This chapter deals with the measurement of solid figures: their surface areas and volumes. Students study solid shapes such as cubes, cuboids, cylinders, cones, spheres, and hemispheres, learning to compute their curved surface areas, total surface areas, and volumes. The chapter also covers combinations of solids, where two or more shapes are joined together, and the conversion of solids from one shape to another through melting and recasting.
The formulas for each solid are systematic and easy to remember when their derivations are understood. A cylinder is formed by rolling a rectangle, a cone is a revolution of a right triangle, and a sphere is the most symmetric of all solids. Recognising the relationship between these shapes and their flat counterparts makes the formulas intuitive rather than a burden to memorise.
Beyond the pure formulas, this chapter's real challenge lies in applications: finding the volume of a vessel made of a cylinder joined to a hemisphere, calculating the number of smaller objects that can be melted to form a larger one, and computing the surface area of a decorative solid. These problems combine mensuration with algebra and are a rich source of long-answer questions in board examinations.
For a cuboid of length l, breadth b, and height h: Total surface area = 2(lb + bh + hl) Lateral surface area = 2h(l + b) Volume = l x b x h Diagonal = sqrt(l^2 + b^2 + h^2)
For a cube of edge a: Total surface area = 6a^2 Lateral surface area = 4a^2 Volume = a^3 Diagonal = a sqrt(3)
These are the simplest solids and serve as the basis for comparing with the curved solids that follow.
For a cylinder of radius r and height h: Curved surface area = 2 pi r h Total surface area = 2 pi r (r + h) Volume = pi r^2 h
The curved surface area is the area of the rectangle obtained by unrolling the curved surface: its length is the circumference 2 pi r and its width is h.
Example: A cylinder of radius 7 cm and height 10 cm. Volume = pi x 49 x 10 = 490 pi cm^3. Total surface area = 2 pi x 7 x 17 = 238 pi cm^2.
For a cone of radius r, height h, and slant height l: l = sqrt(r^2 + h^2) Curved surface area = pi r l Total surface area = pi r (l + r) Volume = (1/3) pi r^2 h
The volume of a cone is one third of the volume of a cylinder with the same base and height, a result that can be demonstrated experimentally.
Example: A cone of radius 6 cm and height 8 cm has slant height l = sqrt(36 + 64) = 10 cm. Curved surface area = pi x 6 x 10 = 60 pi cm^2. Volume = (1/3) pi x 36 x 8 = 96 pi cm^3.
For a sphere of radius r: Surface area = 4 pi r^2 Volume = (4/3) pi r^3
For a hemisphere of radius r: Curved surface area = 2 pi r^2 Total surface area = 3 pi r^2 (curved part plus the flat circular base pi r^2) Volume = (2/3) pi r^3
Example: A sphere of radius 3 cm has volume (4/3) pi x 27 = 36 pi cm^3 and surface area 4 pi x 9 = 36 pi cm^2.
Note the coincidence in this example: for r = 3, the numerical value of the volume equals that of the surface area, but the units differ (cm^3 versus cm^2).
When two or more solids are joined, the total surface area is the sum of the exposed curved and flat surfaces, being careful not to count the hidden joining surfaces. The volume, however, is simply the sum of the volumes of the individual solids.
Common combinations: - A cylinder with a hemisphere on top (capsule, vessel). - A cylinder with a cone on top (tent, tower). - A cuboid with a hemisphere cut out. - A cone with a hemisphere base (ice cream cone).
Example: A toy in the shape of a cone (radius r, height h) surmounted on a hemisphere of radius r: Total volume = (1/3) pi r^2 h + (2/3) pi r^3 Total surface area = pi r l + 2 pi r^2 (excluding the common base).
When a solid is melted and recast into another shape, the volume remains unchanged. This principle gives the equation: Volume of original solid = Volume of new solid
Example: A cone of radius 3 cm and height 12 cm is melted to form a cylinder of radius 6 cm. Let the cylinder height be H. (1/3) pi x 9 x 12 = pi x 36 x H 36 pi = 36 pi H, so H = 1 cm.
Similarly, water filling problems use the volume of the displaced water or the rate of flow through a pipe of given cross-section. These problems are solved by equating volumes and solving for the unknown dimension.
| Solid | Curved/Total Surface Area | Volume |
|---|---|---|
| Cube (edge a) | 6a^2 (total) | a^3 |
| Cuboid (l, b, h) | 2(lb + bh + hl) | l b h |
| Cylinder (r, h) | 2 pi r h (curved), 2 pi r (r + h) | pi r^2 h |
| Cone (r, h, l) | pi r l (curved), pi r (l + r) | (1/3) pi r^2 h |
| Sphere (r) | 4 pi r^2 | (4/3) pi r^3 |
| Hemisphere (r) | 2 pi r^2 (curved), 3 pi r^2 (total) | (2/3) pi r^3 |
| Relation | Formula |
|---|---|
| Cone slant height | l = sqrt(r^2 + h^2) |
| Volume ratio cone:cylinder (same base, height) | 1:3 |
| Recast/melting | Volume conserved |
| Filling with water | Flow volume = cross-section area x length |
Surface areas and volumes bring mensuration to life by measuring the solids we encounter every day. From simple cubes and cylinders to combinations like cones on hemispheres, the formulas provide exact measurements of both the material needed to cover a surface and the space a solid occupies. The principle of volume conservation in melting and recasting links mensuration to algebra, producing rich multi-step problems that reward careful reasoning. This chapter is a high-weightage topic in the board examination, and mastering its formulas and problem-solving strategies ensures substantial marks while building the spatial reasoning needed for higher studies in geometry and calculus.