CBSE Class 9 Maths — Surface Areas and Volumes

Chapter Overview & Weightage

Surface Areas and Volumes is one of the highest-weightage chapters in Class 9 Maths — consistently contributing 10–14 marks in CBSE board/unit exams. Questions are formula-driven and procedural, making this a reliable chapter to score full marks with focused practice.

Year	Marks	Topics Tested
2024	12	Cylinder CSA, cone total SA, sphere volume
2023	11	Cube surface area, cylinder volume, hemisphere
2022	13	Cuboid, cone, sphere combinations

Every formula in this chapter appears in exams — no formula is “safe to skip.” The common exam pattern: given some dimensions, find surface area OR volume. Conversion between units (cm to m, cm³ to litres) also appears.

Key Concepts You Must Know

Cuboid: A box with 6 rectangular faces. Length $l$ , breadth $b$ , height $h$ .

Cube: A special cuboid where $l = b = h = a$ .

Cylinder: Two circular faces + curved lateral surface. Radius $r$ , height $h$ .

Cone: One circular base + curved lateral surface. Radius $r$ , height $h$ , slant height $l = \sqrt{r^2 + h^2}$ .

Sphere: Perfectly round, radius $r$ .

Hemisphere: Half a sphere, radius $r$ .

Important Formulas

\text{LSA (Lateral)} = 2(l+b)h

\text{TSA (Total)} = 2(lb + bh + hl)

\text{Volume} = lbh

\text{Diagonal} = \sqrt{l^2 + b^2 + h^2}

\text{LSA} = 4a^2, \quad \text{TSA} = 6a^2, \quad \text{Volume} = a^3

\text{Diagonal} = a\sqrt{3}

\text{CSA} = 2\pi rh

\text{TSA} = 2\pi r(r + h)

\text{Volume} = \pi r^2 h

l = \sqrt{r^2 + h^2} \quad \text{(slant height)}

\text{CSA} = \pi rl

\text{TSA} = \pi r(r + l)

\text{Volume} = \frac{1}{3}\pi r^2 h

\text{Sphere: TSA} = 4\pi r^2, \quad \text{Volume} = \frac{4}{3}\pi r^3

\text{Hemisphere: CSA} = 2\pi r^2

\text{Hemisphere: TSA} = 3\pi r^2 \quad \text{(curved + flat circular base)}

\text{Hemisphere: Volume} = \frac{2}{3}\pi r^3

Solved Previous Year Questions

PYQ 1 — Cylinder Surface Area (3 marks)

Q: A cylinder has radius 7 cm and height 20 cm. Find its CSA and TSA. (Use $\pi = 22/7$ )

Solution:

CSA = $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 20 = 2 \times 22 \times 20 = \mathbf{880 \text{ cm}^2}$

TSA = $2\pi r(r+h) = 2 \times \frac{22}{7} \times 7 \times (7+20) = 44 \times 27 = \mathbf{1188 \text{ cm}^2}$

PYQ 2 — Cone Volume (3 marks)

Q: Find the volume of a cone of radius 6 cm and height 8 cm. (Use $\pi = 3.14$ )

Solution: First find slant height: $l = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$ cm

Volume $= \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 36 \times 8 = \frac{1}{3} \times 3.14 \times 288 = \frac{904.32}{3} = \mathbf{301.44 \text{ cm}^3}$

PYQ 3 — Sphere Surface Area (2 marks)

Q: The radius of a spherical balloon is 7 cm. Find its surface area. (Use $\pi = 22/7$ )

Solution: SA $= 4\pi r^2 = 4 \times \frac{22}{7} \times 7^2 = 4 \times \frac{22}{7} \times 49 = 4 \times 22 \times 7 = \mathbf{616 \text{ cm}^2}$

PYQ 4 — Cuboid Volume and Surface Area (4 marks)

Q: A rectangular room is 5 m long, 4 m wide, and 3 m high. Find the cost of whitewashing the walls and ceiling at ₹12 per m².

Solution: Walls + ceiling area = LSA + top = $2(l+b)h + lb = 2(5+4) \times 3 + 5 \times 4$

$= 2 \times 9 \times 3 + 20 = 54 + 20 = 74 \text{ m}^2$

Cost $= 74 \times 12 = \mathbf{\text{₹}888}$

Difficulty Distribution

Difficulty	% of Questions	Types
Easy	40%	Direct substitution in single formula
Medium	45%	Two-step problems, finding one dimension from SA or volume
Hard	15%	Combined shapes, unit conversions, cost/rate problems

Expert Strategy

Write the formula before substituting numbers. This earns you method marks even if you make an arithmetic error later. CBSE examiners scan for the formula line first.

For cone problems, always check if slant height or vertical height is given. The CSA formula uses slant height $l$ . The volume formula uses vertical height $h$ . Don’t mix them up. If vertical height is given, calculate $l = \sqrt{r^2 + h^2}$ first.

Memorise which shape has which volume formula by comparing them: cone = 1/3 of cylinder, hemisphere = 2/3 of sphere. If you remember the cylinder and sphere formulas, you can derive the cone and hemisphere formulas immediately.

Common Traps

Trap 1 — Using diameter as radius: When a question says “diameter = 14 cm,” many students substitute 14 directly into the formula. Always halve the diameter: $r = 7$ cm. Check every problem — is the given value radius or diameter?

Trap 2 — Hemisphere TSA: Hemisphere TSA = curved surface + flat circular base = $2\pi r^2 + \pi r^2 = 3\pi r^2$ . Students often write only the curved surface area $2\pi r^2$ , missing the flat base.

Trap 3 — Cone CSA uses slant height, not vertical height: $\pi rl$ uses $l$ (slant height), not $h$ (vertical height). If a problem gives height and radius, you must first compute $l = \sqrt{r^2 + h^2}$ .

Trap 4 — Unit conversion: Volume is in cm³ (cubic centimetres). 1 litre = 1000 cm³. Some questions ask you to convert — e.g., how many litres of water does a cylindrical tank hold? Compute volume in cm³ first, then divide by 1000.