Surface Areas and Volumes — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Maths · Ch 12Class 10 · CBSE 2026–27Medium weightage

Surface Areas and Volumes

What the board actually asks: surface area and volume of a COMBINATION of solids — cone-on-hemisphere toys, capsules, cylinders with hemispherical ends, a cuboid surmounted by a half-cylinder. Break the object into the basic solids you already know (cuboid, cube, cylinder, cone, sphere, hemisphere), add only the EXPOSED areas for surface area, and simply ADD the volumes.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

Break it into shapes you already know.

Every object in this chapter is two basic solids stuck together. Nothing new to memorise — you already know the surface areas and volumes of the cuboid, cube, cylinder, cone, sphere and hemisphere from Class IX. The whole skill is: see the pieces, then combine them by two rules. For surface area, add only the faces you can still see (the join hides some area). For volume, just add the pieces up (nothing disappears).

Formulae

Surface Area and Volume of a Cylinder

Cylinder

CSA = 2 π r h \cdot TSA = 2 π r (r + h) \cdot V = π r^{2} h

Tip: CSA = 2πrh; TSA adds two circles.

Surface Area and Volume of a Cone

Cone

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

Tip: l = √(r² + h²) first.

Surface Area and Volume of a Sphere

Sphere

Surface area = 4 π r^{2} \cdot V = \frac{4}{3} π r^{3}

Surface Area and Volume of a Hemisphere

Hemisphere

CSA = 2 π r^{2} \cdot TSA = 3 π r^{2} \cdot V = \frac{2}{3} π r^{3}

Tip: CSA 2πr², TSA 3πr².

Surface Area and Volume of a Cuboid and Cube

Cuboid & Cube

Cuboid TSA = 2 (l b + bh + h l), V = l bh \cdot Cube TSA = 6 a^{2}, V = a^{3}

Combination rule

Surface: add EXPOSED areas \cdot Volume: add the constituents

Tip: Surface: add only the EXPOSED areas (the joined faces disappear). Volume: add the parts.

Surface Area of a Combination of Solids

Rule and common layouts

Add only the exposed curved and flat areas. Cone on a hemisphere: TSA = CSA of hemisphere + CSA of cone. Cylinder with hemispherical ends: CSA of cylinder + two hemisphere CSAs. If a hemisphere or cone is scooped out of a cylinder, subtract the opening

π r^{2}

and add the inner curved surface.

Volume of a Combination of Solids

Rule

Volume of a combination = sum of the volumes of the parts. If a solid is scooped out, subtract its volume.

Volume Relations

For the same

r

and

h

, a cone is one-third of the cylinder. A hemisphere is half of the sphere.

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Surface area of a combination of solids

NCERT verbatim — memorise this exact wording· Ch 12,

“the total surface area of the new solid is the sum of the curved surface areas of each of the individual parts.”

What it means: add up only the surfaces you can still SEE after joining — for a cylinder capped by two hemispheres, that is the two curved hemispheres plus the curved cylinder; the flat circular faces vanish inside the join.

Volume of a combination of solids

NCERT verbatim — memorise this exact wording· Ch 12,

“The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in the examples below.”

What it means: unlike surface area, volume loses nothing at the join — just ADD the volume of each basic solid. This is the one place where you literally sum the pieces with no subtraction for the shared face.

Why surface areas are not simply added

NCERT verbatim — memorise this exact wording· Ch 12,

“in calculating the surface area, we have not added the surface areas of the two constituents, because some part of the surface area disappeared in the process of joining them.”

What it means: when two solids are stuck together, the faces that touch are hidden inside and no longer count. That is exactly why surface area is NOT the plain sum of the two full surface areas.

Total surface area is not additive

NCERT verbatim — memorise this exact wording· Ch 12,

“You may note that 'total surface area of the top' is not the sum of the total surface areas of the cone and hemisphere.”

What it means: the sharp warning NCERT prints after Example 1. Do not add TSA(cone) + TSA(hemisphere) — that double-counts the hidden circular base. Add only the exposed curved surfaces.

Key terms — NCERT verbatim (state word-for-word)

TSA of a cylinder with two hemispherical ends: TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere (Ch 12, )
TSA of a cone-and-hemisphere toy: Total surface area of the toy = CSA of hemisphere + CSA of cone (Ch 12, )
Surface area of a combination (Summary): To determine the surface area of an object formed by combining any two of the basic solids, namely, cuboid, cone, cylinder, sphere and hemisphere. (Ch 12, )
Volume of a combination (Summary): To find the volume of objects formed by combining any two of a cuboid, cone, cylinder, sphere and hemisphere. (Ch 12, )
Combination of solids: we come across a number of solids made up of combinations of two or more of the basic solids (Ch 12, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Break a combination into basic solids

Every object here is two basic solids joined at a shared face. Name the pieces first — e.g. a medicine capsule is a cylinder with a hemisphere on each end; a toy top is a cone surmounted by a hemisphere. Once you can see the pieces, all the formulas are ones you already know.

NCERT Fig. 12.10Medicine capsule - a cylinder with a hemisphere stuck to each end; length 14 mm, diameter 5 mm.

the setup mark on essentially every combination-solid question

Surface area = only the EXPOSED faces

Add the curved (and any still-visible flat) surfaces that remain after joining; drop the faces hidden inside the join. Never add the two full total surface areas — that double-counts the buried circle. When a solid is scooped out, the removed curved surface is added and the flat opening is handled carefully.

2-3 mark 'find the surface area / area to be painted' problems

Volume = simply ADD the constituents

Volume is the friendly rule: the volume of the joined solid is the plain sum of the pieces' volumes, because no volume is lost at the join. For a scooped/hollow solid, subtract the volume of the removed piece.

3-5 mark volume and capacity problems

The cone's slant height comes first

Any combination with a cone needs its curved surface area

π r l

, which needs the slant height

l = r^{2} + h^{2}

. Compute

l

BEFORE the CSA. And the cone's height is usually the whole solid's height minus the attached hemisphere's radius, not the whole height.

any combination containing a cone (toy top, tent, rocket)

Capacity, and hollowed / scooped solids

When a hemisphere is scooped out of a cylinder (a glass, a bird-bath), the capacity drops by the removed hemisphere's volume, while the surface gains the hemisphere's curved area in place of a flat lid. Read whether the question wants apparent capacity or actual capacity.

capacity and 'surface area of the remaining solid' problems

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 12, Example 1,

“Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)”

Use the combination result.

TSA of the toy = CSA of hemisphere + CSA of cone

— only the two curved surfaces are exposed.Concept: exposed surfaces only

½ mark

Read the dimensions.Radius

r = \frac{3.5}{2} = 1.75

cm. Height of the cone

= 5 - 1.75 = 3.25

cm (subtract the hemisphere's radius from the total height).Concept: cone height = total minus hemisphere radius

½ mark

Slant height first.

l = r^{2} + h^{2} = 1.7 5^{2} + 3.2 5^{2} \approx 3.7

cm.Concept: slant height before CSA

1 mark

Add the two curved areas.

TSA = 2 π r^{2} + π r l = π r (2 r + l) = \frac{11}{2} (3.5 + 3.7) = 39.6 cm^{2}

(approx.).Concept: sum of curved areas

1 mark

Mark-scheme logic (authored): 3-mark shape: 0.5 for the combination result (only the curved surfaces show), 0.5 for the correct cone height (whole height minus the hemisphere's radius), 1 for the slant height, 1 for adding the two curved areas to 39.6 cm². Classic lost mark: using 5 cm as the cone's height.

Ch 12, Example 4,

“Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = 22/7)”

Ch 12, Example 5,

“Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m³, and there are 20 workers, each of whom occupy about 0.08 m³ space on an average. Then, how much air is in the shed? (Take π = 22/7)”

Ch 12, Example 7,

“A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π = 3.14)”

Problem-solving strategies AUTHORED

New shape you can't name?Break it into two basic solids (cuboid, cube, cylinder, cone, sphere, hemisphere) — every object here is a combination of ones you already know.

Surface area?Add only the EXPOSED faces. Drop the circles hidden at the join; never add the two full total surface areas.

Volume?Simply ADD the pieces' volumes. For a scooped/hollow solid, subtract the removed piece.

A cone is involved?Find the slant height

l = r^{2} + h^{2}

first, and take the cone's height as the whole height minus the attached hemisphere's radius.

Units differ (m and cm)?Convert everything to one unit BEFORE substituting — a mixed-unit substitution is a guaranteed lost mark.

Common mistakes that cost marks

Where students lose easy marks

Adding the two full surface areas — TSA of a combination is NOT TSA(part 1) + TSA(part 2); the buried join circle is double-counted. Add only the exposed curved/flat areas.

Wrong cone height — for a cone on a hemisphere, the cone's height is the total height minus the hemisphere's radius, not the whole height.

Mixing units — combining a 1.45 m height with a 30 cm radius without converting. Bring everything to one unit first.

Subtracting for volume — volumes of joined solids simply ADD; nothing disappears at the join. Only SCOOPED-out solids involve subtraction.

Formulae

Surface Area and Volume of a Cylinder

Cylinder

CSA = 2 π r h \cdot TSA = 2 π r (r + h) \cdot V = π r^{2} h

Tip: CSA = 2πrh; TSA adds two circles.

Surface Area and Volume of a Cone

Cone

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

Tip: l = √(r² + h²) first.

Surface Area and Volume of a Sphere

Sphere

Surface area = 4 π r^{2} \cdot V = \frac{4}{3} π r^{3}

Surface Area and Volume of a Hemisphere

Hemisphere

CSA = 2 π r^{2} \cdot TSA = 3 π r^{2} \cdot V = \frac{2}{3} π r^{3}

Tip: CSA 2πr², TSA 3πr².

Surface Area and Volume of a Cuboid and Cube

Cuboid & Cube

Cuboid TSA = 2 (l b + bh + h l), V = l bh \cdot Cube TSA = 6 a^{2}, V = a^{3}

Combination rule

Surface: add EXPOSED areas \cdot Volume: add the constituents

Tip: Surface: add only the EXPOSED areas (the joined faces disappear). Volume: add the parts.

Surface Area of a Combination of Solids

Rule and common layouts

π r^{2}

and add the inner curved surface.

Volume of a Combination of Solids

Rule

Volume of a combination = sum of the volumes of the parts. If a solid is scooped out, subtract its volume.

Volume Relations

For the same

r

and

h

, a cone is one-third of the cylinder. A hemisphere is half of the sphere.

Shortcuts & Tricks

Cone volume = ⅓ cylinder; hemisphere = ½ sphere.

MCQ / check only: don't write this as working

Cone: h = 3V ÷ (πr²).

MCQ / check only: don't write this as working

Write each exposed face before adding.

Safe in written answers

Take πr² as a common factor before computing.

Safe in written answers

Memory Tricks

Do pie r h (bahar ki chadar); do pie r (r+h) (poora); pie r² h (bhar).

2πrh; 2πr(r+h); πr²h.

Pie r l (chadar); pie r (r+l) (poora); tihai pie r² h (bhar).

πrl; πr(r+l); ⅓πr²h.

Chaar pie r² (chadar); chaar-tihai pie r³ (bhar).

4πr²; ⁴⁄₃πr³.

Do, Teen, Do-tihai.

CSA, TSA, volume of a hemisphere. Do = CSA 2πr² · Teen = TSA 3πr² · Do-tihai = V (2/3)πr³.

Jahan do solid jude, sirf jude hue hisse ka area gayab.

combination rule. joined faces are not exposed.

Mindmap

- - Add only the EXPOSED faces
  - Hidden join circles don't count
  - Not TSA(part 1) + TSA(part 2)
- - Simply ADD the constituents
  - Nothing lost at the join
  - Scooped solid → subtract the removed piece
- - Cylinder: $2 π r h$ , $π r^{2} h$
  - Cone: $π r l$ , $\frac{1}{3} π r^{2} h$
  - Sphere: $4 π r^{2}$ , $\frac{4}{3} π r^{3}$
  - Hemisphere: $2 π r^{2}$ , $\frac{2}{3} π r^{3}$
- - Slant height $l = r^{2} + h^{2}$ first
  - Cone height = total - hemisphere radius
- - Adding the two full surface areas
  - Wrong cone height
  - Mixing m and cm
  - Subtracting for volume

Why the surface area of a combination is a sum of curved areas

This is the core reasoning of the chapter, worked on NCERT's cone-on-hemisphere toy (pp. 162-164). It shows WHY you add only the exposed curved surfaces, and why the total surface area is not the plain sum of the two solids' total surface areas.

Whenever we meet a new solid, break it into basic solids we have already solved.

NCERT's method for a combination: reduce the new problem to smaller problems already solved (p.162).

Build the toy by bringing the flat face of a cone and the flat face of a hemisphere together, with equal base radii.

The two flat circular faces meet exactly and are sealed inside the join, so neither is on the outside surface (p.163).

Only the curved surfaces remain visible, so the surface area is their sum.

TSA of the toy = CSA of hemisphere + CSA of cone = 2 π r^{2} + π r l

The exposed surface is exactly the curved hemisphere plus the curved cone (p.163).

Therefore the total surface area is NOT the sum of the two solids' full total surface areas.

TSA(cone) + TSA(hemisphere) would count the shared base circle twice; NCERT prints this warning after Example 1 (p.164).

For volume the same join hides nothing, so volumes simply add.

V_{toy} = \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h

No volume disappears at the join, so the combined volume is the sum of the constituents (p.167).

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

2 marksSection B

A right circular cone has volume 13,600 cm³ and base radius 20 cm. Find the angle which the slant height makes with the base radius. (Take π = 3 and √3 = 1.7.)

Right circular cone with its height dashed to the centre of the base; base radius 20 cm; right angle at the centre; angle between slant height and base radius marked

Step-marked solution

[1 mark] Volume = (1/3)πr²h = (1/3)(3)(20)²h = 400h = 13600 ⟹ h = 34 cm.
[1 mark] tan θ = h/r = 34/20 = 1.7 = √3 ⟹ θ = 60°.

5 marksSection D

Two rectangular sheets, each of dimensions 45 cm × 155 cm, are folded to make hollow right circular cylindrical pipes, with exactly 1 cm overlap when sticking the ends. Sheet 1 is folded along its length (so its height is 155 cm) and Sheet 2 along its width (so its height is 45 cm). Both pipes are closed at both ends. (i) Find the difference in the curved surface areas of the two cylinders. (ii) Find the ratio of the volumes of the two cylinders formed. (Use π = 22/7.)

Step-marked solution

[1 mark] Sheet 1: height h₁ = 155 cm, circumference = 45 − 1 = 44 cm. Sheet 2: height h₂ = 45 cm, circumference = 155 − 1 = 154 cm.
[1 mark] (i) CSA = circumference × height: CSA₁ = 44 × 155 = 6820 cm², CSA₂ = 154 × 45 = 6930 cm²; difference = 6930 − 6820 = 110 cm².
[1 mark] (ii) 2πr₁ = 44 ⟹ r₁ = 7 cm; 2πr₂ = 154 ⟹ r₂ = 49/2 cm.
[1 mark] (ii) V₁ = πr₁²h₁ = (22/7)(49)(155) = 23870 cm³; V₂ = πr₂²h₂ = (22/7)(2401/4)(45) = 84892.5 cm³.
[1 mark] (ii) V₁ : V₂ = 23870 : 84892.5 = 124 : 441.

5 marksSection D

A right-circular cylindrical water tanker (r = 1 m, h = 70 m) supplies water. Each colony has a cuboidal tank (7 × 2 × 3 m). Villages: 400 people fill spherical matkas (r = 21 cm). (i) How many colonies can one full tanker supply? (ii) If a tanker supplies 3 colonies then 400 matkas at a village, how much water (in m^3) does it supply? (Use π = 22/7.)

Step-marked solution

[1 mark] Tanker volume = π·r²·h = (22/7)·1²·70 = 220 m³; cuboidal tank volume = 7·2·3 = 42 m³.
[1 mark] (i) Number of colonies = 220 ÷ 42 = 5.23, so 5 colonies can be supplied completely.
[1 mark] Matka (sphere) volume = (4/3)·π·r³ = (4/3)·(22/7)·(21)³ = 38,808 cm³ ≈ 0.0388 m³.
[1 mark] (ii) Water for 3 colonies = 3 × 42 = 126 m³; water for 400 matkas = 400 × 0.0388 ≈ 16 m³.
[1 mark] (ii) Total water supplied = 126 + 16 = 142 m³ (approx).

Practise Surface Areas and Volumes competency questions

Read NCERT Chapter 12 on LazyTopper (jemh112)

Official NCERT chapter PDF (jemh112) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 12 (jemh112.pdf, printed pages 161-170). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Surface area of a combination of solids	verbatim	NCERT 2026-27 Ch12
Volume of a combination of solids	verbatim	NCERT 2026-27 Ch12
Why surface areas are not simply added	verbatim	NCERT 2026-27 Ch12
Total surface area is not additive	verbatim	NCERT 2026-27 Ch12
TSA of a cylinder with two hemispherical ends	verbatim	NCERT 2026-27 Ch12
TSA of a cone-and-hemisphere toy	verbatim	NCERT 2026-27 Ch12
Surface area of a combination (Summary)	verbatim	NCERT 2026-27 Ch12
Volume of a combination (Summary)	verbatim	NCERT 2026-27 Ch12
Combination of solids	verbatim	NCERT 2026-27 Ch12
Cone surmounted by a hemisphere - area to colour - NCERT Example 1	example	NCERT 2026-27 Ch12 Example 1 -164
Cylinder with a hemispherical depression - TSA - NCERT Example 4	example	NCERT 2026-27 Ch12 Example 4
Cuboid surmounted by a half-cylinder - volume of air - NCERT Example 5	example	NCERT 2026-27 Ch12 Example 5 -168
Hemisphere surmounted by a cone - volume and cylinder difference - NCERT Example 7	example	NCERT 2026-27 Ch12 Example 7
fig_capsule_1210	figure	NCERT 2026-27 Ch12 Fig. 12.10

Surface Areas and Volumes · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 12 (jemh112.pdf, printed pages 161-170)

Practise writing answers the way examiners mark them → Guide