[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-297513-105":3,"detail-sidebar-cat-0-en-105":77,"doc-detail-297513-en":127},{"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":70,"head_meta":72,"extra_data":74,"updated_unix":76},105,"en","surface-area-and-volume-of-frustum-practice-problems","Surface area and volume of frustum - Practice problems","","Practice problems on identifying frustums of cones and pyramids and calculating their surface area and volume. Includes worked solutions for a cone frustum with given radii and depth, a square pyramid frustum with top/bottom side lengths and height, and composite solids involving cylinders, cones, hemispherical bowls, and cube-plus-pyramid volume. Ends with an assignment transferring liquid from a conical funnel into a cylindrical tin, requiring the resulting liquid height.",{"@graph":14,"@context":69},[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/surface-area-and-volume-of-frustum-practice-problems/297513/",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/surface-area-and-volume-of-frustum-practice-problems/297513.png","ImageObject",300,407,{"name":42,"@type":43},"Olivia Brown","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-11-01","2026-09-18",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",35,{"@type":57,"mainEntity":58},"FAQPage",[59,65],{"name":60,"@type":61,"acceptedAnswer":62},"What is the method for finding the surface area of a frustum of a cone?","Question",{"text":63,"@type":64},"Calculate surface area as the area of the larger cone minus the area of the smaller cone, using the slant height derived from right-triangle geometry (often via Pythagoras).","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"In the assignment, how is the liquid height in the cylinder determined?",{"text":68,"@type":64},"Use volume conservation: equate the volume of liquid in the conical funnel (cone frustum/full cone as given by depth and diameter) to the volume of liquid in the cylindrical tin, then solve for the cylinder liquid height.","https://schema.org",{"og:url":32,"og:type":71,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":73,"canonical":32},"index,follow",{"doc_id":75,"site_id":7},297513,1789745233,{"code":4,"msg":78,"data":79},"success",[80,84,88,91,96,101,106,111,116,119,123],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":81,"show_sort_weight":82,"slug":83},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":89,"slug":90},70,"exam",{"id":92,"doc_module":4,"doc_module_name":25,"category_name":93,"show_sort_weight":94,"slug":95},5,"Comic",60,"comic",{"id":97,"doc_module":4,"doc_module_name":25,"category_name":98,"show_sort_weight":99,"slug":100},6,"Technology",50,"technology",{"id":102,"doc_module":4,"doc_module_name":25,"category_name":103,"show_sort_weight":104,"slug":105},7,"Healthcare",40,"healthcare",{"id":107,"doc_module":4,"doc_module_name":25,"category_name":108,"show_sort_weight":109,"slug":110},8,"Research & Report",30,"research-report",{"id":112,"doc_module":4,"doc_module_name":25,"category_name":113,"show_sort_weight":114,"slug":115},9,"Religion & Spirituality",20,"religion-spirituality",{"id":114,"doc_module":4,"doc_module_name":25,"category_name":117,"show_sort_weight":114,"slug":118},"World Cup","world-cup",{"id":120,"doc_module":4,"doc_module_name":25,"category_name":121,"show_sort_weight":120,"slug":122},10,"Lifestyle","lifestyle",{"id":124,"doc_module":4,"doc_module_name":25,"category_name":125,"show_sort_weight":92,"slug":126},19,"General","general",{"code":4,"msg":78,"data":128},{"doc_id":75,"user_id":129,"nickname":42,"user_avatar":130,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":131,"file_id":132,"file_url":133,"file_type":134,"file_size":135,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":33,"language":136,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":137,"faqs":138,"seo_title":139,"seo_description":12,"update_tm":76,"read_time":120},16904993612988,"https://ap-avatar.wpscdn.com/davatar_a8503ba1806abce46bf441b54a3ca4cd","Surface area and volume of frustum. Identifying the frustum of a cone and pyramid.  \nFrustum  \nEvaluating the surface area and volume of frustum of a cone.  \n1. A frustum of a cone has top and bottom radii of 8cm and 4cm respectively and a depth of 16cm. Calculate:  \na. The volume of the frustum.  \nb. The surface area of the frustum.  \n8cm  \nBy similar triangle; 16cm 84 = 1􀯫  \n8􀝔 = 4 (16 + 􀝔) 8􀝔 = 64 + 4􀝔  \n8􀝔 − 4􀝔 = 64 4􀝔 = 64  \n􀝔 = 16􀜿􀝉  \na. Volume of bucket = Volume of big cone − Volume of small cone  \n1 ²  1 ²= 􀟨􀜴 􀜪 − 􀟨􀝎 ℎ  \n3 3 = 13 􀟨 × 8² × (16 + 􀝔) − 13 􀟨 × 4² × 􀝔= 􀟨3 × 64 × (16 + 16) − 􀟨3 × 16 × 16 = 􀟨3 × 64 × 32 − 􀟨3 × 16 × 16  \n2048􀟨 256􀟨= ~~ ~~ − ~~ ~~  \n3 3  \n1792􀟨  \n= ~~ ~~  \n3  \n1792 22  \n= ~~ ~~ × ~~ ~~  \n3 7  \n22 × 256  \n= ~~ ~~  \n3  \n5632  \n= ~~ ~~  \n3 = 1877. 3333􀜿􀝉3 ≅ 1877􀜿􀝉3  \nb. By Pythagoras theorem;  \n􀝈 2 = 􀝔 2 + 42  \n􀜽 2 = 162 + 16  \n= 256 + 16  \n= 272  \n􀜽 = √272  \n􀜽 = 4√17  \n􀜮2 = 82 + (􀝔 + 16)2 (􀜽 + 􀜾)2 = 64 + (16 + 16)2  \n= 64 + 1024  \n= 1088  \n􀜽 + 􀜾 = √1088  \n􀜽 + 􀜾 = 8√17  \nSurface area of frustum = area of big cone – area of small cone.  \n= 􀟨􀜴􀜮 − 􀟨􀝎􀝈  \n= 􀟨 (􀜴􀜮 − 􀝎􀝈)  \n= 􀟨 (8 × 8√17 − 4 × 4√17)  \n= 􀟨 (64√17 − 16√17)  \n= 􀟨 (48√17)  \n= 48􀟨√17􀜿􀝉2  \nCalculating the surface area and volume of frustum of a pyramid.  \n2. A frustum of a pyramid is 3cm square at the top and 6cm square at the bottom and 5cm high. Calculate:  \na. The volume of the frustum.  \nb. The surface area of the frustum.  \n6cm  \nBy similar triangle;􀝔 3  \n~~ ~~ = ~~ ~~  \n􀝔 + 5 6  \n6􀝔 = 3 (􀝔 + 5)  \n6􀝔 = 3􀝔 + 15  \n6􀝔 − 3􀝔 = 15 3􀝔 = 15  \n􀝔 = 5􀜿􀝉  \na. volume of frustum = Volume of big pyramid – Volume of small pyramid = 13 􀜮2 􀜪 − 13 􀝈 2 ℎ  \n= 13 × 6 × 6 × 10 − 13 × 3 × 3 × 5 = 13 × 360 – 13 × 45  \n= 120 − 15  \n= 105􀜿􀝉3  \nb. surface area = area of big pyramid – area of small pyramid  \nSurface area of big pyramid = 4(½BH) + L2 (a + b)2 = 102 + 32  \n= 4(½ X 6 X 10.44) + 6 X 6 = 100 + 9 = 4(3 X 10 .44) + 36 = 109  \n= 4(31 .32) + 36  \n= 125.28 + 36  \n= 161. 28cm2  \nSurface area of small pyramid = 4(½bh) + L2  \n= 4(½ X 3 X 5 .22) + 3 X 3 = 4(7 .83) + 9  \n= 31.32 + 9  \n= 40 .32cm2  \nSurface area of frustum = 161. 28cm2 – 40.32cm2  \n= 120cm2  \n(a + b) = √109  \n(a + b) = 10.44  \na2 = 52 + 1.52 = 25 + 2.25  \n= 27.25  \na = √27 .25  \na = 5.22  \nCalculating compound/composite shapes.  \n3. A pencil is sharpened at both ends as shown below. Calculate the surface area of the whole  \npencil. 3cm    \n~~ ~~ 10cm ~~  ~~Solution:  \nSurface area of pencil = curved surface area of cylinder + 2(curved surface area of a cone)  \n=2􀟨􀝎ℎ + 2(􀟨􀝎􀝈􀟨)  \n= 2 × 227 × 0 . 25 × 4 + 2 (227 × 0 . 25 × 3 .01)  \n= 6.29 + 4.73  \n= 11. 02􀜿􀝉2  \n4. The solid below is a cylinder surmounted by a hemispherical bowl. Calculate its  \na. Total surface area.  \nb. Volume.  \nSolution:  \n7cm  \n10cm  \na. Total surface area = surface area of cylinder + surface area of sphere = 2􀟨􀝎ℎ + 􀟨􀝎2 + 2􀟨􀝎2  \n22  \n= 􀟨􀝎 (2ℎ + 􀝎) + 2 × × 7 × 7  \n7  \n= 227 × 7 (2(10) + 7) + 44 × 7  \n= 22 (27) + 308  \n= 594 + 308  \n= 902􀜿􀝉2  \nb. Volume = volume of cylinder + volume of hemisphere.  \n= 􀟨􀝎2 ℎ + 23 􀟨􀝎3  \n= 227 × 7 × 7 × 10 + 23 × 227 × 7 × 7 × 7 = 1540􀜿􀝉3 + 718. 6667􀜿􀝉3  \n= 2258. 6667􀜿􀝉3  \n= 2259􀜿􀝉3 (􀝊􀝁􀜽􀝎􀝁􀝏􀝐 􀜿􀝉3)  \n------Step 5: Evaluation.  \nMode: whole class.  \nTeacher’s activities: asks the students to calculate the volume of a composite solid consisting of a cube of edge 28cm and a square based pyramid of height 16cm.  \nStudents’ activities: answer the question thus.  \nSolution:  \n28cm  \nVolume = volume of cube + volume of pyramid  \n= 􀜮 × 􀜮 × 􀜪 + 13 × 􀜮 × 􀜮 × 􀜪  \n= (28 × 28 × 28 ) + 13 × 28 × 28 × 16  \n= 21952 + 4181.3333  \n= 26133. 3333􀜿􀝉3  \n= 26133􀜿􀝉3 (􀝊􀝁􀜽􀝎􀝁􀝏􀝐 􀜿􀝉3)  \nAssignment:  \nA conical funnel 12cm deep and 15cm in diameter is full of liquid. It is emptied into a cylindrical tin 10cm in diameter. Calculate the height of the liquid in the tin.  \n\n| \u003Cbr>|  |\n| --- | --- |\n|  | |","cbCaii94lsBNT7YS","https://ap.wps.com/l/cbCaii94lsBNT7YS","pdf",584615,"English","# Surface area and volume of a frustum\n## Frustum of a cone\n## Frustum of a pyramid\n## Composite solids involving cylinders and cones\n## Total surface area and volume for combined shapes\n## Assignment: conical funnel to cylindrical tin","[{\"question\":\"What is the method for finding the surface area of a frustum of a cone?\",\"answer\":\"Calculate surface area as the area of the larger cone minus the area of the smaller cone, using the slant height derived from right-triangle geometry (often via Pythagoras).\"},{\"question\":\"In the assignment, how is the liquid height in the cylinder determined?\",\"answer\":\"Use volume conservation: equate the volume of liquid in the conical funnel (cone frustum/full cone as given by depth and diameter) to the volume of liquid in the cylindrical tin, then solve for the cylinder liquid height.\"}]","Surface area and volume of frustum - Practice problems | PDF"]