[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-297388-105":59,"doc-detail-297388-en":131},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":124,"head_meta":126,"extra_data":128,"updated_unix":130},105,"en","rd-sharma-solution-class-9-maths-chapter-18-surface-area-and-volume-of-cuboid-and-cube-exercise-181","RD Sharma Solution Class 9 Maths Chapter 18 - Surface Area and Volume of Cuboid and Cube - Exercise 18.1","","Exercise 18.1 covers problems on surface areas of cuboids and cubes using standard geometry formulas. It computes lateral surface area and total surface area for given cuboid dimensions, finds lateral and total surface areas for a cube, and derives the ratio between them. Word problems include sheet-covering requirements for a wooden block, cost of whitewashing walls and ceiling, comparisons using combined solids, and surface areas formed by cutting cubes into smaller cubes. It also solves a hall-height equation and a tiled cubical tank cost, concluding with percentage increase in cube surface area.",{"@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/rd-sharma-solution-class-9-maths-chapter-18-surface-area-and-volume-of-cuboid-and-cube-exercise-181/297388/",{"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/rd-sharma-solution-class-9-maths-chapter-18-surface-area-and-volume-of-cuboid-and-cube-exercise-181/297388.png","ImageObject",300,407,{"name":92,"@type":93},"Ivy","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-21","2026-09-18",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",16,{"@type":107,"mainEntity":108},"FAQPage",[109,115,119],{"name":110,"@type":111,"acceptedAnswer":112},"How do you find the total surface area of a cuboid with given length, breadth, and height?","Question",{"text":113,"@type":114},"Use TSA = 2(lb + bh + hl). Substitute the given dimensions and simplify to get the total area in square centimeters.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"What is the lateral surface area of a cube, and how is it related to total surface area?",{"text":118,"@type":114},"For a cube of side a, LSA = 4a² and TSA = 6a². The ratio TSA:LSA becomes 6a²:4a² = 3:2.",{"name":120,"@type":111,"acceptedAnswer":121},"How do you calculate the number of paper sheets needed to cover a box surface?",{"text":122,"@type":114},"Compute the box surface area, then divide by the area of one sheet. For each sheet of side 40 cm, its area is 40×40 = 1600 cm², so sheets = (surface area)/(1600).","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},297388,1789744201,{"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":105,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":130,"read_time":36},549758252649,"https://ap-avatar.wpscdn.com/avatar/8000253669c5317157?_k=1778319167496531819","Exercise 18.1 Page No: 18.14  \nQuestion 1: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.  \nSolution:  \nGiven, Dimensions of a cuboid: Length (l) = 80 cm  \nBreadth (b) = 40 cm Height (h) = 20 cm  \nWe know that, Total Surface Area = 2[lb + bh + hl]  \nBy substituting the values, we get  \n= 2[(80)(40)+(40)(20)+(20)(80)]  \n= 2[3200+800+1600]  \n= 2[5600]  \n= 11200  \nTherefore, Total Surface Area = 11200 cm2  \nNow,  \nLateral Surface Area = 2[l + b]h  \n= 2[80+40]20  \n= 40[120]  \n= 4800  \nThus, Lateral Surface Area is 4800 cm2.  \nQuestion 2: Find the lateral surface area and total surface area of a cube of edge 10 cm.  \nSolution:  \nSide of a Cube = 10 cm (Given)  \nFormula for Cube Lateral Surface Area = 4 side2 Cube Lateral Surface Area = 4(10 x 10)= 400 cm2  \nTotal Surface Area = 6 Side2 = 6(102)  \n= 600 cm2  \nQuestion 3: Find the ratio of the total surface area and lateral surface area of a cube.  \nSolution:  \nTotal Surface Area of the Cube (TSA) = 6 Side2  \nLateral surface area of the Cube (LSA) = 4 Side2  \nNow,  \nRatio of TSA and LSA = (6 Side2)/ (4 Side2) = 3/2 or 3:2 .  \nQuestion 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?  \nSolution:  \nThe dimensions of the wooden block are: Length (l) = 80cm  \nBreadth (b) = 40cm Height (h) = 20cm  \nSurface Area of the wooden box = 2[lb + bh + hl]= 2[(80x40)+(40x20)+(20x80)]  \n= 2[5600]  \n= 11200  \nSurface Area of the wooden box is 11200 cm2  \nThe Area of each sheet of the paper = 40×40 cm2 = 1600 cm2  \nNow,  \nThe total number of sheets required = (Surface area of the box )/(Area of one sheet of paper)  \n= 11200/1600  \n= 7  \nTherefore, Marry would require 7 sheets. Answer!!  \nQuestion 5: The length, breadth, and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 m2.  \nSolution:  \nFormula: Total Area to be washed = lb + 2(l + b)h   (1)  \nWhere, l = length , b = breadth and h = height.  \nFrom given: Length = l = 5 m  \nBreadth = b = 4 m  \nHeight = h = 3 m  \nTotal area to be white washed = (5x4) + 2(5 + 4)3 (using (1))  \n= 74  \nTotal area to be white washed is 74 m2  \nNow, cost of white washing 1 m2 is Rs. 7.50 (Given) Therefore, the cost of white washing 74 m2 = (74 x 7 .50)= Rs. 555  \nQuestion 6: Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.  \nSolution:  \nLet breadth of the cuboid = a  \nThen, length of the new cuboid = 3a and Height of the new cuboid = a  \nNow,  \nTotal surface area of the new cuboid (TSA) = 2(lb+bh+hl)  \n= 2(3a x a + a x a + a x 3a)  \n= 14a2  \nAgain,  \nTotal Surface area of three cubes = 3 x (6 side2 )  \n= 3 x 6a2  \n= 18a2  \nTherefore, ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a2/18a2 = 7/9 or 7:9  \nTherefore, required ratio is 7:9 . Answer.  \nQuestion 7: A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.  \nSolution:  \nEdge of the cube = 4 cm (Given)  \nVolume of the cube = Side3 = 43 = 64  \nVolume of the cube is 64cm3  \nAgain,  \nEdge of the cube = 1 cm3  \nSo, Total number of small cubes = 64cm3/1cm3 = 64  \nAnd, total surface area of all the cubes = 64 x 6 x 1 = 384 cm2  \nQuestion 8: The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.  \nSolution:  \nDimensions of the hall are: Length = 18m  \nWidth = 12m  \nFrom statement:  \nArea of the floor and the ","cbCaibzDuLww8QFT","https://ap.wps.com/l/cbCaibzDuLww8QFT","pdf",195842,"English","# Exercise 18.1\n## Lateral and total surface area of a cuboid\n## Lateral and total surface area of a cube\n## Ratio of total to lateral surface area for a cube\n## Surface-area-based paper covering for a wooden block\n## Whitewashing walls and ceiling cost\n## Ratio after forming a new cuboid from cubes\n## Total surface area after cutting a cube into small cubes\n## Finding hall height from area relationships\n## Tile cost for a cubical tank excluding base\n## Percentage increase in surface area when cube edge increases 50%","[{\"question\":\"How do you find the total surface area of a cuboid with given length, breadth, and height?\",\"answer\":\"Use TSA = 2(lb + bh + hl). Substitute the given dimensions and simplify to get the total area in square centimeters.\"},{\"question\":\"What is the lateral surface area of a cube, and how is it related to total surface area?\",\"answer\":\"For a cube of side a, LSA = 4a² and TSA = 6a². The ratio TSA:LSA becomes 6a²:4a² = 3:2.\"},{\"question\":\"How do you calculate the number of paper sheets needed to cover a box surface?\",\"answer\":\"Compute the box surface area, then divide by the area of one sheet. For each sheet of side 40 cm, its area is 40×40 = 1600 cm², so sheets = (surface area)/(1600).\"}]","RD Sharma Solution Class 9 Maths Chapter 18 - Surface Area and Volume of Cuboid and Cube - Exercise 18.1 | PDF"]