[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-297573-105":59,"doc-detail-297573-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","mensuration-of-solid-shapes-surface-area-of-solid-shapes-lesson-plan","Mensuration of Solid Shapes - Surface Area of Solid Shapes - Lesson Plan","","Lesson plan for teaching mensuration focused on the surface area of solid shapes. The objectives cover identifying solid shapes, recalling and applying surface-area formulas, recognizing cross-sections for each solid, and computing total surface areas. Instructional flow includes eliciting students’ prior ideas, guided recall of formulas for cube, cuboid, cylinder, cone, triangular prism, and rectangular pyramid, followed by worked examples (cube, cuboid, cylinder cases, and a composite prism/pyramid shape). Includes application, evaluation problems, and an assignment on pyramid total 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/mensuration-of-solid-shapes-surface-area-of-solid-shapes-lesson-plan/297573/",{"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/mensuration-of-solid-shapes-surface-area-of-solid-shapes-lesson-plan/297573.png","ImageObject",300,407,{"name":92,"@type":93},"Kyle","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-29","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},"What are the specific objectives of the lesson on surface area of solid shapes?","Question",{"text":113,"@type":114},"Students should identify solid shapes, recall surface-area formulas, identify cross-sections for each solid, and calculate the surface area of solid shapes.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"How is the surface area of a cube with edge 10.5 cm calculated?",{"text":118,"@type":114},"Use Surface area = 6a²: 6 × (10.5 × 10.5) = 661.5 cm².",{"name":120,"@type":111,"acceptedAnswer":121},"Why does the cylinder problem consider multiple cases?",{"text":122,"@type":114},"Curved and total surface areas depend on whether the cylinder has open ends or closed ends; therefore calculations are done for hollow with two open ends, one open end, and two closed ends.","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},297573,1789745860,{"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":24,"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":144},3985741905716,"https://ap-avatar.wpscdn.com/davatar_994ba38a5ba835b3df7d355c54d3ed8d","Date: 22-04-2024  \nTheme: Mensuration of solid shapes.  \nUnit topic: Mensuration.  \nLesson topic: Surface area of solid shapes.  \nTime: 11:10-12:30  \nDuration: 80 minutes  \nPeriod: 5th  \nSpecific objectives: At the end the lesson, the students should be able to:  \n1. Identify a solid shape;  \n2. Recall the formulae for calculate the surface area of solid shape;  \n3. Identify the cross section for each solid shape;  \n4. Calculate the surface area of solid shapes.  \nInstructional resources: Some solid shapes.  \nPresentation:  \n------Step 1: Identification of students’ prior ideas.  \nMode: whole class.  \nTeacher’s activities: leads the students in identifying the surface area of the solid shapes.  \nCube Surface area = 6􀝈2  \nCuboids Surface area = 2 (􀝈􀜾 + 􀝈ℎ + 􀜾ℎ)  \nCylindercurved surface area = 2􀟨􀝎ℎ Total surface area = 2􀟨􀝎ℎ + 2􀟨􀝎2  \n= 2􀟨􀝎(ℎ + 􀝎)  \nSurface  \narea = sum of the area of the 3 squares + the areas of the 2 triangles.  \nTriangular prism  \nCone Curved surface area = 􀟨􀝎􀝈 Total surface area = 􀟨􀝎􀝈 + 􀟨􀝎2  \n= 􀟨􀝎 (􀝈 + 􀝎)  \nRectangular pyramid  \nTotal surface area = sum of the 4 congruent triangular faces and the base.  \nStudents’ activities: recall and take note.  \nMode: whole class.  \nTeacher’s activities: leads the students in calculating the surface area of the mentioned solids.  \n1. Calculate the surface area of a cube with edge of 10.5cm.  \nSolution:  \nSurface area = 6􀝈2  \n= 6 (10 .5 × 10. 5)  \n= 6 (110 . 25)  \n= 661. 5􀜿􀝉2  \n2. The total surface area of a rectangular solid is 130° . If the solid is 7cm long and 5cm wide, calculate the height .  \nSolution: Let the height be h;  \nSurface area = 2 (􀝈􀜾 + 􀜾ℎ + 􀝈􀜾)  \n130 = 2 (7 × 5 + 5 × ℎ + 7 × ℎ)  \n130 = 2 (35 + 5ℎ + 7ℎ)  \n130 = 2 (35 + 12ℎ)  \n130 = 70 + 24ℎ  \n130 − 70 = 24ℎ  \n60 = 24ℎℎ = 60  \n24  \nℎ = 2. 5􀜿􀝉  \nStudents’ activities: discover that the formula can be applied to find any missing dimension.------Step 3: Discussion.  \nMode: whole class.  \nTeacher’s activities: leads the students in calculating all the possible cases of a cylinder.  \n3. A cylindrical container has a radius of 7cm and height of 6cm. Calculate curved surface area  \nusing all the possible cases Solution:  \na. Case 1: Hollow with two open ends; Curved surface area = 2􀟨􀝎ℎ  \n= 2 × 272 × 7 × 6 􀜥􀜵􀜣 = 264􀜿􀝉2  \n6cm  \nb. Case 2: One open end;  \nCurved surface area = 2􀟨􀝎ℎ + 􀟨􀝎2  \n= 􀟨􀝎 (2ℎ + 􀝎) 7cm  \n= 22 × 7 (2 × 6 + 7) 6cm  \n7  \n= 22 (19)  \n􀜥􀜵􀜣 = 418􀜿􀝉2  \nc. Case 3: Two closed ends;  \nCurved surface area = 2􀟨􀝎ℎ + 2􀟨􀝎2  \n= 2􀟨􀝎ℎ + 􀝎)  \n= 2 × 272 × 7 (6 + 7) 6cm  \n= 44 (13)  \n= 572􀜿􀝉2  \n7cm  \n------Step 4: Application.  \nMode: whole class.  \nTeacher’s activities: leads the students in calculating the areas of prism and pyramid.  \n4. Calculate the surface area of the shape below.  \nSolution:  \nBy Pythagoras theorem;  \n􀝔 2 = 52 − 32  \n= (5 + 3)(5 − 3)  \n= 8 × 2  \n= 16  \n􀝔 = 4􀜿􀝉  \nStudents’ activities: calculate the surface area of the above shape.  \nSurface area = 2{½(􀜽 + 􀜾)ℎ} + 􀝈􀜾 + 􀝈􀜾 + 􀝈􀜾 + 􀝈􀜾  \n= 2{½(9 + 12)4} + 5 × 13 + 12 × 13 + 9 × 13 + 4 × 13  \n= 84 + 65 + 156 + 117 + 52  \n= 474􀜿􀝉2 ------Step 5: Evaluation.  \nMode: whole class.  \nTeacher’s activities: asks the students to solve this.  \n5. A square – based pyramid with height of 13cm and square base size of 6cm is given below. Calculate the surface area of the pyramid.  \nStudents’ activities: answer the questions thus:  \nSolution:  \n13􀜿􀝉 􀝔  \n3cm  \n6cm  \nSurface area = area of square base + sum of 4 triangles.  \nArea of square base = 6􀜿􀝉 × 6􀜿􀝉  \n= 36􀜿􀝉2 Area of one triangle = ½􀜾􀝔  \nBut 􀝔 2 = 132 + 32 (􀝑􀝏􀝅􀝊􀝃 􀜲􀝕􀝐ℎ􀜽􀝃􀝋􀝎􀜽􀝏 􀝐ℎ􀝁􀝋􀝎􀝁􀝉)  \n= 169 + 9  \n= 178  \n􀝔 = √178  \n􀝔 = 13. 34􀜿􀝉  \nArea of 1 triangle = ½ × 6􀜿􀝉 × 13. 34􀜿􀝉  \n= 40. 08􀜿􀝉2  \nArea of 4 congruent triangles = 4 × 40.02  \n= 160. 08􀜿􀝉2  \nAssignment:  \nCalculate the total surface area of a pyramid with square base 7cm long and height 10cm.  \nReferences:  \n􀀹 New General Mathematics for SS 1 by J B Channon etal.  \n􀀹 New Concept Mathematics for SS 1 by H N Odogwu etal.","cbCailgLTwJqXvLm","https://ap.wps.com/l/cbCailgLTwJqXvLm","pdf",610778,"English","# Presentation\n## Step 1: Identification of students’ prior ideas\n## Step 3: Discussion\n## Step 4: Application\n## Step 5: Evaluation\n# Assignment\n# References","[{\"question\":\"What are the specific objectives of the lesson on surface area of solid shapes?\",\"answer\":\"Students should identify solid shapes, recall surface-area formulas, identify cross-sections for each solid, and calculate the surface area of solid shapes.\"},{\"question\":\"How is the surface area of a cube with edge 10.5 cm calculated?\",\"answer\":\"Use Surface area = 6a²: 6 × (10.5 × 10.5) = 661.5 cm².\"},{\"question\":\"Why does the cylinder problem consider multiple cases?\",\"answer\":\"Curved and total surface areas depend on whether the cylinder has open ends or closed ends; therefore calculations are done for hollow with two open ends, one open end, and two closed ends.\"}]","Mensuration of Solid Shapes - Surface Area of Solid Shapes - Lesson Plan | PDF",13]