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Frustum of a cone.  \n¾ Problems involving converting one type of metallic solid into another and other mixed problems. Problems with combination of not more than two different solids.  \nTrend Analysis  \n\n| List of Concepts | 2018 |  | 2019 |  | 2020 |  |\n| --- | --- | --- | --- | --- | --- | --- |\n|  | Delhi\u003Cbr>2 Q (3 M)\u003Cbr>1 Q (4 M) | Outside\u003Cbr>Delhi | Delhi\u003Cbr>1 Q (4 M) | Outside Delhi\u003Cbr>1 Q (3 M)\u003Cbr>1 Q (4 M) | Delhi\u003Cbr>1 Q (1 M) | Outside Delhi\u003Cbr>3 Q (2 M)\u003Cbr>1 Q (4 M) |\n| Surface Areas and volumes |  |  |  |  |  |  |\n| Conversion of one type of metallic solid into another |  |  |  |  | 1 Q (1 M) |  |\n| Frustum of a cone | 1 Q (4 M) |  | 1 Q (4 M) |  | 2 Q (4 M) | 1 Q (1 M) |\n\nTOPIC-1  \nSurface Areas and Volumes  \n  Revision Notes   \n¾ A sphere is a perfectly round geometrical object in three-dimensional space .  \n¾ A Cone is a three dimensional geometric shape tapers smoothly from a flat base to a point called the apex or vertex.  \n¾ A cylinder is a solid or a hollow object that has a circular base and a circular top of the same size.  \n¾ A hemisphere is half of a sphere.  \nKnow the Formulae   \n¾ Cuboid:  \nLateral surface area or area of four walls = 2(l + b)h  \nTotal surface area = 2(lb + bh + hl)  \nVolume = l × b × h  \nDiagonal = l2 + b2 + h2  \nHere, l is length, b is breadth and h is height of the cuboid  \n¾ Cube:  \nLateral surface area or area of four walls = 4 × (edge)2  \nTotal surface area = 6 × (edge)2  \nVolume = (edge)3  \nDiagonal of a cube = 3 × edge.  \n¾ Right circular cylinder:  \nArea of base or top face = πr²  \nArea of curved surface or curved surface area = perimeter of the base × height  \n= 2πrh  \nTotal surface area (including both ends) = 2πrh + 2πr² = 2πr(h + r)  \nVolume = (Area of the base × height)= πr²h  \nHere, r is the radius of base and h is the height of the right circular cylinder.  \n¾ Right circular hollow cylinder:  \nTotal surface area = (External surface area + internal surface area) + (Area of brim)  \n= (2πRh + 2πrh) + 2(πR² – πr²)  \n= [2πh(R+ r) + 2π(R² – r²)]  \n= [2π(R + r)(h + R – r)]  \nCurved surface area = (2πRh + 2πrh)= 2πh(R + r)  \nVolume of the material used = (External volume)–(Internal volume)  \n= πR²h – πr²h = πh(R² – r²)  \nHere, R and r are the external and internal radii and h is the right circular height of the right circular hollow cylinder.  \n¾ Right circular cone:  \nSlant height, l = h2 +r2  \nArea of curved surface = πrl = πr h2 + r 2  \nTotal surface area = Area of curved surface + Area of base  \n= πrl + πr² = πr(l + r)  \nVolume = 13 πr²h  \nHere, r, h and l are the radius, vertical height and slant height respectively of the right circular cone.  \n¾ Sphere:  \nSurface area = 4πr²  \nVolume = 4 πr3  \n3  \nHere, r is the radius of the sphere.  \n¾ Spherical shell:  \nSurface area (outer) = 4πR2  \nVolume of material = 43 πR3 – 43 πr3 = 43 π(R3 – r3)  \nHere, R and r are the external and internal radii of the spherical shell.  \n¾ Hemisphere:  \nArea of curved surface = 2πr²  \nTotal surface area = Area of curved surface + Area of base = 2πr² + πr²  \n= 3πr²  \nVolume = 2 πr3 3  \nHere, r is the radius of the hemisphere.  \nKnow the Terms   \n¾ The platonic solids also called the regular solids or regular polyhedra. 5 such solids are : dodecahedron, icosahedron, octahedron and tetrahedron.  \n¾ Greek mathematician Plato equated tetrahedron with the ‘element’ fire, the cube with earth, the icosahedron with water, the octahedron with air and dodecahedron with the stuff of which the constellations and heavens were made.  \n¾ The stone of platonic solids are kept in Ashmolean Museum in Oxford.  \n¾ The fonds of Archimedes carried a sculpture consisting of a sphere and cylinder circumscribing it.  \nHow is it done on the GREENBOARD?  \nQ.1. The circumference of the base of a conical tent is 44 m. If the height of tent i","cbCaigsb7DYzFvpY","https://ap.wps.com/l/cbCaigsb7DYzFvpY","pdf",2460583,35,"English","# Surface Areas and Volumes\n## Revision Notes\n## Know the Formulae\n## Know the Terms\n## GREENBOARD Questions and Solutions","[{\"question\":\"How does the document handle problems and examples?\",\"answer\":\"It includes worked solutions and question sets, such as finding canvas length for a conical tent and ratio-based volume or surface-area problems using standard geometry relations.\"}]","Surface Areas and Volumes - Revision Notes - Know the Formulae - Q&A | PDF",88]