Class 10 · Chapter 12 · Mensuration unit (10 of 80 marks)

Surface Areas and Volumes Class 10: notes and important questions

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 46 board-style questions.

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46 questions
13 multiple choice, 3 assertion–reason, 10 very short answer, 11 short answer, 6 long answer, 3 case study
About 11 hours to master

Mensuration unit: 10 of 80 theory marks (Areas Related to Circles and Surface Areas and Volumes).

NCERT textbook: Chapter 12 on ncert.nic.in (PDF)

Revision notes

Formulae (radius $r$ , height $h$ , slant height $l$ )

Solid	Curved/lateral SA	Total SA	Volume
Cuboid $l \times b \times h$	$2 h (l + b)$	$2 (l b + b h + h l)$	$l b h$
Cube (edge $a$ )	$4 a^{2}$	$6 a^{2}$	$a^{3}$
Cylinder	$2 π r h$	$2 π r (r + h)$	$π r^{2} h$
Cone ( $l = \sqrt{r^{2} + h^{2}}$ )	$π r l$	$π r (l + r)$	$\frac{1}{3} π r^{2} h$
Sphere	$4 π r^{2}$	$4 π r^{2}$	$\frac{4}{3} π r^{3}$
Hemisphere	$2 π r^{2}$	$3 π r^{2}$	$\frac{2}{3} π r^{3}$

Combinations of solids

Volume of a combined solid $=$ sum of the volumes of the parts. If a part is scooped out, subtract its volume.
Surface area is not simply the sum of the parts' surface areas. Add only the surfaces that are actually exposed. Where two solids are joined, the common face disappears from both.
Scooping a hemisphere (or cone) out of a face: remove the circular base area $π r^{2}$ from the face, and add the curved surface of the cavity.
Placing a hemisphere on a face: face area loses $π r^{2}$ , and $2 π r^{2}$ is gained. Net change $= + π r^{2}$ .

Worked example 1

A toy is a cone of radius 4.2 cm on a hemisphere of the same radius. The total height is 9.8 cm. Find its total surface area $(π = \frac{22}{7})$ .

Cone height $= 9.8 - 4.2 = 5.6$ , $l = \sqrt{{5.6}^{2} + {4.2}^{2}} = 7$ . TSA $= π r l + 2 π r^{2} = 92.4 + 110.88 = 203.28 {cm}^{2}$ .

Worked example 2

A solid is a cylinder of radius 7 cm and height 10 cm with a hemisphere on top. Find its volume.

$π r^{2} h + \frac{2}{3} π r^{3} = 1540 + 718.67 = 2258.67 {cm}^{3}$ .

Worked example 3

Two cubes of edge 6 cm are joined face to face. Find the surface area of the cuboid formed.

Cuboid $12 \times 6 \times 6$ : $2 (72 + 36 + 72) = 360 {cm}^{2}$ . Check: $2 \times 6 \times 36 - 2 \times 36 = 360$ .

Common errors

Adding the total surface areas of the parts and so counting the hidden joined faces.
Using the vertical height $h$ in $π r l$ instead of the slant height.
Using the diameter as the radius, or mixing units (cm and m, mm and cm).
For a vessel, "inner surface area" excludes the open top.

Board-exam tips

Syllabus limit: combinations of any two of cubes, cuboids, spheres, hemispheres, right circular cylinders and cones.
List which surfaces are exposed before calculating. This is where most marks are lost.
Take out $π$ as a common factor to simplify arithmetic: e.g. $π r (2 h + l)$ .
$1 m^{3} = 1000$ litres and $1000 {cm}^{3} = 1$ litre.

Topics in this chapter: Volume of basic solids · Surface area of basic solids · Surface area of combined solids · Volume of combined solids.

Formulas for this chapter: Chapter 12 on the Class 10 formula sheet (free PDF and A3 poster).

Surface Areas and Volumes Class 10: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 5 pages · 164 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceVolume of combined solids

A cylinder and a cone have the same base radius and the same height. The ratio of the volume of the cylinder to the volume of the cone is

Show worked answer

$π r^{2} h : \frac{1}{3} π r^{2} h = 3 : 1$ A1
Correct option: (d) $3 : 1$

Q2·1 mark·Multiple choiceSurface area of combined solids

A solid hemisphere has radius $14$ cm. Taking $π = \frac{22}{7}$ , its total surface area is

Q3·1 mark·Assertion–reasonSurface area of combined solids

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): Two identical cubes of edge $a$ are joined face to face. The total surface area of the cuboid formed is $10 a^{2}$ .

Reason (R): When two solids are joined, the surface area of the new solid is the sum of their surface areas.

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerSurface area of combined solids

A conical cap is $12$ cm tall with a circular base of radius $5$ cm. How much material, in terms of $π$ , is needed for its curved surface? (Find the slant height first.)

Show worked answer

Answer:

13

cm;

65 π {cm}^{2}

$ℓ = \sqrt{5^{2} + 12^{2}} = 13$ cm A1
Curved surface area $= π r ℓ = 65 π {cm}^{2}$ A1

Q5·2 marks·Short answerVolume of combined solids

A solid is made of a cylinder of radius $3$ cm and height $10$ cm with a hemisphere of radius $3$ cm fixed to one end. Find its volume in terms of $π$ .

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerSurface area of combined solids

A solid is a cylinder of radius $6$ cm and height $8$ cm with a cone of the same radius and slant height $10$ cm on top. Find its total surface area in terms of $π$ .

Show worked answer

Answer:

192 π {cm}^{2}

Curved surface of the cone $= π \times 6 \times 10 = 60 π$ M1
Curved surface of the cylinder $= 2 π \times 6 \times 8 = 96 π$ ; base $= π \times 36 = 36 π$ A1
Total $= 60 π + 96 π + 36 π = 192 π {cm}^{2}$ (the top of the cylinder is covered by the cone) A1

Q7·3 marks·Short answerVolume of combined solids

From a solid cylinder of height $14$ cm and base radius $6$ cm, a cone of the same height and base radius is hollowed out. Find the volume of the solid left. (Take $π = \frac{22}{7}$ .)

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerVolume of combined solids

A grain silo is a cylinder of radius $3$ m and height $8$ m with a conical roof of the same radius and height $4$ m. Find (a) the slant height of the roof, (b) the volume of the silo, and (c) the area of metal sheet used for the curved wall and the roof. Give (b) and (c) in terms of $π$ .

Show worked answer

Answer:

5

84 π m^{3}

;

63 π m^{2}

(a) $ℓ = \sqrt{3^{2} + 4^{2}} = 5$ m A1
(b) Cylinder $= π \times 9 \times 8 = 72 π$ ; cone $= \frac{1}{3} π \times 9 \times 4 = 12 π$ M1
Volume $= 84 π m^{3}$ A1
(c) Wall $= 2 π \times 3 \times 8 = 48 π$ ; roof $= π \times 3 \times 5 = 15 π$ M1
Metal sheet $= 63 π m^{2}$ A1

Q9·5 marks·Long answerSurface area of combined solids

A solid metal block is a cuboid $12$ cm by $10$ cm by $8$ cm. A hemispherical hollow of radius $3$ cm is scooped out of its top face. Find the volume of metal left and the total surface area of the block now. (Take $π = 3.14$ .)

Section E style · Case study (4 marks)

Q10·4 marks·Case studySurface area of combined solids

Spinning top. A wooden spinning top is a cone of radius $3$ cm and height $4$ cm, joined at its base to a hemisphere of radius $3$ cm.

How long is the slant edge of the conical part? [1 mark]
What is the area of the conical surface, in terms of $π$ ? [1 mark]
Find the total surface area of the top to be painted, in terms of $π$ . [2 marks]

Show worked answer

$ℓ = \sqrt{9 + 16} = 5$ cm A1
$π r ℓ = 15 π {cm}^{2}$ A1
Hemisphere's curved surface $= 2 π r^{2} = 18 π$ M1
Total $= 15 π + 18 π = 33 π {cm}^{2}$ A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 10 assertion–reason questions · the Class 10 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can use the surface area and volume formulae of the cube, cuboid, cylinder, cone, sphere and hemisphere.

Read first: Formulae (radius r, height h, slant height l) 10 practice questions · checkpoint: 4 questions, 6 marks, pass 80%

Practise step 1

Step 2

Board standard

You can find the surface area and volume of solids made from two shapes, deciding which faces are exposed.

Read first: Formulae; Combinations of solids; Worked examples 1-3 15 practice questions · checkpoint: 4 questions, 8 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can write complete 5-mark answers on combined solids and real objects (tents, toys, vessels) with units and conversions.

Read first: Combinations of solids; Board-exam tips 7 practice questions · checkpoint: 3 questions, 13 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can handle drilled holes, hollow objects and the 'largest cone in a cube' questions that decide 95+.

Read first: Combinations of solids; Common errors 5 practice questions · checkpoint: 3 questions, 9 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 12 of the 46 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceSurface area of basic solids

The total surface area of a solid hemisphere of radius $r$ is

(a) $2 π r^{2}$
(b) $4 π r^{2}$
(c) $3 π r^{2}$
(d) $π r^{2}$

Q2·1 mark·Multiple choiceSurface area of basic solids

The slant height of a cone with base radius 5 cm and height 12 cm is

(a)13 cm
(b)17 cm
(c) $\sqrt{119}$ cm
(d)7 cm

Q3·1 mark·Multiple choiceVolume of basic solids

The volume of a hemisphere of radius 21 cm is $(π = \frac{22}{7})$

(a) $38808 {cm}^{3}$
(b) $19404 {cm}^{3}$
(c) $2772 {cm}^{3}$
(d) $9702 {cm}^{3}$

Worked solution·2 marks·Very short answerSurface area of basic solids

A solid cone stands on a circular base of radius $14$ cm and is $48$ cm tall. Taking $π = \frac{22}{7}$ , work out the area of its sloping surface and its whole surface area.

Full worked solution Free, no sign-in

$l = \sqrt{196 + 2304} = 50$ cm; curved surface $= π r l = 2200 {cm}^{2}$ A1
Total $= 2200 + π r^{2} = 2200 + 616$ $= 2816 {cm}^{2}$ A1

Every question here has a full worked solution. See them all →

Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I84 · Problem I91 · Problem I98 · Problem I16. All 28 →

Themed lesson: Halloween maths lesson · Diwali maths lesson · Holi maths lesson (starter, activity and worked answers). All themed maths

Knowledge organiser: Surface Areas and Volumes on one page (key facts, formulas and mistakes to avoid; printable PDF)

Formulas in this topic: Volume of a cone · Volume of a cylinder · Volume of a sphere. All formulas A–Z

Where marks are lost in Surface Areas and Volumes

Adding total surface areas of the parts and counting hidden joined faces. Fix: list exposed surfaces in words (curved surface of cone + curved surface of hemisphere) before any numbers.
Using the vertical height h in πrl. Fix: find l = √(r² + h²) as a separate line.
Mixing units (cm with m, cm³ with litres). Fix: convert at the start; 1000 cm³ = 1 litre and 1 m³ = 1000 litres.
Using the diameter as the radius. Fix: write r = d/2 on the first line.
Inner surface area of a vessel including the open top. Fix: read 'open', 'hollow', 'inner' carefully and drop that face.
Long arithmetic slips with π. Fix: factor out π (e.g. πr(2h + l)) and simplify before multiplying.

Exam strategy: how answers are step-marked · competency-based questions

Surface Areas and Volumes in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 14, 31
Sample paper 2 (Board standard): questions 15, 16, 34
Sample paper 3 (Above board): questions 15, 20, 35
Sample paper 4 (Above board): questions 15, 16, 25, 36
Sample paper 5 (95+ challenge): questions 15, 16, 34

Want it against the clock? Take a timed 30-mark chapter test on Surface Areas and Volumes, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Surface Areas and Volumes: common questions

What is in CBSE Class 10 Maths Chapter 12, Surface Areas and Volumes?

The topics are: Volume of basic solids, Surface area of basic solids, Surface area of combined solids, Volume of combined solids. This page has revision notes for each, and 46 board-style practice questions with step-marked solutions.

How many marks is Surface Areas and Volumes worth?

It is in the Mensuration unit, which carries 10 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Surface Areas and Volumes?

Adding total surface areas of the parts and counting hidden joined faces. Fix: list exposed surfaces in words (curved surface of cone + curved surface of hemisphere) before any numbers. Using the vertical height h in πrl. Fix: find l = √(r² + h²) as a separate line.

Are the Surface Areas and Volumes important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

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Revision notes

Formulae (radius r, height h, slant height l)

Combinations of solids

Worked example 1

Worked example 2

Worked example 3

Common errors

Board-exam tips

Surface Areas and Volumes Class 10: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Surface Areas and Volumes

Surface Areas and Volumes in our sample papers

Surface Areas and Volumes: common questions

Work through the Route to 95 with your progress saved

Formulae (radius $r$ , height $h$ , slant height $l$ )