Surface Area and Volume is one of the most important and frequently examined chapters in Class 9 Mathematics (NCERT). This topic introduces students to three-dimensional (3D) solid figures cuboids, cubes, cylinders, cones, spheres, and hemispheres and teaches how to calculate both the area of their surfaces and the space they occupy (volume).
Learn concepts not only strengthens your foundation for Class 10 and competitive exams like JEE and NTSE, but also builds essential mathematical reasoning and spatial thinking skills. This comprehensive study guide covers every key formula with clear derivations, step-by-step solved examples directly from the syllabus, and conceptual explanations designed to help students truly understand not just memorise.
What are Solid (Three-Dimensional) Figures?
A two-dimensional figure, like a rectangle, has only two measurements: length and width. A three-dimensional (solid) figure possesses three measurements length, width, and height — and occupies physical space. These solids have both a surface area (the total area of their outer surfaces) and a volume (the amount of space enclosed within them).
- Two key measurements for any solid
- Surface Area measured in square units (cm², m²).
- Volume measured in cubic units (cm³, m³).
The solids studied in Class 9 are: Cuboid, Cube, Cylinder, Cone, Sphere, and Hemisphere — along with their hollow variants.
Cuboid
A cuboid (also called a rectangular prism) is a box-shaped solid with 6 rectangular faces, 8 vertices, and 12 edges. Its three dimensions are length (ℓ), breadth (b), and height (h). Think of a brick, a book, or a room these are all cuboids.
Cuboid Formulas — where ℓ = length, b = breadth, h = height
Total Surface Area (TSA)
2(ℓb + bh + hℓ)
Lateral Surface Area (LSA)
2h(ℓ + b)
Volume
ℓ × b × h
Diagonal
√(ℓ² + b² + h²)
LSA vs TSA: The Lateral Surface Area is the area of just the 4 side walls (excluding top and bottom). The Total Surface Area includes all 6 faces. For room-painting problems, always use LSA (walls only) and remember to subtract door and window areas.
Three equal cubes (side = a) are placed in a row. Find the ratio of TSA of the new cuboid to the sum of TSAs of the three cubes.
- TSA of one cube = 6a². Sum for three cubes = 18a²
- New cuboid: ℓ = 3a, b = a, h = a
- TSA of new cuboid = 2(3a·a + a·a + a·3a) = 2(3a² + a² + 3a²) = 14a²
- Ratio = 14a² : 18a² = 7 : 9
- Answer: 7 : 9
Cube
A cube is a special cuboid where all three dimensions are equal (side = x). It has 6 square faces, 8 vertices, and 12 equal edges. Examples: a Rubik's cube, a dice, a sugar cube.
Cube Formulas where x = side length
Total Surface Area (TSA): 6x²
Lateral Surface Area (LSA): 4x²
Volume: x³
Diagonal: x√3
Cylinder
A cylinder has two circular bases and a curved lateral surface. When the curved surface is unrolled, it forms a rectangle of length 2πr and height h. The variables are: r = base radius, h = height.
Solid Cylinder — r = radius, h = height
- Curved Surface Area (CSA): 2πrh
- Total Surface Area (TSA): 2πr(h + r)
- Volume: πr²h
Hollow Cylinder
A hollow cylinder (like a pipe) has an outer radius R and inner radius r.
Hollow Cylinder — R = outer radius, r = inner radius, h = height
- CSA: 2π(R + r)h
- TSA: π(R+r)(2h + R − r)
- Volume: π(R² − r²)h
A cylindrical vessel (no lid) is tin-coated on both sides. Radius = ½ m, height = 1.4 m. Find cost at Rs. 50 per 1000 cm².
- r = 50 cm, h = 140 cm
- Area = 2(2πrh + πr²) = 2[2×3.14×50×140 + 3.14×50²]
- = 2[43960 + 7850] = 2 × 51810 = 103620 cm²
- Cost = (50/1000) × 103620 = Rs. 5181
Answer: Rs. 5,181
Cone
A cone has a circular base and tapers to a point (apex). Key dimensions: base radius r, vertical height h, and slant height ℓ, where ℓ = √(r² + h²).
Cone Formulas — r = radius, h = height, ℓ = slant height
- Slant Height: ℓ = √(r² + h²)
- Curved Surface Area (CSA): πrℓ
- Total Surface Area (TSA): πr(ℓ + r)
- Volume: (1/3)πr²h
Remember: Volume of a cone is exactly one-third the volume of a cylinder with the same base radius and height. This relationship is a common exam question.
Solved Example
A conical tent has vertical height = 12 m and base radius = 16 m. How many metres of 1.1 m wide cloth is needed? Find cost at Rs. 14/metre.
- ℓ = √(16² + 12²) = √(256 + 144) = √400 = 20 m
- CSA = πrℓ = (22/7) × 16 × 20 = 7040/7 m²
- Length of cloth = (7040/7) ÷ 1.1 = 70400/77 = 6400/7 m
- Cost = (6400/7) × 14 = Rs. 12,800
- Answer: Rs. 12,800
Sphere & Hemisphere
A sphere is a perfectly round 3D solid where every point on the surface is equidistant from the centre. A hemisphere is exactly half of a sphere, with a flat circular face and a curved surface.
Sphere — r = radius
Total Surface Area: 4πr²
Volume: (4/3)πr³
Hemisphere — r = radius
Curved Surface Area: 2πr²
Total Surface Area: 3πr²
Volume: (2/3)πr³
Hollow Hemisphere
Hollow Hemisphere — R = outer radius, r = inner radius
CSA: 2π(R² + r²)
TSA: 2π(R²+r²) + π(R²−r²)
Volume: (2/3)π(R³ − r³)
Solved Example
How many balls of radius 1 cm can be made from a solid lead sphere of radius 8 cm?
- Volume of large sphere = (4/3)π × 8³
- Volume of small ball = (4/3)π × 1³
- Number of balls (n) = 8³ / 1³ = 512
- Answer: 512 balls
Surface Area And Volume Class 9 Maths Revision Notes PDF Download
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Surface Area & Volume Formula Table
Use this table for rapid revision before exams. All formulas at a glance:
| Solid | TSA / SA | CSA / LSA | Volume |
|---|---|---|---|
| Cuboid | 2(ℓb + bh + hℓ) | 2h(ℓ + b) | ℓ × b × h |
| Cube | 6x² | 4x² | x³ |
| Cylinder | 2πr(h + r) | 2πrh | πr²h |
| Hollow Cylinder | π(R+r)(2h+R−r) | 2π(R+r)h | π(R²−r²)h |
| Cone | πr(ℓ + r) | πrℓ | (1/3)πr²h |
| Sphere | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere | 3πr² | 2πr² | (2/3)πr³ |
| Hollow Hemisphere | 2π(R²+r²)+π(R²−r²) | 2π(R²+r²) | (2/3)π(R³−r³) |
A classroom is 7 m × 6.5 m × 4 m. It has 1 door (3 m × 1.4 m) and 3 windows (2 m × 1 m each). Find cost of colour-washing walls at Rs. 15/m².
LSA of 4 walls = 2(ℓ + b)h = 2(7 + 6.5) × 4 = 108 m²
Area of door = 3 × 1.4 = 4.2 m²; Area of 3 windows = 3 × 2 × 1 = 6 m²
Net area = 108 − (4.2 + 6) = 108 − 10.2 = 97.8 m²
Cost = 97.8 × 15 = Rs. 1,467
Answer: Rs. 1,467
Ratio of volumes of two cones = 4 : 5; ratio of base radii = 2 : 3. Find ratio of vertical heights.
V₁/V₂ = (r₁²h₁)/(r₂²h₂) = 4/5
⟹ h₁/h₂ = (4/5) × (r₂/r₁)² = (4/5) × (3/2)² = (4/5) × (9/4) = 9/5
Answer: h₁ : h₂ = 9 : 5
3r_cone = 2R_cylinder and H_cylinder : h_cone = 4 : 3. How many cones can be made?
R = 3r/2 and H = 4h/3
Volume of cylinder = πR²H = π × (9r²/4) × (4h/3) = 3πr²h
Volume of one cone = (1/3)πr²h
n = 3πr²h ÷ (1/3)πr²h = 9 cones
Answer: 9 cones
A cinema hall is 100 m × 50 m × 18 m. Each person needs 150 m³ of air. How many can sit?
Volume = 100 × 50 × 18 = 90,000 m³
Number of persons = 90,000 ÷ 150 = 600
Answer: 600 persons