Surface Areas & Volumes – CBSE Class 10 Mathematics
Goal: Master the concepts, formulae, derivations, and problem‑solving techniques required for IIT‑JEE and other competitive exams.
Scope: All 3‑D solids that appear in the CBSE syllabus – cubes, cuboids, cylinders, cones, spheres, pyramids, frustums, torus, etc.
Style: Clear, rigorous, exam‑oriented explanations, with derivations, key points, and a hard practice question at the end.
1. Basic Notions
| Quantity | Symbol | Units | Notes |
|---|---|---|---|
| Surface area | (S) | ( \text{cm}^2,\ \text{m}^2) | Total area of all faces. |
| Lateral surface area | (S_{\text{lat}}) | – | Area of side faces only. |
| Volume | (V) | ( \text{cm}^3,\ \text{m}^3) | Capacity of the solid. |
| Radius | (r) | – | Distance from centre to the boundary. |
| Height | (h) | – | Perpendicular distance between parallel faces. |
| Slant height | (l) | – | Distance from apex to a point on the base (for right pyramids & cones). |
Tricky Point: In many problems the height of a cone or a pyramid is the perpendicular distance from the apex to the base, not the slant height.
2. Surface Area – General Strategy
- Find the area of each face (rectangles, squares, triangles, semicircles, etc.).
- Add all areas.
- For regular solids (e.g., cube, cube‑like pyramid), use symmetry to multiply a single face area by the number of identical faces.
- If asked for “total surface area” use (S = S_{\text{lat}} + S_{\text{base}}).
- For solids of revolution (cylinders, cones, spheres) use the general formulae derived below.
3. Surface Area & Volume of Basic Solids
| Solid | Surface Area | Volume |
|---|---|---|
| Cube (edge (a)) | (S = 6 a^{2}) | (V = a^{3}) |
| Cuboid (edges (a,b,c)) | (S = 2(ab + bc + ca)) | (V = abc) |
| Sphere (radius (r)) | (S = 4\pi r^{2}) | (V = \frac{4}{3}\pi r^{3}) |
| Right Circular Cylinder (radius (r), height (h)) | (S = 2\pi r(h + r)) | (V = \pi r^{2}h) |
| Right Circular Cone (radius (r), height (h), slant height (l = \sqrt{r^{2}+h^{2}})) | (S = \pi r(r + l)) | (V = \frac{1}{3}\pi r^{2}h) |
| Right Circular Cone – Lateral Only | (S_{\text{lat}} = \pi r l) | – |
| Right Pyramid (regular base (n)-gon of side (a), height (h)) | (S = \frac{n a}{2}\cot\frac{\pi}{n} + n\frac{a}{2}\sqrt{h^{2} + \frac{a^{2}}{4}\cot^{2}\frac{\pi}{n}}) | (V = \frac{1}{3} \times \text{(Base area)} \times h) |
| Frustum of a Right Circular Cone (top radius (r_1), bottom radius (r_2), height (h)) | (S = \pi (r_{1}+r_{2})\sqrt{(r_{2}-r_{1})^{2}+h^{2}} + \pi (r_{1}^{2}+r_{2}^{2})) | (V = \frac{1}{3}\pi h (r_{1}^{2}+r_{1}r_{2}+r_{2}^{2})) |
Tricky Point: For a frustum, the slant height (l = \sqrt{(r_{2}-r_{1})^{2}+h^{2}}) appears in the lateral area.
4. Derivations (Proofs)
4.1 Cylinder Surface Area
Lateral area: Unroll the curved surface → rectangle of width (2\pi r) (circumference) and height (h).
[
S_{\text{lat}} = 2\pi r h
]
Base area: Two circles of radius (r).
[
S_{\text{base}} = 2\pi r^{2}
]
Total surface area:
[
S = S_{\text{lat}} + S_{\text{base}} = 2\pi r(h + r)
]
4.2 Cone Surface Area
Lateral area: Unroll the lateral surface → sector of a circle with radius (l) (slant height) and arc length equal to circumference of base (2\pi r).
Area of sector:
[
S_{\text{lat}} = \frac{2\pi r}{2\pi l}\cdot \pi l^{2} = \pi r l
]
Base area: (\pi r^{2}).
Total surface area:
[
S = \pi r l + \pi r^{2} = \pi r(r + l)
]
4.3 Sphere Surface Area
A sphere can be considered as the limit of a regular polyhedron with (n) faces as (n