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The content covers a temple formed by a right circular cylinder and cone, including cost calculation using a given plastering rate. It then works multiple examples involving hemispheres, cones, cylinders, and tents with specified dimensions and height-to-radius or ratio constraints, concluding with “board question(s)” on total surface area of combined solids and remaining shapes after removing hemispheres.",{"@graph":69,"@context":123},[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/surface-areas-and-volumes-curved-surface-area-of-combination-of-solids/297251/",{"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/surface-areas-and-volumes-curved-surface-area-of-combination-of-solids/297251.png","ImageObject",300,407,{"name":92,"@type":93},"Ava Thompson","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-20","2026-09-18",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",13,{"@type":107,"mainEntity":108},"FAQPage",[109,115,119],{"name":110,"@type":111,"acceptedAnswer":112},"How do you find the curved surface area of a temple made of a cylinder and a cone?","Question",{"text":113,"@type":114},"Add the curved surface areas of the cone and the cylinder. Use the given height and radius relationships to compute both curved areas, then sum them to get the temple’s curved surface area before applying the plastering rate.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"In the hemisphere mounted by a cone example, how are the radii and slant height determined?",{"text":118,"@type":114},"The hemisphere diameter equals 21 cm, so the hemisphere radius is 10.5 cm. The cone shares the same base radius, and the cone height is found from the total toy height minus the hemisphere radius. The slant height is then used to compute the cone’s curved surface area.",{"name":120,"@type":111,"acceptedAnswer":121},"How is the cost determined when a tent is made from a cylinder with a conical top?",{"text":122,"@type":114},"Let the cylinder height, radius, and cone height follow the given ratio 8:4:3. Use these to compute the cone slant height and the cylinder and cone curved surface areas, which together give the canvas area. Multiply by the canvas rate per square meter to get the cost and then solve for the required tent height.","https://schema.org",{"og:url":83,"og:type":125,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":127,"canonical":83},"index,follow",{"doc_id":129,"site_id":62},297251,1789743377,{"code":4,"msg":5,"data":132},{"doc_id":129,"user_id":133,"nickname":92,"user_avatar":134,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":140,"language":141,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":142,"faqs":143,"seo_title":144,"seo_description":67,"update_tm":130,"read_time":145},1649267921044,"https://us-avatar.wpscdn.com/avatar/1800007509477c92dfb?_k=1786009248482753345","Surface Areas and Volumes  \nCurved Surface Area Of Combination Of Solids  \nWe know how to find the curved surface areas of three-dimensional figures such as cone, cylinder, sphere, etc. Now, we will learn to find out the curved surface area of a combination of solids.  \nLet us consider the following figure.  \nThe given figure shows a temple. In this figure, the base of the temple is a right circular cylinder and the top is a right circular cone. The height of the conical part is half the height of the cylindrical part and the radius of the base of the temple is 2 ft more than the height of the conical part. The temple has to be plastered from outside at the rate of Rs 5 per square foot.  \nCan we find the approximate cost of plastering the temple from outside?  \nIn order to find the cost, we have to find the curved surface area of the temple. By adding the curved surface areas of cone and cylinder, we can obtain the curved surface area of the temple.  \nFirst of all, let us review the formulae to find the curved surface areas of different solids. The formulae for different solids are given in the following table.  \n\n| Name of Solid | Figure | Formula |\n| --- | --- | --- |\n\n\n| Cuboid | | Lateral or curved surface area =\u003Cbr>2h (l + b)\u003Cbr>where l = length, b = breadth, and h = height of the cuboid |\n| --- | --- | --- |\n| Cube | | Lateral or curved surface area = 4l2 where l = length of the edge of the cube |\n| Right Circular Cylinder | | C. S.A. = 2πrh\u003Cbr>where r = radius and h = height of the cylinder |\n| Right Circular Cone | | C. S.A. = πrl\u003Cbr>where r = radius, l =  = slant height, and h = height of the cone |\n| Sphere and\u003Cbr>Hemisphere | | C.S.A. of sphere = 4πr2\u003Cbr>C.S.A. of hemisphere = 2πr2 where r = radius of the sphere |\n\nIn this way, we can find the curved surface area of a combination of solids. Now, let us solve some more examples to understand the concept better.  \nExample 1: A toy is in the form of a hemisphere mounted by a cone. The diameter of the hemisphere is 21 cm and height of the whole toy is 24.5 cm. If the surface of the  \ntoy is painted at the rate ofRs 1 per 5 cm2, then find the cost required to paint the entire toy.  \nSolution:  \nIt is given that the diameter of the hemisphere is 21 cm.∴ Radius of the hemisphere, r = 10.5 cm  \nRadius of the base of the cone, r = 10.5 cm  \nFrom the figure, height of cone, h = 24.5 − 10.5 = 14 cm  \nNow, l =  =  cm  \n∴ Surface area of the toy = C.S.A. of hemisphere + C.S.A. of cone = 2πr2 + πrl  \n= πr (2r + l)  \n=  \n= 22 × 1.5 × 38.5 = 1270.5 cm2  \n∴ Cost required for painting = 1270.5 ×  = Rs 254 approximately  \nExample 2: The given figure shows a post office box. It was painted with red colour and it is observed that the ratio between the cost of painting the hemispherical part  \nand the cylindrical part is 2:7. What is the ratio between the height of the box and the radius of the hemispherical part?  \nSolution:  \nAs the box is painted with a constant cost, irrespective of the hemispherical part and the cylindrical part, we obtain  \n, where r is the radius of the hemispherical part and h is height of the cylindrical part  \nThus, the ratio between the height of the box and the radius of the hemispherical part is 9:2 .  \nExample 3: A tent is in the shape of a cylinder surmounted by a conical top. If the ratio of the height of the cylinder, the radius of cylinder, and the height of the cone is  \n8:4:3 and the cost of canvas required for making the tent is Rs 10560 at the rate ofRs  \n10 per m2, then find the height of the tent.  \nSolution:  \nLet us consider, height of cylinder = h’, radius of cylinder = r, and height of cone = h Let h’ = 8x, r = 4x, and h = 3x  \nThen, l = = = 5x  \nAmount of canvas =  m2  \nHowever, amount of canvas = C.S.A. of cylinder + C.S.A. of cone 1056 = 2πrh’ + πrl  \n1056 = πr (2h’ + l)  \n1056 =  \n1056 =  \n264x2 = 1056  \nx2 =  x =  m  \nHeight oftent = height of cylindrical part + height of conical part = 8x + 3x  \n= 11x  \n= 11× 2  \n= 22  \nThus, the height ","cbCaibN9sPkbCWBT","https://ap.wps.com/l/cbCaibN9sPkbCWBT","pdf",823332,33,"English","# Curved Surface Area of Combination of Solids\n## Curved surface area of common solids\n## Example 1: Toy (hemisphere + cone)\n## Example 2: Post office box (ratio of costs)\n## Example 3: Tent (cylinder + conical top)\n# Board question(s): Total Surface Areas of Combination of Solids\n## Remaining solid after removing a hemisphere from a cube\n## Example 1: Cylinder minus cone with different painting rates\n## Example 2: Surface from revolving a right triangle","[{\"question\":\"How do you find the curved surface area of a temple made of a cylinder and a cone?\",\"answer\":\"Add the curved surface areas of the cone and the cylinder. Use the given height and radius relationships to compute both curved areas, then sum them to get the temple’s curved surface area before applying the plastering rate.\"},{\"question\":\"In the hemisphere mounted by a cone example, how are the radii and slant height determined?\",\"answer\":\"The hemisphere diameter equals 21 cm, so the hemisphere radius is 10.5 cm. The cone shares the same base radius, and the cone height is found from the total toy height minus the hemisphere radius. The slant height is then used to compute the cone’s curved surface area.\"},{\"question\":\"How is the cost determined when a tent is made from a cylinder with a conical top?\",\"answer\":\"Let the cylinder height, radius, and cone height follow the given ratio 8:4:3. Use these to compute the cone slant height and the cylinder and cone curved surface areas, which together give the canvas area. Multiply by the canvas rate per square meter to get the cost and then solve for the required tent height.\"}]","Surface Areas and Volumes - Curved Surface Area of Combination of Solids | PDF",83]