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The document provides expressions for total, lateral/curved, and volume calculations, then applies them through 1-mark, 2-mark, and 3-mark problem sets with step-by-step solutions. Exercises cover joining solids to form new cuboids, melting/recasting shapes, computing slant height, inner surface area of combined vessels, volume left after carving a sphere from a cube, and ratios of volumes and surface areas.",{"@graph":69,"@context":115},[70,84,106],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/ch_12_surface_areas_and_volumes-surface-areas-and-volumes/297676/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/ch_12_surface_areas_and_volumes-surface-areas-and-volumes/297676.png","ImageObject",300,407,{"name":92,"@type":93},"Aria Callaghan","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-11-04","2026-09-18",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",16,{"@type":107,"mainEntity":108},"FAQPage",[109],{"name":110,"@type":111,"acceptedAnswer":112},"What type of question appears in the 1-mark and 2-mark sections?","Question",{"text":113,"@type":114},"The 1-mark section includes direct calculation problems (e.g., resulting cuboid surface area after joining cubes, cylinder curved surface area). The 2-mark section extends to conversion/melting scenarios and relationships involving dimensions and derived lengths.","Answer","https://schema.org",{"og:url":83,"og:type":117,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":119,"canonical":83},"index,follow",{"doc_id":121,"site_id":62},297676,1789746873,{"code":4,"msg":5,"data":124},{"doc_id":121,"user_id":125,"nickname":92,"user_avatar":126,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":127,"file_id":128,"file_url":129,"file_type":130,"file_size":131,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":29,"language":132,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":133,"faqs":134,"seo_title":135,"seo_description":67,"update_tm":122,"read_time":136},962084926284,"https://ap-avatar.wpscdn.com/davatar_29158cc5080c5b710cf443261637dec0","Mathematics Kolabari Tutorial CBSE X  \nComprehensive Study material  \nCh12. SURFACE AREAS AND VOLUMES  \nFormulae  \n1) Total Surface area of cube = 6  \n2) Lateral Surface area of cube = 4  \n3) Volume of cube =  \n4) Total Surface area of cuboid = 2(lb+bh+hl)  \n5) Lateral Surface area of cuboid = 2(l+b)h  \n6) Volume of cuboid = l×b×h  \n7) Curved Surface area of Cylinder = 2πrh  \n8) Total Surface area of Cylinder = + 2π = 2πr r+h  \n9) Volume of Cylinder = π h  \n10) Total Surface area of Cone = πr l+r  \n11) Lateral Surface area of Cone = πrl  \n12) Volume of Cone = ~~ ~~ π h  \n13) Surface area of Sphere = 4π  \n14) Volume of Sphere = ~~ ~~ π  \n15) Total Surface area of Hemisphere = 3  \n16) Curved Surface area of Hemisphere = 2π  \n17) Volume of Hemisphere = ~~ ~~ π  \n18) Total Surface area of Frustum = + + π +π = π R+r l +π + ),  \nwhere l = + ,R & r are the radii of circular bases ofthe frustum.  \n19) Curved Surface area of Frustum = π R+r l  \n20) Volume of a Frustum = ~~ ~~ πh + +Rr)  \n1 MARKS QUESTIONS  \nQ1. Two cubes each with 12cm. edge are joined end to end. Find the surface area of the resulting cuboid.  \nSolution: Length of resulting cuboid = 12+12 =24cm. Breadth= 12cm. Height = 12cm  \nS.A. of resulting cuboid = 2(lb+bh+hl) =2(288+144+288) =2 x 720 =1440 cm.2  \nQ2. Three cubes of metal whose edges are 3cm., 4cm. and 5cm. are melted and formed into a single cube. Find its edge.  \nSolution: Volume of new cube= Volume of 3 cubes= 33+ 43 +53=27+64+125=216 cm.3, then edge of new cube = 2 =6 cm.  \nQ3 . Find the curved surface area of a right circular cylinder of height 13.5 cm. and radius of whose base is 7 cm.  \nSolution: height of cylinder(h)=13.5 cm. Radius of base(r)= 7 cm. C.S.A of cylinder = 2πrh =2 x (22/7) x 7 13.5 = 704 cm.  \nQ4.The radii of end of frustum of a cone 40 cm high are 20 cm and 11 cm. Find its slant height.  \nSolution: l= + = 4 + 2 cm = cm = 41 cm  \nQ5. A vessel is in the form of hollow hemisphere mounted by a hollow cylinder .The diameter of hemisphere is 14 cm and total height of vessel is 13 cm. Find the inner surface area of vessel.  \nSolution: Here the radius of hemisphere = radius of cylinder = r cm =7 cm and height of cylinder, h= 13-7=6 cm Now, inner S.A of vessel = curved surface area of cylindrical part + curved surface area of hemispherical part  \n= (2 + 2 2 ) = 2 + = 2 x ~~ ~~ x 7 x (6 +7) =2 x 22 x 13 =572cm2  \nQ6. The largest possible sphere is curved out of a wooden solid cube of side 7 cm. Find the volume of wood left. Sol. Diameter of sphere carved out = side of cube = 7 cm Thus, R= 3.5 cm  \nMathematics Kolabari Tutorial CBSE X  \nComprehensive Study material  \n=> Volume of sphere carved out = (4/3) 3 = 179.66 cm3  \nAlso, Volume of cube = a3=73 =343 cm3 => Volume of wood left = 343-179.66 =163 .34 cm3  \nQ7. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of wire.  \nSolution: The volume ofthe rod = π(1/2)2 x 8 cm3 = 2 cm3 The length of new wire of the same volume = 18 m =1800 cm If r is the radius of cross-section of wire its volume = r2 x 1800  \nThen r2 x 1800=2 => r =1/30 => The thickness of wire of cross section is 1/15 cm. = 0.67 mm  \n2 MARKS QUESTIONS  \nQ8. A solid sphere of radius 12cm. is melted and recast into spherical balls of radius 1 cm. Find the number of balls made? Solution: Vol. of sphere = (4/3) π x 23 ; Vol. of spherical ball = (4/3) π x 3  \nLet n is the no. of balls then, n x (4/3) π x 3 = (4/3) π x 23 => n = 1728 balls  \nQ9. The material of a cone is converted into the shape of a cylinder of equal radius. If the height of a cylinder is 5cm. , find the height of a cone.  \nSolution: As per given condition,(1/3) πr2h = π r2 H => h = 3H = 3 x 5 =15cm.  \nQ10. A cone, a hemisphere and a cylinder stand on equal bases and have the same heights. Show that their volumes are in the ratio 1:2:3 .  \nSolution: Ratio of volumes = (1/3) π x 2 x h : (2/3) π x r3 : π x r2 x h = (1/3) h : (2/3) r : h = (1/3) h : (2/3) h : ","cbCaiohp09HTIHTV","https://ap.wps.com/l/cbCaiohp09HTIHTV","pdf",838330,"English","# Ch12. Surface Areas and Volumes\n## Formulae\n## 1 Marks Questions\n## 2 Marks Questions\n## 3 Marks Questions","[{\"question\":\"What type of question appears in the 1-mark and 2-mark sections?\",\"answer\":\"The 1-mark section includes direct calculation problems (e.g., resulting cuboid surface area after joining cubes, cylinder curved surface area). The 2-mark section extends to conversion/melting scenarios and relationships involving dimensions and derived lengths.\"}]","ch_12_surface_areas_and_volumes - Surface Areas and Volumes | PDF",15]