>= 2 (16cm2 + 32cm2 + 32cm2 )\u003Cbr>= 2􀁵80cm2\u003Cbr>= 160cm2\u003Cbr>Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.\u003Cbr>Difficulty Level:\u003Cbr>Medium\u003Cbr>Known/given:\u003Cbr>The diameter of the hemisphere is 14 cm and total height of the vessel is 13 cm.\u003Cbr>Unknown:\u003Cbr>The inner surface area of the vessel.\u003Cbr>Reasoning:\u003Cbr>Create a figure of the vessel according to the given description\u003Cbr>|  |\n\nFrom the figure it’s clear that the inner surface area of the vessel includes the CSA of the hemisphere and the cylinder.  \nInner surface area of the vessel = CSA of the hemisphere + CSA of the cylinder  \nWe will find the area of the vessel by using formulae;  \nCSA of the hemisphere = 2􀁓r2  \nwhere r is the radius of the hemisphere  \nCSA of the cylinder = 2􀁓rh  \nwhere r and h are the radius and height of the cylinder respectively.  \nHeight of the cylinder = Total height of the vessel – height of the hemisphere  \nSolution:  \nDiameter of the hemisphere, d = 14cm  \n14cm  \nRadius of the hemisphere, r = = 7cm  \n2  \nHeight of the hemisphere = radius of the hemisphere, r = 7cm  \nRadius of the cylinder, r = 7cm  \nHeight of the cylinder = Total height of the vessel – height of the hemisphere  \nh = 13cm − 7cm = 6cm  \nInner surface area of the vessel = CSA of the hemisphere + CSA of the cylinder  \n= 2􀁓 r2 + 2􀁓 rh  \n= 2􀁓 r (r + h )  \n= 2 􀁵 227 􀁵 7cm(7cm + 6cm )  \n= 2􀁵 22􀁵13cm2  \n= 572cm2  \nQ3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.  \nDifficulty Level:  \nMedium  \nKnown/given:  \nThe toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere with the same radius. The total height of the toy is 15.5 cm.  \nUnknown:  \nThe total surface area of the toy.  \nReasoning:  \nWe can create the figure of the toy as per given information  \nFrom the figure it’s clear that total surface area of the toy includes CSA of the cone and hemisphere.  \nTotal surface area of the toy = CSA of the hemisphere + CSA of the cone  \nWe will find the total area of the toy by using formulae;  \nCSA of the hemisphere = 2􀁓r2  \nwhere r is the radius of the hemisphere  \nCSA of the cone = 􀁓rl  \nwhere r and l are the radius and slant height of the cone respectively.  \nSlan","cbCais6G2iQysKKD","https://ap.wps.com/l/cbCais6G2iQysKKD","pdf",1027489,"English","# Exercise 13.1\n## Q1. Surface area of a cuboid formed by two joined cubes\n## Q2. Inner surface area of a hollow hemisphere and cylinder vessel\n## Q3. Total surface area of a cone mounted on a hemisphere toy\n## Q4. Greatest hemisphere diameter on a cubical block and surface area of the solid","[{\"question\":\"How is the surface area of the cuboid in Q1 determined?\",\"answer\":\"First compute the cube edge length from the given cube volume, then treat the joined shape as a cuboid. Use the cuboid surface area formula 2(lb + bh + lh).\"},{\"question\":\"What surfaces are included when finding the inner surface area in Q2?\",\"answer\":\"The inner surface consists of the curved surface area of the hemisphere plus the curved surface area of the cylinder. The cylinder height is the total vessel height minus the hemisphere height.\"},{\"question\":\"In Q3, which areas contribute to the total surface area of the toy?\",\"answer\":\"The total surface area includes the curved surface areas of the hemisphere and the cone. The cone’s slant height is found using l = r^2 + h^2 as used in the solution.\"},{\"question\":\"For Q4, why is the maximum hemisphere diameter limited by the cube?\",\"answer\":\"The hemisphere’s rim cannot extend beyond the top surface of the cube. The greatest diameter equals the cube’s edge, otherwise part of the hemisphere would lie outside the cube.\"}]","Chapter-13: Surface Area and Volumes - Exercise 13.1 | PDF",38]