Chapter 12 of 14

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Rajasthan (RBSE)Class 10 Mathematics★ 4–6 marks · RBSE Class 10 Maths board (Unit — Mensuration)

Surface Areas and Volumes

RBSE Class 10 Maths on 3-D mensuration — surface areas and volumes of cuboid, cube, cylinder, cone, sphere and hemisphere, plus combinations of solids and conversion of one solid into another. Learn the frustum formulas and solve board-pattern word problems with full solutions.

⏱ 30 min readMedium difficulty✓ Free · NCERT-aligned

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

1Recall surface area and volume formulas for the six basic solids
2Find surface areas of combinations by counting only exposed faces
3Find volumes of combinations by adding component volumes
4Solve recasting problems by conserving volume
5Apply the frustum formulas to buckets, glasses and lampshades

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Why this chapter matters

A high-weight mensuration chapter that almost always carries a combined-solid or recasting word problem (3–4 marks) plus shorter formula questions. Careful visualisation and unit handling make it very scorable.

Before you start — revise these

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

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Surface Areas and Volumes (Class 9)

Basic solids: cuboid, cylinder, cone, sphere formulas.

Surface Areas and Volumes — RBSE Class 10 (Mathematics)

An ice-cream cone topped with a scoop, a capsule, a circus tent, a metal toy — real objects are rarely one clean shape. This chapter equips you to handle combinations of the basic solids: add surface areas, add volumes, and remember that when one solid is melted and recast into another, the volume stays the same.

1. The basic solids (quick reference)

For radius $r$ , height $h$ , slant height $l$ :

Solid	Curved/Lateral SA	Total SA	Volume
Cuboid ( $l, b, h$ )	$2 h (l + b)$	$2 (l b + bh + h l)$	$l bh$
Cube (side $a$ )	$4 a^{2}$	$6 a^{2}$	$a^{3}$
Cylinder	$2 π r h$	$2 π r (h + r)$	$π r^{2} h$
Cone	$π r l$	$π r (l + r)$	$\frac{1}{3} π r^{2} h$
Sphere	—	$4 π r^{2}$	$\frac{4}{3} π r^{3}$
Hemisphere	$2 π r^{2}$	$3 π r^{2}$	$\frac{2}{3} π r^{3}$

For a cone, the slant height $l = r^{2} + h^{2}$ .

2. Combinations of solids

Most exam objects are two basic solids joined:

Surface area: add only the exposed surfaces. When a hemisphere sits on a cylinder, the joined circular faces disappear — do not count them.
Volume: simply add the component volumes (nothing is lost inside).

Examples: a capsule = cylinder + two hemispheres; an ice-cream cone = cone + hemisphere; a tent = cylinder + cone (curved surfaces only, no base for the fabric).

3. Conversion / recasting — volume is conserved

When a solid is melted, recast, or reshaped, its material — hence its volume — is unchanged: $Volume before = Volume after$

This single equation solves "a sphere is melted into wire," "how many small cones from a big cylinder," or "a well is dug and the earth spread into an embankment." Set up one volume equation and solve for the unknown.

4. The frustum of a cone

Cut a cone with a plane parallel to the base and remove the small top cone — what remains is a frustum (think of a bucket or a lampshade). With radii $R$ (bottom) and $r$ (top) and height $h$ , slant $l = h^{2} + (R - r)^{2}$ :

$Volume = \frac{1}{3} π h (R^{2} + r^{2} + R r)$ $Curved SA = π l (R + r), Total SA = π l (R + r) + π R^{2} + π r^{2}$

5. Method that never fails

Identify the component solids and list their given dimensions.
Decide whether you need surface area (which faces are exposed?) or volume (just add) — or a recasting equation (equate volumes).
Substitute, keep units consistent (convert everything to cm or m first), and choose $π$ as specified.

6. Closing thought

There are only a handful of formulas here, but the marks come from visualising the object and knowing which surfaces to include. Volume always simply adds; surface area needs care about hidden faces; recasting means equal volumes. The RBSE board reliably sets one combined-solid or conversion word problem worth 3–4 marks — practise picturing the solid before touching a formula.

Key formulas & results

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

Cylinder

CSA 2πrh; TSA 2πr(h+r); V πr²h

Right circular cylinder.

Cone

CSA πrl; TSA πr(l+r); V ⅓πr²h; l = √(r²+h²)

l is slant height.

Sphere / Hemisphere

Sphere SA 4πr², V (4/3)πr³; Hemisphere CSA 2πr², TSA 3πr², V (2/3)πr³

Hemisphere TSA includes the flat face.

Combination SA

add only EXPOSED surfaces

Joined faces vanish.

Recasting

Volume before = Volume after

Material conserved.

Frustum

V = ⅓πh(R²+r²+Rr); CSA = πl(R+r), l=√(h²+(R−r)²)

Bucket/lampshade shape.

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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

✗ Counting hidden joined faces in surface area

✓ When two solids are joined, the faces in contact are NOT exposed — exclude them (e.g. cylinder+hemisphere loses two circles).

WATCH OUT

✗ Using height instead of slant height for cone CSA

✓ Cone CSA uses slant height l = √(r²+h²), not the vertical height h.

WATCH OUT

✗ Mixing units

✓ Convert all dimensions to the same unit before substituting; volume in cm³ needs cm throughout.

WATCH OUT

✗ Adding surface areas in recasting problems

✓ Recasting conserves VOLUME, not surface area. Equate volumes to find the unknown.

WATCH OUT

✗ Forgetting the flat face of a hemisphere

✓ Curved SA = 2πr²; TOTAL SA = 3πr² (curved + the flat circular face).

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 combinations of solids

Questions

Ex 12.2

Exercise 12.2

Volume of combinations of solids

Questions

Ex 12.3 / frustum

Exercise 12.3 / frustum

Recasting problems and the frustum of a cone

Questions

Practice problems

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

Q1EASY· Volume

Find the volume of a sphere of radius 3 cm (π = 22/7).

▶ Show solution

Step 1 — V = (4/3)πr³ = (4/3)(22/7)(27) = (4/3)(594/7) = 792/7 ≈ 113.1 cm³.

✦ Answer: ≈ 113.1 cm³.

Q2EASY· Slant height

A cone has radius 5 cm and height 12 cm. Find its slant height.

▶ Show solution

Step 1 — l = √(5² + 12²) = √169 = 13 cm.

✦ Answer: 13 cm.

Q3EASY· TSA

Find the total surface area of a hemisphere of radius 7 cm (π = 22/7).

▶ Show solution

Step 1 — TSA = 3πr² = 3 × 22/7 × 49 = 462 cm².

✦ Answer: 462 cm².

Q4MEDIUM· Combination SA

A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. If the cone's slant height is 6 cm, find the total surface area (π = 22/7).

▶ Show solution

Step 1 — Exposed surfaces = cone CSA + hemisphere CSA.

Step 2 — πrl + 2πr² = 22/7 × 3.5 × 6 + 2 × 22/7 × 3.5² = 66 + 77 = 143 cm².

✦ Answer: 143 cm².

Q5MEDIUM· Combination volume

A capsule is a cylinder of length 5 cm (radius 0.5 cm) with a hemisphere on each end. Find its total length and volume (π = 22/7).

▶ Show solution

Step 1 — Total length = 5 + 0.5 + 0.5 = 6 cm.

Step 2 — V = πr²h + 2·(2/3)πr³ = 22/7×0.25×5 + (4/3)(22/7)(0.125) ≈ 3.93 + 0.52 = 4.45 cm³.

✦ Answer: length 6 cm, volume ≈ 4.45 cm³.

Q6MEDIUM· Recasting

A metallic sphere of radius 6 cm is melted and drawn into a wire of radius 0.2 cm. Find the length of the wire (π = 22/7).

▶ Show solution

Step 1 — Conserve volume: (4/3)π(6³) = π(0.2)²·L.

Step 2 — (4/3)(216) = 0.04·L ⇒ 288 = 0.04L.

Step 3 — L = 7200 cm = 72 m.

✦ Answer: 72 m.

Q7MEDIUM· How many

How many spherical lead shots of radius 1 cm can be made from a cuboid of lead 8 cm × 6 cm × 4 cm? (π = 22/7)

▶ Show solution

Step 1 — Cuboid volume = 8×6×4 = 192 cm³.

Step 2 — One shot = (4/3)π(1³) = 88/21 ≈ 4.19 cm³.

Step 3 — Number = 192 ÷ (88/21) = 192 × 21/88 = 45.8 ⇒ 45 whole shots.

✦ Answer: 45 shots.

Q8HARD· Frustum

A bucket (frustum) has radii 20 cm and 10 cm and height 24 cm. Find its volume (π = 3.14).

▶ Show solution

Step 1 — V = ⅓πh(R² + r² + Rr) = ⅓ × 3.14 × 24 × (400 + 100 + 200).

Step 2 — = ⅓ × 3.14 × 24 × 700 = 3.14 × 8 × 700 = 17584 cm³.

✦ Answer: 17584 cm³ ≈ 17.58 litres.

Q9HARD· Well & embankment

A well of diameter 3 m is dug 14 m deep. The earth is spread evenly to form a 4 m wide embankment around it. Find the embankment's height (π = 22/7).

▶ Show solution

Step 1 — Well (cylinder) volume: r = 1.5, V = 22/7 × 1.5² × 14 = 99 m³.

Step 2 — Embankment is a ring: inner r = 1.5, outer R = 1.5+4 = 5.5; area = π(R²−r²) = 22/7(30.25−2.25) = 88 m².

Step 3 — Height = Volume/Area = 99/88 = 1.125 m.

✦ Answer: 1.125 m.

Q10HARD· Cone from cylinder

A solid cylinder of radius 6 cm and height 15 cm is melted and recast into cones of radius 2 cm and height 3 cm. How many cones are formed? (π cancels)

▶ Show solution

Step 1 — Cylinder V = πr²h = π×36×15 = 540π.

Step 2 — Cone V = ⅓πr²h = ⅓π×4×3 = 4π.

Step 3 — Number = 540π / 4π = 135.

✦ Answer: 135 cones.

5-minute revision

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

•Cylinder V πr²h, CSA 2πrh; Cone V ⅓πr²h, CSA πrl (l=√(r²+h²)).
•Sphere V (4/3)πr³, SA 4πr²; Hemisphere V (2/3)πr³, TSA 3πr².
•Combination surface area adds only exposed faces.
•Combination volume simply adds component volumes.
•Recasting/melting conserves volume: equate volumes.
•Frustum V = ⅓πh(R²+r²+Rr); CSA = πl(R+r).
•Keep units consistent and choose π as specified.

Rajasthan (RBSE) marks blueprint

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

Typical chapter weightage: 4–6 marks

Question type	Marks each	Typical count	What it tests
MCQ / very short	1	1	Single-solid volume/SA; slant height
Short answer	2–3	1	Combination SA/volume or recasting
Long answer	4–5	1	Frustum or well/embankment word problem

Prep strategy

Tabulate all basic-solid formulas and memorise them
For combinations, decide exposed surfaces vs added volumes
For recasting, always write Volume before = Volume after
Convert every dimension to one unit before computing

Where this shows up in the real world

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

Packaging and containers

Tins, capsules and buckets are cylinders, spheres and frustums — volume sets capacity.

Manufacturing and casting

Melting and recasting metal conserves volume — the basis of moulding.

Civil engineering

Wells, embankments and tanks are direct volume/earthwork problems.

Everyday capacity

Litres of a bucket or glass come from frustum volumes.

Exam strategy

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

Draw the solid and label all dimensions before choosing a formula.
For surface area, explicitly note which faces are exposed.
For recasting, write the volume-conservation equation first.
Keep π consistent and convert units up front.
Give the final answer with correct units (cm², cm³, or litres if asked).

Going beyond the textbook

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

Deriving the frustum volume by subtracting two cones.
Volume of a sphere by Cavalieri's principle.
Optimising surface area for a fixed volume (isoperimetric ideas).
Solids of revolution and links to integration.

Where else this chapter is tested

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

RBSE Class 10 Board (BSER Ajmer)High — a combined-solid or recasting problem almost every year

NTSE / state scholarshipMedium — mensuration MCQs

SSC / competitive examsHigh — 3-D mensuration is a core quantitative topic

JEE FoundationMedium — volumes feed into integration applications

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Questions students ask

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

Yes — RBSE (BSER, Ajmer) prescribes the NCERT Mathematics textbook; chapters match the national syllabus while RBSE sets its own exam pattern.

Add only the surfaces that are exposed. Faces where two solids join are hidden and must not be counted.

The volume. Set 'volume before = volume after' and solve for the unknown dimension or count.

The part of a cone left after slicing off the top with a plane parallel to the base — like a bucket or a glass. It has two circular faces of different radii.

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