Surface Areas and Volumes - Cones Spheres Class 9 Maths Chapter 13 CBSE, NCERT

Surface Areas and Volumes: Cones, Spheres, and Hemispheres

This chapter in CBSE Class 9 Mathematics extends the study of surface areas and volumes from simple solids (cubes, cuboids, cylinders) to more complex shapes — right circular cones, spheres, and hemispheres. These shapes appear frequently in everyday life: ice cream cones are cones, cricket balls and footballs are spheres, and cupolas on buildings are often hemispheres. Understanding the formulas for their curved surface areas, total surface areas, and volumes — and knowing when to apply each — is essential for solving practical problems in engineering, architecture, packaging, and construction.

For a right circular cone with base radius r, vertical height h, and slant height l = √(r² + h²), the curved surface area (CSA) is πrl, and the total surface area (TSA) is πr² + πrl = πr(r + l). The volume of a cone is (1/3)πr²h — exactly one-third the volume of a cylinder with the same base and height. For example, a cone with radius 7 cm and slant height 25 cm has a CSA = π×7×25 = 550 cm² and a TSA = π×7(7+25) = 704 cm². The volume = (1/3)π×7²×√(25²−7²) = (1/3)π×49×24 = 1232 cm³. Note that the height h = √(l² − r²) = √(625−49) = √576 = 24 cm. For a sphere of radius r, the surface area is 4πr², and the volume is (4/3)πr³. The surface area formula can be derived by imagining a sphere is composed of many small pyramids meeting at the centre — the volume of each small pyramid plus up to the surface area gives 4πr². For a hemisphere (half of a sphere with a flat circular base), the curved surface area is 2πr², the total surface area is 3πr² (CSA + base area = 2πr² + πr²), and the volume is (2/3)πr³.

Practical problems involve combining these solids. For example, a toy consists of a cone mounted on a hemisphere of the same radius. The total surface area is the sum of the cone's CSA and the hemisphere's CSA (the base areas are not exposed). If a sphere and a cone have the same volume, we can find unknown dimensions by equating (4/3)πR³ = (1/3)πr²h → 4R³ = r²h. The volume of material in a hollow sphere (spherical shell) is the difference between the volumes of the outer and inner spheres: V = (4/3)π(R³ − r³). When a solid is melted and recast into a different shape, the volume remains constant — this principle is used to solve problems such as how many small spheres can be made from a larger sphere, or what dimensions a cone or cylinder will have if a cube is melted and recast into it. For example, how many lead shots of diameter 2 cm can be made from a solid sphere of diameter 14 cm? Volume of large sphere = (4/3)π×7³ = 1436.03 cm³. Volume of one lead shot = (4/3)π×1³ = 4.19 cm³. Number = 1436.03/4.19 = 342.7, so approximately 342 shots. The ratio of similar solids is important: if two similar cones have their radii in the ratio 2:3, their surface areas are in the ratio 4:9 (square of the ratio) and their volumes are in the ratio 8:27 (cube of the ratio).

• Cone: CSA = πrl, TSA = πr(r+l), V = (1/3)πr²h where l = √(r²+h²). Sphere: SA = 4πr², V = (4/3)πr³.
• Hemisphere: CSA = 2πr², TSA = 3πr², V = (2/3)πr³. Use CSA when the base is not included.
• Combined solids: add only the exposed surface areas; for melted-and-recast problems, volume is conserved.
• The volume of a cone equals 1/3 the volume of a cylinder with the same base and height.
• Ratios: corresponding lengths in similar solids scale as k, areas as k², and volumes as k³.