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Updated on 23 Jul 2026, 15:38 IST
Surface Areas and Volumes Class 10 Notes PDF explains the formulas, concepts and methods used to calculate the surface area, volume and capacity of three-dimensional solids. These notes cover individual shapes, combinations of solids, hidden surfaces, unit conversions, solved examples, important questions and common mistakes.
The current CBSE Class 10 Mathematics curriculum includes surface areas and volumes of combinations of two solids selected from cubes, cuboids, spheres, hemispheres, right circular cylinders and cones.
Also Check: CBSE Class 10 Maths notes
Surface Areas and Volumes teaches students to calculate the outer covering, occupied space and capacity of simple and combined three-dimensional solids.
The chapter uses formulas for cubes, cuboids, cylinders, cones, spheres and hemispheres. Its main challenge is not formula memorisation; it is deciding which surfaces or volumes should be included in a particular situation.
By the end of the chapter, students should be able to:
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The Surface Areas and Volumes Class 10 Notes PDF contains formulas, diagrams, solved examples, combined-solid methods, important questions and a practice worksheet.
Current CBSE-facing textbooks and resources commonly place Surface Areas and Volumes as Chapter 12. Older textbook editions and study resources may label it Chapter 13.
Students should follow the chapter number shown in their prescribed textbook. The chapter title is more reliable than the number when comparing new and old resources.
Surface area measures the specified outer surfaces of a solid, volume measures the space occupied by it, and capacity measures the internal space available in a container.
Surface area: The total area of the specified surfaces of a three-dimensional object.
Surface area is required when a question asks about:
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Surface area is expressed in square units such as cm² or m².
Volume: The three-dimensional space occupied by a solid.
Volume is required when a question asks about:
Volume is expressed in cubic units such as cm³ or m³.
Capacity: The internal volume available inside a container.
Capacity is commonly measured in:
A water tank may have a large outer volume but a smaller capacity because the walls occupy space.
| Measurement | Meaning | Common clue words | Units |
| Surface area | Area of specified surfaces | Paint, cover, polish, wrap | cm², m² |
| Volume | Space occupied by a solid | Material, solid, occupied space | cm³, m³ |
| Capacity | Internal space available | Fill, hold, water, litres | mL, L |
Surface area combines two dimensions. For example, the area of a rectangle is length × breadth, so its unit is cm × cm = cm².
Volume combines three dimensions. The volume of a cuboid is length × breadth × height, so its unit is cm × cm × cm = cm³.
Correctly identifying the radius, diameter, height, slant height and exposed surfaces is necessary before choosing a formula.
| Symbol | Meaning |
| l | Length of a cuboid or slant height of a cone |
| b | Breadth |
| h | Perpendicular height |
| r | Radius |
| R | Larger or outer radius |
| d | Diameter |
| π | Pi |
| CSA | Curved surface area |
| LSA | Lateral surface area |
| TSA | Total surface area |
The meaning of l depends on the solid. It may represent the length of a cuboid or the slant height of a cone.
The radius is half the diameter:
r = d ÷ 2
If the diameter of a sphere is 14 cm:
r = 14 ÷ 2 = 7 cm
A common mistake is substituting the diameter directly into a formula that requires the radius.
A right circular cone has:
These measurements form a right-angled triangle:
l² = r² + h²
Therefore:
l = √(r² + h²)
Use:
Also Check: Triangles Class 10 Notes
A container with thick walls has different inner and outer dimensions.
For a cylindrical vessel:
Inner radius = Outer radius − Wall thickness
For an open cuboidal box with uniform wall thickness:
Inner length = Outer length − 2 × Thickness
Inner breadth = Outer breadth − 2 × Thickness
Use inner dimensions to calculate capacity and outer dimensions to calculate exterior surface area.
CSA or LSA excludes bases, while TSA includes all the surfaces of a closed solid.
Curved surface area: The area of only the curved part of a solid.
CSA is used for:
The curved surface area of a cylinder excludes its two circular bases.
Lateral surface area: The area of the side faces of a flat-sided solid, excluding its top and bottom.
LSA is used for:
The LSA of a cuboid represents the area of its four walls.
Total surface area: The area of every exposed surface of a closed solid.
For a closed cylinder, TSA includes:
| Solid | CSA or LSA includes | TSA additionally includes |
| Cube | Four side faces | Top and bottom |
| Cuboid | Four side faces | Top and bottom |
| Cylinder | Curved wall | Two circular bases |
| Cone | Curved sloping surface | Circular base |
| Hemisphere | Curved dome | Circular base |
| Sphere | Complete outer surface | Nothing additional |
The Class 10 formula table covers the surface areas and volumes of cubes, cuboids, cylinders, cones, spheres and hemispheres.
For a cube with edge a:
LSA = 4a²
TSA = 6a²
Volume = a³
Diagonal = a√3
For a cuboid with length l, breadth b and height h:
LSA = 2h(l + b)
TSA = 2(lb + bh + hl)
Volume = lbh
Diagonal = √(l² + b² + h²)
For a cylinder with radius r and height h:
CSA = 2πrh
TSA = 2πr(h + r)
Volume = πr²h
If the curved surface of a cylinder is cut vertically and opened, it forms a rectangle with:
Length = Circumference of base = 2πr
Breadth = Height of cylinder = h
Therefore:
CSA = 2πr × h = 2πrh
For a cone with radius r, height h and slant height l:
Slant height = √(r² + h²)
CSA = πrl
TSA = πr(l + r)
Volume = ⅓πr²h
Use l for surface area and h for volume.
For a sphere with radius r:
Surface area = 4πr²
Volume = 4/3πr³
A sphere has no flat base, so its entire surface is curved.
For a hemisphere with radius r:
CSA = 2πr²
TSA = 3πr²
Volume = 2/3πr³
The TSA of a hemisphere is:
Curved area + Circular base
= 2πr² + πr²
= 3πr²
| Solid | CSA or LSA | TSA | Volume |
| Cube | 4a² | 6a² | a³ |
| Cuboid | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cylinder | 2πrh | 2πr(h + r) | πr²h |
| Cone | πrl | πr(l + r) | ⅓πr²h |
| Sphere | 4πr² | 4πr² | 4/3πr³ |
| Hemisphere | 2πr² | 3πr² | 2/3πr³ |
Also Check: Class 10 Maths Formula Sheet
Do not use both 22/7 and 3.14 in the same calculation.
The correct formula is selected by identifying whether the question asks about covering, filling, exposed surfaces or conserved volume.
| Wording in the question | Required calculation |
| Paint, polish, cover or wrap | Surface area |
| Fill, hold or capacity | Volume |
| Curved wall only | CSA |
| Four walls of a cuboid | LSA |
| Complete closed exterior | TSA |
| Open at the top | Exclude the top |
| Open at both ends | Exclude both bases |
| Two solids joined | Exclude hidden joining faces |
| Melted and recast | Equate volumes |
| Material removed | Subtract removed volume |
The surface area of a combined solid is the total area of its exposed surfaces after hidden joining faces have been removed.
A combined solid is formed by joining or removing two or more simple solids.
Examples include:
Exposed surface: A surface that remains visible or accessible from outside.
Hidden surface: A joining surface covered by another solid.
If a cone and hemisphere are joined along equal circular bases, both circular faces become hidden. They must not be included in the exposed surface area.
Exposed surface area = Relevant component surfaces − Hidden joining surfaces
Blindly adding the TSAs of two joined solids usually counts their common faces twice.
A toy consists of a cone mounted on a hemisphere. Both have radius 3 cm. The cone has slant height 5 cm. Find the exposed surface area.
The circular base of the cone and the flat base of the hemisphere are joined, so both are hidden.
Exposed area = CSA of cone + CSA of hemisphere
= πrl + 2πr²
= π × 3 × 5 + 2π × 3²
= 15π + 18π
= 33π cm²
Using π = 22/7:
Exposed area = 726/7 cm²
Approximate exposed area = 103.71 cm²
Answer: The exposed surface area is 33π cm² or approximately 103.71 cm².
A capsule consists of a cylinder with two hemispherical ends. Its radius is 2 cm, and the cylindrical part is 8 cm long.
Two hemispheres form one complete sphere.
Exposed area = CSA of cylinder + Surface area of sphere
= 2πrh + 4πr²
= 2π × 2 × 8 + 4π × 2²
= 32π + 16π
= 48π cm²
Answer: The exposed surface area is 48π cm².
A hemisphere of radius r is mounted on the top face of a cube with edge a.
The circular region below the hemisphere is hidden.
Exposed area = TSA of cube − Hidden circular area + CSA of hemisphere
= 6a² − πr² + 2πr²
= 6a² + πr²
The hidden circular face is subtracted once, while the exposed curved hemisphere is added.
The volume of a combined solid is found by adding joined volumes or subtracting the volume of a removed part.
When solids are joined:
Combined volume = Volume of Solid 1 + Volume of Solid 2
When material is removed:
Remaining volume = Original volume − Removed volume
Hidden faces do not affect volume. They affect only exposed surface area.
A toy consists of a cone joined to a hemisphere. Both have radius 3 cm, and the cone has a perpendicular height of 4 cm.
Volume of cone:
= ⅓πr²h
= ⅓π × 3² × 4
= 12π cm³
Volume of hemisphere:
= 2/3πr³
= 2/3π × 3³
= 18π cm³
Total volume:
= 12π + 18π
= 30π cm³
Answer: The volume of the toy is 30π cm³.
A capsule has radius 2 cm and a cylindrical length of 8 cm.
Volume = Volume of cylinder + Volume of sphere
= πr²h + 4/3πr³
= π × 2² × 8 + 4/3π × 2³
= 32π + 32/3π
= 128/3π cm³
Answer: The volume is 128π/3 cm³.
A cuboid measures 20 cm × 14 cm × 10 cm. A cylindrical hole of radius 3 cm is drilled through its height of 10 cm.
Volume of cuboid:
= 20 × 14 × 10
= 2,800 cm³
Volume removed:
= π × 3² × 10
= 90π cm³
Remaining volume:
= 2,800 − 90π cm³
Using π = 22/7:
Remaining volume ≈ 2,517.14 cm³
When a solid is melted and recast without material loss, its volume remains constant even though its shape and surface area may change.
Original volume = New volume
Surface area is not conserved because the shape and exposed dimensions change.
A solid sphere of radius 6 cm is melted to form spheres of radius 2 cm. Find the number of smaller spheres.
Number of smaller spheres:
= Volume of large sphere ÷ Volume of one small sphere
= (4/3π × 6³) ÷ (4/3π × 2³)
= 6³ ÷ 2³
= 216 ÷ 8
= 27
Answer: 27 smaller spheres are formed.
A cylinder of radius 3 cm and height 8 cm is melted to form a sphere of radius R.
Volume of cylinder = Volume of sphere
π × 3² × 8 = 4/3πR³
72π = 4/3πR³
R³ = 54
R = ∛54 cm
Approximate radius = 3.78 cm
Surface area depends on the shape and dimensions of the exposed boundary. Recasting changes that boundary even when the amount of material remains constant.
All measurements must be converted to the same linear unit before applying a surface-area or volume formula.
| Conversion | Value |
| 1 m | 100 cm |
| 1 m² | 10,000 cm² |
| 1 m³ | 1,000,000 cm³ |
| 1 litre | 1,000 cm³ |
| 1 m³ | 1,000 litres |
| 1 mL | 1 cm³ |
Because:
1 m = 100 cm
Therefore:
1 m² = 100 cm × 100 cm
1 m² = 10,000 cm²
Because:
1 m = 100 cm
Therefore:
1 m³ = 100 cm × 100 cm × 100 cm
1 m³ = 1,000,000 cm³
Convert 25,000 cm³ into litres.
1 litre = 1,000 cm³
Capacity = 25,000 ÷ 1,000
= 25 litres
Answer: 25,000 cm³ equals 25 litres.
Tank and container problems require students to distinguish inner capacity, exterior area, open faces and wall thickness.
| Container | Surfaces included |
| Open cuboidal tank | Four walls and base |
| Closed cuboidal tank | All six faces |
| Cylinder open at one end | Curved wall and one base |
| Cylinder open at both ends | Curved wall only |
| Closed cylinder | Curved wall and both bases |
An open tank has length 2 m, breadth 1.5 m and height 1 m. Find the sheet metal required.
The tank has:
Required area:
= lb + 2lh + 2bh
= 2 × 1.5 + 2 × 2 × 1 + 2 × 1.5 × 1
= 3 + 4 + 3
= 10 m²
Answer: The tank requires 10 m² of sheet metal.
Volume:
= lbh
= 2 × 1.5 × 1
= 3 m³
Because 1 m³ = 1,000 litres:
Capacity = 3,000 litres
Answer: The tank can hold 3,000 litres.
A cuboidal tank has inner measurements of 180 cm × 140 cm × 160 cm.
Capacity:
= 180 × 140 × 160
= 4,032,000 cm³
Convert into litres:
= 4,032,000 ÷ 1,000
= 4,032 litres
Answer: The tank’s capacity is 4,032 litres.
CBSE competency material includes practical tank questions involving wall thickness, outer surface area, capacity and lids.
The following solved examples progress from direct formula use to reverse and application-based questions.
Find the TSA and volume of a cube with edge 5 cm.
TSA = 6a²
= 6 × 5²
= 150 cm²
Volume = a³
= 5³
= 125 cm³
Answer: TSA = 150 cm² and volume = 125 cm³.
A cylinder has radius 7 cm and height 10 cm.
CSA:
= 2πrh
= 2 × 22/7 × 7 × 10
= 440 cm²
Volume:
= πr²h
= 22/7 × 7² × 10
= 1,540 cm³
Answer: CSA = 440 cm² and volume = 1,540 cm³.
A cone has radius 3 cm and height 4 cm.
Slant height:
l = √(r² + h²)
= √(3² + 4²)
= √25
= 5 cm
CSA:
= πrl
= π × 3 × 5
= 15π cm²
Answer: Slant height = 5 cm and CSA = 15π cm².
The surface area of a sphere is 616 cm². Find its radius using π = 22/7.
4πr² = 616
4 × 22/7 × r² = 616
r² = 49
r = 7 cm
Answer: The sphere’s radius is 7 cm.
Find the TSA of a hemisphere with radius 7 cm.
TSA = 3πr²
= 3 × 22/7 × 7²
= 462 cm²
Answer: The TSA is 462 cm².
A closed cylindrical tank has radius 3.5 m and height 4 m. Find the cost of painting its exterior at ₹20 per m².
TSA:
= 2πr(h + r)
= 2 × 22/7 × 3.5 × (4 + 3.5)
= 165 m²
Cost:
= 165 × 20
= ₹3,300
Answer: The painting cost is ₹3,300.
The most common errors are counting hidden faces, confusing radius with diameter and using the wrong units or cone measurement.
| Mistake | Why it is wrong | Correct approach |
| Adding TSAs of joined solids | Counts hidden faces | Add only exposed surfaces |
| Using diameter as radius | Makes the input twice as large | Divide diameter by two |
| Using h in cone CSA | CSA requires slant height | Use πrl |
| Using l in cone volume | Volume requires perpendicular height | Use ⅓πr²h |
| Using outer measurements for capacity | Capacity uses inner space | Use inner dimensions |
| Equating surface areas during recasting | Only volume is conserved | Equate volumes |
| Writing cm² for volume | Volume is three-dimensional | Use cm³ |
| Counting the top of an open tank | The top does not exist | Omit the top |
Incorrect solution:
A sphere has diameter 14 cm, so its volume is:
4/3π × 14³
Error: The formula requires radius, not diameter.
Correction:
r = 14 ÷ 2 = 7 cm
Volume = 4/3π × 7³
Incorrect solution:
A student adds the TSA of a cone and the TSA of a hemisphere joined at equal circular bases.
Error: The two circular joining faces are hidden.
Correction:
Exposed area = CSA of cone + CSA of hemisphere
Incorrect solution:
A student uses πrh to calculate cone CSA.
Error: Cone CSA requires slant height.
Correction:
CSA of cone = πrl
Important questions should test formula selection, exposed-surface reasoning, unit conversion and practical modelling.
Q1. The TSA of a hemisphere with radius r is:
A. 2πr²
B. 3πr²
C. 4πr²
D. πr²
Answer: B. 3πr²
Q2. The volume of a cone is:
A. πr²h
B. 2πrh
C. ⅓πr²h
D. πrl
Answer: C. ⅓πr²h
Q3. One litre equals:
A. 100 cm³
B. 1,000 cm³
C. 10,000 cm³
D. 1 m³
Answer: B. 1,000 cm³
Q4. When a solid is melted and recast without wastage:
A. Surface area remains constant
B. Radius remains constant
C. Volume remains constant
D. Height remains constant
Answer: C. Volume remains constant
Q5. The CSA of a cylinder open at both ends is:
A. 2πr(h + r)
B. πr²h
C. 2πrh
D. 2πr²
Answer: C. 2πrh
Assertion: The common circular faces of two joined solids are not included in the exposed surface area.
Reason: The common faces become hidden after the solids are joined.
Answer: Both statements are true, and the reason correctly explains the assertion.
A company manufactures a closed capsule-shaped container consisting of a cylinder with two hemispherical ends. The radius is 3 cm, and the cylindrical section is 12 cm long.
Answer the following:
Which simple solids form the container?
What is the total length of the container?
What is its exposed surface area?
What is its volume?
Why are the circular bases of the cylinder not included separately?
Answer outline:
The container consists of one cylinder and two hemispheres.
Total length:
= Cylindrical length + Two radii
= 12 + 6
= 18 cm
Exposed area:
= CSA of cylinder + Surface area of sphere
= 2π × 3 × 12 + 4π × 3²
= 72π + 36π
= 108π cm²
Volume:
= π × 3² × 12 + 4/3π × 3³
= 108π + 36π
= 144π cm³
The circular bases are hidden where the hemispheres join the cylinder.
Also Check: Class 10 Maths Important Questions
Students can revise the chapter efficiently by learning formula meaning, practising surface identification and then attempting mixed applications.
NCERT questions should form the foundation of preparation because they cover the prescribed concepts and standard applications. Students should also practise current official competency-based questions and sample papers to improve interpretation and multi-step problem solving.
Also Check: NCERT Class 10 Maths solutions
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The main formulas cover the CSA, LSA, TSA and volume of cubes, cuboids, cylinders, cones, spheres and hemispheres. The complete formula table is provided near the beginning of these notes.
CSA measures only the curved part of a solid, while LSA measures the side faces of a flat-sided solid. TSA includes all exposed surfaces of a closed solid.
Use surface area for covering, painting or wrapping. Use volume for filling, capacity, material or occupied space.
Separate the object into simple solids and count only the surfaces exposed after joining. Common joining faces must not be included.
Yes, if the face becomes hidden where two solids are joined. It should not be counted as part of the exposed surface area.
Add the volumes of joined components. Subtract a component’s volume when that part has been removed or hollowed out.
No, volume remains constant when there is no material loss. Surface area may change because the new solid has different dimensions.
Divide the volume in cubic centimetres by 1,000. For example, 5,000 cm³ equals 5 litres.
Volume is the space occupied by a solid. Capacity is the internal space available inside a container.
Use slant height in the cone’s curved and total surface-area formulas. Use perpendicular height in the cone’s volume formula.
The current 2026–27 curriculum statement names cubes, cuboids, spheres, hemispheres, right circular cylinders and cones in its combinations-of-solids scope. Students should confirm frustum coverage using their current prescribed textbook and school guidance.
Current CBSE-facing editions commonly label it Chapter 12, while older editions may use Chapter 13. Follow the number in your prescribed textbook.
Students, download the CBSE Class 10 Surface Areas and Volumes Class 10 Notes PDF from Infinity Learn websites. A separate formula sheet and practice worksheet are also available.