Chapter 11 of 12

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Rajasthan (RBSE)Class 9 Mathematics★ 7 marks · CBSE Class 9 board (high-yield)

Surface Areas and Volumes

Formulas for the surface area and volume of cuboids, cubes, cylinders, cones, spheres and hemispheres — and how to mix and match them in composite-solid problems.

⏱ 14 min readEasy difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Recall CSA, TSA, volume and diagonal formulas for cuboid and cube
2Apply cylinder, cone, sphere and hemisphere formulas to direct problems
3Compute the slant height of a cone using ℓ = √(r² + h²)
4Decompose composite solids (cone-on-hemisphere, cylinder-with-hemispheres) and find total exposed surface area
5Use volume conservation in melting or recasting problems

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Why this chapter matters

Mensuration is the single highest-mark, lowest-skill chapter in Class 9. Memorise the formula table once and the problems become arithmetic. The same set of formulas appears in Class 10 with frustums added, so the work you do now compounds. A 4-mark composite solid question is almost guaranteed in the board exam.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Mensuration (Class 8)

Cube, cuboid, cylinder basics — extended here

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Heron's Formula (Class 9)

Triangular faces in 3D shapes

Surface Areas and Volumes

This chapter is a formula-heavy mensuration chapter. Memorise the table once and the problems become arithmetic. Common 4-mark question: a composite shape (e.g. cylinder + hemisphere). The trick is knowing exactly which surfaces are exposed.

1. The complete formula sheet

For all formulas, $π \approx \frac{22}{7}$ unless stated.

Cuboid (length $l$ , breadth $b$ , height $h$ )

$Lateral surface area (LSA) = 2 h (l + b)$ $Total surface area (TSA) = 2 (l b + bh + h l)$ $Volume = l bh$ $Diagonal = l^{2} + b^{2} + h^{2}$

Cube (side $a$ )

$LSA = 4 a^{2}, TSA = 6 a^{2}, Volume = a^{3}, Diagonal = a 3$

Right circular cylinder (radius $r$ , height $h$ )

$Curved surface area (CSA) = 2 π r h$ $TSA = 2 π r (r + h)$ $Volume = π r^{2} h$

Right circular cone (radius $r$ , height $h$ , slant $ℓ$ )

$ℓ = r^{2} + h^{2}$ $CSA = π r ℓ$ $TSA = π r (r + ℓ)$ $Volume = \frac{1}{3} π r^{2} h$

Sphere (radius $r$ )

$Surface area = 4 π r^{2}$ $Volume = \frac{4}{3} π r^{3}$

Hemisphere (radius $r$ )

$CSA = 2 π r^{2}$ $TSA = 3 π r^{2} (adds the flat circular face)$ $Volume = \frac{2}{3} π r^{3}$

2. Worked example — cylinder

A cylindrical pillar has radius 0.7 m and height 4 m. Find the curved surface area and volume.

CSA $= 2 π r h = 2 \times \frac{22}{7} \times 0.7 \times 4 = 17.6$ m².

Volume $= π r^{2} h = \frac{22}{7} \times 0.49 \times 4 = 6.16$ m³.

3. Worked example — composite shape

A toy is in the shape of a cone mounted on a hemisphere. Both have the same radius 3 cm, and the cone's height is 4 cm. Find the total surface area.

The exposed surfaces are:

The cone's curved surface (not the base — it's joined to the hemisphere).
The hemisphere's curved surface (not the flat face — it's joined to the cone).

Cone slant $ℓ = 3^{2} + 4^{2} = 5$ .

$CSA_{cone} = π r ℓ = π \times 3 \times 5 = 15 π$ . $CSA_{hemisphere} = 2 π r^{2} = 2 π \times 9 = 18 π$ .

Total $= 33 π \approx 103.7$ cm².

4. The "conservation of volume" trick

When you melt one solid and recast it as another, volume is conserved (mass = volume × density and density is unchanged).

Example. A solid cylindrical iron rod of radius 2 cm and height 16 cm is melted and recast into a sphere. Find the sphere's radius.

$π r_{cyl}^{2} h = \frac{4}{3} π r_{sph}^{3}$ $π \cdot 4 \cdot 16 = \frac{4}{3} π r^{3}$ $r^{3} = \frac{3 \cdot 64}{4} = 48 \Rightarrow r = 348 \approx 3.63 cm .$

5. Tips for marks

Always state the formula first before substituting. Examiners give 1 mark for the formula alone.
Mind the units. Convert everything to the same unit before computing. Volume answers go in cm³ or m³.
For composite shapes, draw a sketch and mark which surfaces are exposed. The figure earns 1 mark.

What's next

Statistics — the final chapter — switches gears entirely into data: mean, median, mode, bar graphs, histograms and frequency polygons.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Cuboid — LSA

LSA = 2h(l + b)

Only the four side faces.

Cuboid — TSA

TSA = 2(lb + bh + hl)

All six faces.

Cuboid — Volume

V = lbh

Cuboid — Diagonal

d = √(l² + b² + h²)

Cube

TSA = 6a², V = a³, Diagonal = a√3

Special case of cuboid with l = b = h = a.

Cylinder — CSA

CSA = 2πrh

Just the curved (lateral) face.

Cylinder — TSA

TSA = 2πr(r + h)

CSA + 2 circular ends.

Cylinder — Volume

V = πr²h

Cone — Slant height

ℓ = √(r² + h²)

Derived from Pythagoras.

Cone — CSA

CSA = πrℓ

Curved face only.

Cone — TSA

TSA = πr(r + ℓ)

CSA + circular base.

Cone — Volume

V = (1/3)πr²h

One-third of the corresponding cylinder.

Sphere — Surface area

SA = 4πr²

Sphere — Volume

V = (4/3)πr³

Hemisphere — CSA

CSA = 2πr²

Curved part only.

Hemisphere — TSA

TSA = 3πr²

Adds the flat circular face (πr²).

Hemisphere — Volume

V = (2/3)πr³

Half of a sphere.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing CSA (curved) with TSA (total)

✓ CSA = only the curved surface; TSA includes the flat end(s). Read the question carefully — 'curved surface area' vs 'total surface area' is a 1-mark trap in every exam.

WATCH OUT

✗ Using the wrong π — sometimes 22/7, sometimes 3.14

✓ Use whichever value the question specifies. If unspecified, default to 22/7 (gives clean integer answers for typical values like 7, 14, 21).

WATCH OUT

✗ Adding ALL surfaces of a composite solid without excluding the joined faces

✓ For a hemisphere on top of a cylinder, EXCLUDE the cylinder's top circle AND the hemisphere's flat face — they are joined and sealed. Draw the solid and shade the exposed surfaces first.

WATCH OUT

✗ Mismatched units (cm and m mixed)

✓ Convert everything to ONE UNIT BEFORE plugging into the formula. 1 m = 100 cm; 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 11.1

Exercise 11.1

Right circular cone — CSA, TSA, volume; slant height computation

Questions

Ex 11.2

Exercise 11.2

Sphere and hemisphere — surface area and volume problems

Questions

Ex 11.3

Exercise 11.3

Mixed composite-solid problems and melting/recasting

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Cylinder

Find the curved surface area of a cylinder with radius 3.5 cm and height 10 cm. (Use π = 22/7.)

▶ Show solution

Step 1 — Formula: CSA = 2πrh.

Step 2 — Substitute: CSA = 2 × (22/7) × 3.5 × 10.
  = 2 × (22/7) × 35 = 2 × 110 = 220 cm².

✦ Answer: CSA = 220 cm².

Q2EASY· Sphere

Find the volume of a sphere of radius 7 cm. (Use π = 22/7.)

▶ Show solution

Step 1 — Formula: V = (4/3)πr³.

Step 2 — Substitute: V = (4/3) × (22/7) × 7³ = (4/3) × (22/7) × 343.
  = (4/3) × (22 × 343/7) = (4/3) × (22 × 49) = (4/3) × 1078 = 4312/3 ≈ 1437.3 cm³.

✦ Answer: Volume = 4312/3 cm³ ≈ 1437.3 cm³.

Q3MEDIUM· Cone

A cone has height 12 cm and radius 5 cm. Find its slant height and total surface area. (Use π = 22/7.)

▶ Show solution

Step 1 — Slant height: ℓ = √(r² + h²) = √(25 + 144) = √169 = 13 cm.

Step 2 — TSA = πr(r + ℓ) = (22/7) × 5 × (5 + 13) = (22/7) × 5 × 18 = (22 × 90)/7 = 1980/7 ≈ 282.86 cm².

✦ Answer: Slant height = 13 cm. TSA ≈ 282.86 cm².

Q4MEDIUM· Composite solid

A solid is a cylinder (radius 7 cm, height 10 cm) with a hemisphere of the same radius on top. Find the total surface area. (Use π = 22/7.)

▶ Show solution

Step 1 — Identify exposed surfaces:
  • Cylinder's curved surface (side)
  • Cylinder's bottom circular face
  • Hemisphere's curved surface (the hemisphere's flat face joins the cylinder top — NOT counted)

Step 2 — Compute each:
  Cylinder CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².
  Cylinder bottom = πr² = (22/7) × 49 = 154 cm².
  Hemisphere CSA = 2πr² = 2 × (22/7) × 49 = 308 cm².

Step 3 — Total = 440 + 154 + 308 = 902 cm².

✦ Answer: Total surface area = 902 cm².

Q5HARD· Recasting

A metallic sphere of radius 10.5 cm is melted and recast into small cones of radius 3.5 cm and height 3 cm. How many cones are formed?

▶ Show solution

Step 1 — Volume of sphere = (4/3)πr³ = (4/3) × π × (10.5)³.
  10.5³ = 1157.625.
  V_sphere = (4/3) × π × 1157.625 = 1543.5π cm³.

Step 2 — Volume of one cone = (1/3)πr²h = (1/3) × π × (3.5)² × 3.
  = (1/3) × π × 12.25 × 3 = 12.25π cm³.

Step 3 — Number of cones = V_sphere / V_cone = 1543.5π / 12.25π = 1543.5 / 12.25 = 126 cones.

✦ Answer: 126 small cones are formed.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Cube: TSA = 6a², V = a³, Diagonal = a√3.
•Cuboid: TSA = 2(lb + bh + hl), V = lbh, Diagonal = √(l² + b² + h²).
•Cylinder: CSA = 2πrh, TSA = 2πr(r + h), V = πr²h.
•Cone: slant ℓ = √(r² + h²), CSA = πrℓ, TSA = πr(r + ℓ), V = (1/3)πr²h.
•Sphere: SA = 4πr², V = (4/3)πr³.
•Hemisphere: CSA = 2πr², TSA = 3πr², V = (2/3)πr³.
•Volume is conserved in melting/recasting problems.
•For composite solids: identify which faces are exposed (NOT joined) and sum only those.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 7 marks

Question type	Marks each	Typical count	What it tests
MCQ	1	1	Definition or formula identification
Short answer (2-mark)	2	1–2	Direct formula application
Long answer (3-mark)	3	1–2	Multi-step problem solving
Long answer (4-mark)	4	1	Proof or complex application

Prep strategy

Write all key formulas on a revision sheet — recall speed matters in board exams
Show ALL working steps clearly — partial marks are awarded for method even if final answer is wrong
Practise past 5 years of CBSE board questions for this chapter
For proof questions: state the theorem/property used at each step by name
Time management: allocate time based on marks — 1 min per mark is a good rule

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Architecture and construction

Calculating how much paint is needed for a cylindrical water tank, or how much concrete to pour for a spherical dome.

Ice cream wafer cones

A classic composite — the cone (volume of ice cream it can hold) topped with a hemisphere of ice cream. Exactly the composite solid formula.

Tanks and pipes

Cylindrical water storage tanks — their capacity is V = πr²h. Engineering students use this daily.

Sports equipment

Basketballs, footballs (spheres), cricket pitch rollers (cylinders) — all involve these formulas for manufacturing specifications.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

ALWAYS write the formula before substituting. Examiners give 1 mark for the formula alone — even if arithmetic goes wrong, you get the formula mark.
For composite shapes: draw a sketch and mark which surfaces are exposed in a different colour. This earns a diagram mark AND prevents you from missing a surface.
State all conversions explicitly: '1 m = 100 cm, so radius = 0.7 m = 70 cm'. Mixed units cost full marks.
For recasting problems, write 'Volume is conserved: V₁ = n × V₂' as the first equation — this signals you know the principle.
Use π = 22/7 unless the question says otherwise — it usually gives cleaner numbers.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Frustum of a cone (Class 10): If a cone is cut by a plane parallel to the base, the lower portion is a frustum. Volume = (πh/3)(R² + Rr + r²). LSA = π(R + r)ℓ where ℓ = √(h² + (R − r)²).
Euler's formula for polyhedra: V − E + F = 2 (vertices − edges + faces). Verify for cube: 8 − 12 + 6 = 2. Works for all convex polyhedra.
Golden sphere in a cube: if a sphere of radius r fits exactly in a cube, then side of cube = 2r and the sphere's volume / cube's volume = π/6 ≈ 0.5236 — the 'packing fraction'.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 9 Board~7 marks. High-yield chapter — always study this.

NTSEMedium. Mensuration formulas appear in MAT and SAT.

Class 10 (same chapter)Very high — Class 10 extends this chapter with frustums and more complex composites.

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

About 7 marks of Unit V (Mensuration), which is 13 marks total. Expect at least one 3-mark and one 4-mark question from this chapter in the board exam.

Use whichever the question specifies. If unspecified, default to 22/7 (gives clean integer answers for values like 7, 14, 21, 3.5).

No — frustums (truncated cones) appear in Class 10. Class 9 covers only the six basic solids: cuboid, cube, cylinder, cone, sphere, hemisphere.

Slant height ℓ is the distance from the apex of the cone to any point on the circular edge of the base. It is NOT the vertical height h. Relationship: ℓ = √(r² + h²) by Pythagoras.

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