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

  • 1Calculate surface area and volume of basic 3D shapes
  • 2Handle combinations of solids
  • 3Apply conversion principle (volume conserved)
  • 4Solve frustum problems
💡
Why this chapter matters
Practical and high-mark chapter. Used in engineering, architecture, packaging, daily life.

Before you start — revise these

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

Surface Areas and Volumes — Class 10 Mathematics

"From cricket ball to ice-cream cone, from water tank to pencil — every shape can be measured."

1. About the Chapter

This chapter calculates surface areas and volumes of 3D shapes:

  • Cube, Cuboid
  • Cylinder
  • Cone
  • Sphere, Hemisphere
  • Frustum (cone with top cut off)
  • Combinations of these
  • Conversions between shapes

Why Important

  • Engineering (tanks, pipes, machines)
  • Architecture (buildings, domes)
  • Daily life (containers, packaging)
  • Industrial (storage, transport)

2. Recap — Basic 3D Shapes

Cube (side a)

  • Surface Area = 6a²
  • Volume = a³

Cuboid (L × B × H)

  • Surface Area = 2(LB + BH + HL)
  • Volume = L × B × H

Cylinder (radius r, height h)

  • Curved Surface Area (CSA) = 2πrh
  • Total Surface Area (TSA) = 2πr(r + h)
  • Volume = πr²h

Cone (radius r, height h, slant l)

  • Slant height l = √(r² + h²)
  • CSA = πrl
  • TSA = πr(l + r)
  • Volume = (1/3)πr²h

Sphere (radius r)

  • Surface Area = 4πr²
  • Volume = (4/3)πr³

Hemisphere (radius r)

  • Curved Surface Area = 2πr²
  • Total Surface Area = 3πr² (curved + flat circle)
  • Volume = (2/3)πr³

Frustum (radius R, r at top and bottom, height h, slant l)

  • Slant l = √(h² + (R−r)²)
  • CSA = π(R + r)l
  • TSA = π[(R + r)l + R² + r²]
  • Volume = (1/3)πh(R² + r² + Rr)

3. Combination of Solids

Strategy

Many real objects combine 2+ shapes:

  • Ice-cream cone (cone + hemisphere)
  • Capsule (cylinder + 2 hemispheres)
  • Building (cuboid + half-cylinder)
  • Pencil (cylinder + cone)

Procedure

  1. Identify the simple shapes
  2. Calculate each part separately
  3. ADD or SUBTRACT appropriately

Example: Ice-Cream Cone

A cone with radius 5 cm and height 12 cm, topped by hemisphere of radius 5 cm.

Surface Area (assuming flat base of cone exposed):

  • Cone slant = √(25+144) = 13 cm
  • Cone CSA = π × 5 × 13 = 65π
  • Hemisphere CSA = 2π(5)² = 50π
  • Total (cone CSA + hemisphere CSA) = 115π ≈ 361 cm²

Volume (full cone + hemisphere):

  • Cone: (1/3)π(25)(12) = 100π
  • Hemisphere: (2/3)π(125) = 250π/3
  • Total: 100π + 250π/3 = 550π/3 ≈ 575.5 cm³

Example: Capsule

A medicine capsule is a cylinder of radius 5 mm and length 14 mm, with hemispheres at both ends.

Total length = 14 + 2(5) = 24 mm Volume:

  • Cylinder: π(25)(14) = 350π
  • 2 hemispheres = 1 sphere = (4/3)π(125) = 500π/3
  • Total: 350π + 500π/3 = 1550π/3 ≈ 1623 mm³

4. Conversion of Solids

Concept

When one solid is recast into another, volume is conserved.

Volume of original = Volume of new shape.

Examples

Example 1: A metal sphere of radius 5 cm is melted and recast into a cylinder of radius 2 cm. Find the height of the cylinder.

  • Volume of sphere = (4/3)π(125) = 500π/3
  • Volume of cylinder = π(4)h
  • Set equal: 4πh = 500π/3
  • h = 500/12 = 41.67 cm

Example 2: How many lead shots of diameter 1 cm can be made from a lead cuboid 1 m × 50 cm × 25 cm?

  • Cuboid volume = 100 × 50 × 25 = 125,000 cm³
  • Each lead shot (sphere) radius = 0.5 cm
  • Volume of each shot = (4/3)π(0.5)³ = (4/3)π(0.125) = π/6 cm³
  • Number of shots = 125,000 / (π/6) = 750,000/π ≈ 238,732

Example 3: Cube of side 6 cm melted to form 4 cuboids. If cuboid has 3 cm × 2 cm base, find height of cuboid.

  • Cube volume = 216 cm³
  • 4 cuboids = 216 → each cuboid volume = 54 cm³
  • Cuboid volume = 3 × 2 × h = 6h
  • 6h = 54 → h = 9 cm

5. Frustum of a Cone

What is a Frustum?

When a cone is cut PARALLEL to its base, the remaining part below the cut is a frustum (also called truncated cone).

Used For

  • Buckets, drinking glasses, lamp shades
  • Industrial containers

Formulas

For a frustum with radii R (lower) and r (upper), height h:

Slant height (l): l = √(h² + (R − r)²)

CSA: CSA = π(R + r)l

TSA: TSA = π[(R + r)l + R² + r²]

Volume: V = (1/3)πh(R² + r² + Rr)

Example

A bucket has radii 25 cm (lower) and 15 cm (upper), height 30 cm.

  • Slant l = √(900 + 100) = √1000 ≈ 31.6 cm
  • Volume = (1/3)π(30)(625 + 225 + 375) = 10π × 1225 = 12250π ≈ 38,500 cm³

6. Worked Examples

Example 1: Cylinder and Cone

A cylinder of height 14 cm and radius 7 cm has a cone of same dimensions on top. Find total volume.

  • Cylinder: πr²h = (22/7) × 49 × 14 = 2156 cm³
  • Cone: (1/3)πr²h_cone — but height of cone needed. Assume same height = 14.
  • Cone: (1/3) × (22/7) × 49 × 14 = 2156/3 ≈ 718.67 cm³
  • Total: 2156 + 718.67 ≈ 2874.67 cm³

Example 2: Hemisphere on Cylinder

A solid is a cylinder of height 8 cm, radius 5 cm, with hemispheres at each end. Find total surface area.

  • Length of cylinder + 2 hemispheres (cylindrical part): 8 cm
  • TSA = CSA of cylinder + 2 × CSA of hemisphere
  • = 2πr × 8 + 2 × 2πr²
  • = 2π × 5 × 8 + 4π × 25
  • = 80π + 100π = 180π
  • ≈ 565.5 cm²

Example 3: Hollow Cylinder

A hollow cylinder has outer radius 10 cm, inner radius 8 cm, height 15 cm. Find volume of material.

  • Outer cylinder volume = π(100)(15) = 1500π
  • Inner cylinder volume = π(64)(15) = 960π
  • Material volume = 1500π − 960π = 540π ≈ 1696.5 cm³

7. Common Mistakes

  1. Wrong formula confusion

    • Cone volume has (1/3) factor; cylinder doesn't.
    • Sphere has (4/3) factor.
  2. Slant height vs height

    • Slant height: along the slope (cone, frustum)
    • Vertical height: straight up
  3. CSA vs TSA

    • CSA = curved only (no top/bottom)
    • TSA = curved + flat ends
  4. Hemisphere TSA

    • Hemisphere TSA = 3πr² (curved 2πr² + flat πr²)
    • Not 4πr² (that's full sphere)
  5. Forgetting to add/subtract

    • In combinations: think carefully which surfaces are exposed.

8. Real-World Applications

Engineering

  • Water tank (cylinder)
  • Storage silos (cone + cylinder)
  • Pipes (hollow cylinder)
  • Capsule design

Architecture

  • Domes (hemispheres)
  • Spires (cones)
  • Indian temple gopuram (pyramid + cone-like)

Manufacturing

  • Bottle and can design
  • Packaging optimisation
  • Material costs

Daily Life

  • Ice cream cones
  • Tents (cone, dome)
  • Water tanks
  • Cylindrical containers

9. Indian Context

Ancient Indian Measurements

  • Sulba Sutras gave precise volume formulas
  • Used for fire-altar construction
  • Aryabhata computed volumes accurately

Modern Indian

  • Water tank manufacturing (Vajra, Sintex)
  • LPG cylinder manufacturing
  • ISRO rocket design uses precise volume calculations

10. Conclusion

Surface Areas and Volumes connect geometry to real-world objects:

  • Cube to cylinder to cone to sphere — basic shapes
  • Combinations reflect real objects
  • Conversions are everywhere in industry

Master:

  • ALL formulas for 6 basic shapes + frustum
  • COMBINATION strategies
  • CONVERSION principle (volume conserved)

Practice 20+ problems. This is HIGH-MARK chapter for board exam.

Every 3D shape can be measured. This chapter gives you the tools.

Key formulas & results

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

Cube
SA = 6a²; V = a³
Cuboid
SA = 2(LB+BH+HL); V = LBH
Cylinder
CSA = 2πrh; TSA = 2πr(r+h); V = πr²h
Cone
CSA = πrl; TSA = πr(l+r); V = (1/3)πr²h
l = √(r²+h²)
Sphere
SA = 4πr²; V = (4/3)πr³
Hemisphere
CSA = 2πr²; TSA = 3πr²; V = (2/3)πr³
Frustum
V = (1/3)πh(R² + r² + Rr); CSA = π(R+r)l
l = √(h² + (R-r)²)
Conservation
V_original = V_new (in melting/recasting)
⚠️

Common mistakes & fixes

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

WATCH OUT
Forgetting (1/3) in cone
V_cone = (1/3)πr²h. The 1/3 is crucial.
WATCH OUT
Forgetting (4/3) in sphere
V_sphere = (4/3)πr³.
WATCH OUT
Hemisphere TSA wrong
Hemisphere TSA = 3πr² (curved 2πr² + flat circle πr²).
WATCH OUT
Slant height vs height
Slant l = √(r² + h²) for cone, NOT just h.

Practice problems

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

Q1EASY· Cylinder
Find volume of cylinder with radius 7 cm and height 10 cm.
Show solution
✦ Answer: V = πr²h = (22/7)(49)(10) = 1540 cm³.
Q2MEDIUM· Cone
Find slant height, CSA, and volume of cone with radius 6 cm and height 8 cm.
Show solution
Step 1 — Slant height. l = √(r² + h²) = √(36 + 64) = √100 = 10 cm Step 2 — CSA. CSA = πrl = (22/7)(6)(10) = 1320/7 ≈ 188.57 cm² Step 3 — Volume. V = (1/3)πr²h = (1/3)(22/7)(36)(8) = 6336/21 = 301.71 cm³ ✦ Answer: l = 10 cm, CSA ≈ 188.57 cm², V ≈ 301.71 cm³.
Q3HARD· Conversion
A solid sphere of radius 10.5 cm is melted and recast into smaller cones of radius 3.5 cm and height 3 cm. Find the number of cones.
Show solution
Step 1 — Volume of sphere. V_sphere = (4/3)π(10.5)³ 10.5³ = 1157.625 V = (4/3)(22/7)(1157.625) = (88/21)(1157.625) ≈ 4851 cm³ Step 2 — Volume of one cone. V_cone = (1/3)π(3.5)²(3) = (1/3)(22/7)(12.25)(3) = (1/3)(22/7)(36.75) = 808.5/7 = 115.5 cm³ Or: V_cone = (1/3)π × 12.25 × 3 = π × 12.25 = (22/7)(12.25) = 269.5/7 = 38.5 cm³ Wait, let me recompute: (1/3)(22/7)(12.25)(3) = (22/7)(12.25) = 269.5/7 = 38.5 cm³ ✓ Step 3 — Number of cones. N = V_sphere / V_cone = 4851 / 38.5 = 126 Step 4 — Verify. 126 × 38.5 = 4851 ✓ ✦ Answer: Number of cones = 126.

5-minute revision

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

  • Cube SA=6a², V=a³
  • Cuboid V=LBH
  • Cylinder CSA=2πrh, V=πr²h
  • Cone l=√(r²+h²), CSA=πrl, V=(1/3)πr²h
  • Sphere SA=4πr², V=(4/3)πr³
  • Hemisphere CSA=2πr², TSA=3πr², V=(2/3)πr³
  • Frustum V=(1/3)πh(R²+r²+Rr)
  • Conversion: volume conserved when shape changes

CBSE marks blueprint

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

Typical chapter weightage: 10-12 marks

Question typeMarks eachTypical countWhat it tests
MCQ13Formulas
Short2-32Single shape
Long51Combinations or conversions
Prep strategy
  • Memorise ALL 7 shape formulas
  • Practice 20+ problems
  • Master conversion (volume conserved) principle
  • Frustum is tricky — practice extra

Where this shows up in the real world

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

Water tanks

Indian homes use cylindrical or spherical tanks. Volume calculations for capacity.

LPG cylinders

Cylindrical, sometimes with hemispherical ends. Volume measured.

Ice cream cones

Cone + scoop hemisphere = combination of solids.

ISRO rocket fuel tanks

Complex shapes optimised by volume calculations.

Exam strategy

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

  1. Memorise all formulas COLD
  2. Identify shapes in combinations
  3. Use π = 22/7 if r is multiple of 7
  4. For conversions, set volumes equal

Going beyond the textbook

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

  • Pappus's centroid theorem
  • Cavalieri's principle
  • Volume of irregular shapes (calculus)

Where else this chapter is tested

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

CBSE Class 10 BoardVery High
Maths OlympiadHigh
JEE FoundationVery High

Questions students ask

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

If you pour water from a cone of radius r and height h into a cylinder of same radius r and height h, the cone fills EXACTLY 1/3 of the cylinder. So V_cone = (1/3)V_cylinder = (1/3)πr²h. This was proved by Greek mathematicians (Eudoxus, Archimedes) using exhaustion method.
Verified by the tuition.in editorial team
Last reviewed on 20 May 2026. Written and reviewed by subject-matter experts — read about our process.
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