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Bihar (BSEB)Class 10 Mathematics★ Unit VI — Mensuration: 10 marks, shared with Areas Related to Circles · CBSE Class 10 Maths (Chapter 12)

Surface Areas and Volumes

Surface Areas and Volumes — surface area and volume of solids made by combining a cube, cuboid, cylinder, cone, sphere and hemisphere. CBSE Class 10 Maths Chapter 12, updated to the current two-exercise structure.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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

1Identify which two basic solids a combined object is built from
2Find the surface area of a combination by adding only the surfaces that stay exposed
3Find the volume of a combination by adding (or subtracting, for a hollowed-out solid) the individual volumes
4Handle a solid enclosed inside another (a toy inside its circumscribing cylinder) as a subtraction
5Convert a real description — a capsule's length, a tent's slant height — into the radius and height each formula needs

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Why this chapter matters

Almost nothing you actually hold is a pure cone or a pure cylinder — a test tube, a medicine capsule, an ice-cream cone, a bucket-shaped glass. This chapter is entirely about that gap: taking the six shapes you already know and combining them the way real objects do.

Before you start — revise these

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

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Areas Related to Circles

What CBSE examines here (2026-27). Surface area and volume of solids formed by combining two of the basic shapes — a cone on a hemisphere, a cylinder with hemispherical ends, a cuboid with conical holes, and so on. The chapter runs to two exercises, 12.1 and 12.2. Both the frustum of a cone and conversion of one solid into another ("melt and recast") were removed; both are in the appendix, marked as background.

"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 solids built by combining:

Cube, Cuboid
Cylinder
Cone
Sphere, Hemisphere

Every question is one of these shapes stuck to, or hollowed out of, another.

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

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

Identify the simple shapes
Calculate each part separately
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. 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

5. Common Mistakes

Wrong formula confusion
- Cone volume has (1/3) factor; cylinder doesn't.
- Sphere has (4/3) factor.
Slant height vs height
- Slant height: along the cone's slope, l = √(r² + h²)
- Vertical height: straight up from the base to the apex
CSA vs TSA
- CSA = curved only (no top/bottom)
- TSA = curved + flat ends
Hemisphere TSA
- Hemisphere TSA = 3πr² (curved 2πr² + flat πr²)
- Not 4πr² (that's full sphere)
Forgetting to add/subtract
- In combinations: think carefully which surfaces are exposed.

6. 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

7. 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

8. Conclusion

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

Cube to cylinder to cone to sphere — basic shapes
Combinations reflect real objects — nearly everything you can hold is two of these joined together

Master:

ALL formulas for the six basic shapes
Surface area of a combination: add the curved surfaces that stay exposed, and remember that the joined faces vanish
Volume of a combination: simply add (or subtract, for a hollowed-out solid) — nothing disappears the way surface area does

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

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

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. Two topics left this chapter during rationalisation, along with the exercises that practised them: the frustum of a cone, and conversion of one solid into another (the classic "melt and recast" problem). The rationalised chapter is 12.1 Introduction → 12.2 Surface Area of a Combination of Solids → 12.3 Volume of a Combination of Solids → 12.4 Summary, with exercises 12.1 and 12.2 only. Both are kept here because they are common in older guidebooks, PYQs, and general mensuration practice.

Frustum of a cone

When a cone is cut parallel to its base and the top piece is removed, what remains is a frustum — the shape of a bucket, a lampshade, or a drinking glass that is narrower at the bottom than the top.

For a frustum with radii R (larger, bottom) and r (smaller, top), and height h:

Slant height: l = √(h² + (R − r)²)
Curved surface area: π(R + r)l
Total surface area: π[(R + r)l + R² + r²]
Volume: (1/3)πh(R² + r² + Rr)

Example. A bucket has radii 25 cm (bottom) and 15 cm (top), and height 30 cm.

Slant height = √(30² + 10²) = √1000 ≈ 31.6 cm
Volume = (1/3)π(30)(25² + 15² + 25×15) = 10π(625 + 225 + 375) = 10π(1225) = 12250π ≈ 38,485.7 cm³

Conversion of solids ("melt and recast")

The principle here is simple: volume is conserved when a solid is melted and recast into a different shape. Only the shape changes; the amount of material does not.

Volume of the original solid = Volume of the new solid

Example — sphere recast into a cylinder. 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(5)³ = 500π/3
Volume of cylinder = π(2)²h = 4πh
Equate: 4πh = 500π/3, so h = 500/12 ≈ 41.67 cm

Example — lead shots from a cuboid. How many spherical lead shots of radius 0.5 cm can be made from a lead cuboid 100 cm × 50 cm × 25 cm?

Cuboid volume = 100 × 50 × 25 = 125,000 cm³
Volume of one shot = (4/3)π(0.5)³ = π/6 cm³
Number of shots = 125,000 ÷ (π/6) = 750,000/π ≈ 238,732

If you meet either topic in an older paper, the formulas and the volume-conservation principle above are exactly right — they are simply no longer part of this chapter's own exercise.

Key formulas & results

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

Cylinder

CSA = 2πrh; TSA = 2πr(r + h); Volume = πr²h

Cone

l = √(r² + h²); CSA = πrl; Volume = (1/3)πr²h

The 1/3 factor is easy to drop under exam pressure

Sphere

Surface area = 4πr²; Volume = (4/3)πr³

Hemisphere

CSA = 2πr²; TSA = 3πr²; Volume = (2/3)πr³

TSA includes the flat circular base; CSA does not

Surface area of a combination

sum of the curved surfaces that remain exposed

Never add the two solids' full TSAs — the joined faces are hidden, not doubled

Volume of a combination

sum (or difference) of the individual volumes

Unlike surface area, nothing is hidden here — every bit of volume counts

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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

✗ Adding the full TSA of both constituent solids

✓ Where two solids join, that face is hidden inside the combination — it contributes to neither solid's exposed surface. Add only the curved parts (and any flat part that stays on the outside).

WATCH OUT

✗ Using the same reasoning for surface area and volume

✓ Volume behaves simply: it is always additive (or subtractive for a hollow), because nothing about the material's *amount* changes when two solids are joined. Surface area is the one where joining hides area — the two ideas are genuinely different and the chapter's own section 12.3 says so explicitly.

WATCH OUT

✗ Forgetting a hemisphere's flat base is part of its TSA but not its CSA

✓ CSA = 2πr² (curved only). TSA = 3πr² (curved + the flat circle). Which one you need depends on whether that flat circle is exposed or joined to something else.

WATCH OUT

✗ Not working out the missing dimension first

✓ A toy's 'total height' is not the cone's height — subtract the hemisphere's radius first. A capsule's 'length' is not the cylinder's height — subtract both hemispherical ends first.

NCERT exercises (with solutions)

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

Ex 12.1

Exercise 12.1

Surface area of a combination of solids — vessels, toys, capsules, tents, hollowed cylinders

Questions

Ex 12.2

Exercise 12.2

Volume of a combination of solids — models, sweets, pen stands, lead shots, an iron pole, water displacement

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Surface Areas and Volumes?

3 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

3 questions~2 min worth ~12 marks in Bihar (BSEB) exams

5-minute revision

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

•This chapter is entirely about solids built from two of: cuboid, cylinder, cone, sphere, hemisphere
•Surface area of a combination: add only the surfaces still exposed after joining — the shared face disappears
•Volume of a combination: simply add (or subtract for a hollow) — volume is never hidden the way surface area is
•Toy = cone + hemisphere is the most common combination in the whole exercise
•When a solid is enclosed by a circumscribing shape, the 'difference of volumes' is (enclosing shape) − (solid)
•This chapter has two exercises, 12.1 (9 Q, surface area) and 12.2 (8 Q, volume)
•The frustum of a cone and conversion of solids ('melt and recast') were removed from this chapter

Bihar (BSEB) marks blueprint

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

Typical chapter weightage: 10-12 marks

Question type	Marks each	Typical count	What it tests
MCQ	1	3	Formulas
Short	2-3	2	Single shape
Long	5	1	Combinations 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.

Name the two solids out loud before writing any formula — most marks are lost from misidentifying the shape, not from arithmetic

For surface area, ask which faces are still on the outside; for volume, just add or subtract

Work out any 'hidden' dimension (the cone's own height inside a toy, the cylinder's own height inside a capsule) before substituting into a formula

Keep π as 22/7 or 3.14 exactly as the question specifies, and do not switch partway through

Going beyond the textbook

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

STRETCH

Pappus's centroid theorem

STRETCH

Cavalieri's principle

STRETCH

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

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Questions students ask

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

No. The frustum — and conversion of one solid into another — were both removed from this chapter when the syllabus was rationalised, along with their exercises. The rationalised chapter is Introduction → Surface Area of a Combination of Solids → Volume of a Combination of Solids → Summary, and covers only combinations of the six basic solids: cuboid, cylinder, cone, sphere, hemisphere. Both removed topics are on the chapter page in the appendix, marked non-examinable, since they still appear in older guidebooks and general mensuration problems.

Because joining two solids hides a face — where a hemisphere sits on a cylinder, that flat circle is inside the object, not on its surface, so it must not be counted. Volume has no such hidden part: the total amount of material is simply the sum of what each piece contributes (or the difference, if one solid is scooped out of another). The chapter's own section 12.3 makes this contrast explicit before starting the volume examples.

Seventeen: 9 in Exercise 12.1 (surface area) and 8 in Exercise 12.2 (volume). The older edition had 29 across five exercises, with frustum and conversion problems making up the other 12.

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