CBSE Class 10 Mathematics Surface Areas and Volumes Notes

1. Introduction

Surface Areas and Volumes is one of the most important and high-weightage chapters in CBSE Class 10 Mathematics. Almost every board exam paper carries at least one 5-mark question and one 3-mark question from this chapter alone. The chapter builds on your knowledge of 3D shapes from earlier classes and takes it further by introducing combinations of solids — shapes formed when two or more basic 3D shapes are joined or carved out from each other.

In real life, this chapter is applied constantly — from designing containers, tanks, tents, and buildings, to calculating the amount of material needed to manufacture a product. Understanding these concepts deeply not only helps you score well in your board exam but also builds a strong foundation for competitive exams like JEE and NEET, where mensuration appears regularly.

What This Chapter Covers

• Surface Area of Combinations of Solids (Cuboid, Cylinder, Cone, Sphere, Hemisphere)

• Volume of Combinations of Solids

• Conversion of Solid from One Shape to Another

• Frustum of a Cone — Surface Area and Volume

• Real-life application problems — a CBSE board exam favourite

2. Revision — Basic 3D Shapes

Before studying combinations, it is essential to have all formulas of individual solids on your fingertips. This section provides a complete recap of every shape you will encounter in this chapter. Make sure to memorise all these — they appear directly in calculations.

2.1 Cuboid

Definition: A cuboid is a three-dimensional box-shaped figure with six rectangular faces, twelve edges, and eight vertices. All angles are right angles. A cube is a special cuboid where all sides are equal.

Total Surface Area (TSA) = 2(lb + bh + lh)

Lateral Surface Area (LSA) = 2(l + b) × h

Volume = l × b × h

Diagonal = √(l² + b² + h²)

where l = length, b = breadth, h = height

The Lateral Surface Area (LSA) is useful when calculating the area of only the four side walls — for example, painting the walls of a room (excluding floor and ceiling).

2.2 Cylinder

Definition: A cylinder is a solid with two circular bases of equal radius connected by a curved surface. Common examples include pipes, drums, cans, and pillars.

Curved Surface Area (CSA) = 2πrh

Total Surface Area (TSA) = 2πr(r + h)

Volume = πr²h

where r = radius of base, h = height

A hollow cylinder (like a pipe) has both an inner radius (r₁) and outer radius (r₂). Its CSA = 2πh(r₁ + r₂) and Volume = πh(r₂² − r₁²).

2.3 Cone

Definition: A cone is a solid with a circular base and a pointed top called the apex. The distance from the apex to the centre of the base is the height (h), and the slant height (l) is the distance from the apex to any point on the base circumference.

Slant Height = l = √(r² + h²)

Curved Surface Area (CSA) = πrl

Total Surface Area (TSA) = πr(r + l) = πr(r + √(r² + h²))

Volume = (1/3)πr²h

where r = base radius, h = vertical height, l = slant height

Important — Slant Height vs Vertical Height

Students often confuse h (vertical height) and l (slant height).

Always check what the question gives you:

• If h is given → calculate l = √(r² + h²) before using CSA formula

• If l is given → calculate h = √(l² − r²) before using Volume formula

These two intermediate steps cause most errors in cone problems.

2.4 Sphere

Definition: A sphere is a perfectly round 3D shape where every point on the surface is equidistant from the centre. There is no flat base — the entire surface is curved.

Surface Area = 4πr²

Volume = (4/3)πr³

where r = radius of sphere

2.5 Hemisphere

Definition: A hemisphere is exactly half a sphere. It has a curved surface and one flat circular base. You encounter it frequently in problems involving bowls, domes, and ice cream scoops placed on cones.

Curved Surface Area (CSA) = 2πr²

Total Surface Area (TSA) = 3πr²

Volume = (2/3)πr³

Note: TSA = CSA + base circle area = 2πr² + πr² = 3πr²

2.6 All Formulas at a Glance

Shape	CSA / LSA	TSA	Volume
Cuboid	2(l+b)h	2(lb+bh+lh)	l×b×h
Cube	4a²	6a²	a³
Cylinder	2πrh	2πr(r+h)	πr²h
Cone	πrl	πr(r+l)	(1/3)πr²h
Sphere	4πr²	4πr²	(4/3)πr³
Hemisphere	2πr²	3πr²	(2/3)πr³

3. Surface Area of Combinations of Solids

Most real-life objects are not simple solids — they are combinations of two or more basic shapes joined together. For example, a toy rocket may be a cone placed on top of a cylinder, or a capsule may be a cylinder with two hemispheres at each end. When two solids are joined, the surface area of the resulting solid is NOT simply the sum of surface areas of the individual solids. Instead, you must subtract the areas of the joined faces (which are now internal and no longer exposed).

Golden Rule for Surface Area of Combinations

Surface Area of Combination = Sum of exposed curved/lateral surfaces of each solid

Do NOT add the base area where two shapes are joined — that part is hidden inside.

Step-by-step approach:

1. Identify each solid forming the combination.

2. List the surfaces of each solid.

3. Remove surfaces that are joined (hidden) between solids.

4. Add up only the visible/exposed surfaces.

3.1 Cone on a Cylinder

This is one of the most commonly tested combinations — a cone placed on top of a cylinder (like a pencil or a tent). The base of the cone and the top face of the cylinder are joined, so both are excluded from the surface area.

TSA = CSA of Cylinder + Base of Cylinder + CSA of Cone

= 2πrh + πr² + πrl

Note: The top circle of the cylinder (joined to cone base) is NOT included.

Note: The base circle of the cone is also NOT included (hidden).

Worked Example 3.1 — Pencil Shape (Cone on Cylinder)

Problem: A solid toy is in the form of a cylinder with a conical top. The height of the cylinder is

12 cm and its radius is 3 cm. The height of the cone is 4 cm. Find the total surface area. (π = 3.14)

Step 1: Find slant height of cone

l = √(r² + h²) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm

Step 2: Identify exposed surfaces

• CSA of cylinder = 2πrh = 2 × 3.14 × 3 × 12 = 226.08 cm²

• Base of cylinder = πr² = 3.14 × 9 = 28.26 cm²

• CSA of cone = πrl = 3.14 × 3 × 5 = 47.1 cm²

Step 3: Total Surface Area = 226.08 + 28.26 + 47.1 = 301.44 cm²

3.2 Hemisphere on a Cylinder or Cone

Another common combination is a hemisphere placed on top of a cylinder — like a capsule tablet or a water storage tank with a dome roof. Here, the flat circular base of the hemisphere coincides with the top face of the cylinder, so both are excluded.

TSA (Hemisphere on Cylinder) = CSA of Cylinder + Base of Cylinder + CSA of Hemisphere

= 2πrh + πr² + 2πr²

= πr(2h + 3r)

Worked Example 3.2 — Medicine Capsule (Cylinder + 2 Hemispheres)

Problem: A medicine capsule is shaped like a cylinder with two hemispheres at each end.

Total length = 14 mm, diameter = 5 mm. Find the surface area. (π = 22/7)

r = 2.5 mm

Height of cylinder = 14 − 2r − 2r = 14 − 5 = 9 mm

Wait: length = cylinder height + 2 × hemisphere radius

∴ Cylinder height = 14 − 2(2.5) = 14 − 5 = 9 mm

CSA of cylinder = 2πrh = 2 × (22/7) × 2.5 × 9 = 141.43 mm²

CSA of 2 hemispheres = 2 × 2πr² = 4πr² = 4 × (22/7) × 6.25 = 78.57 mm²

Total Surface Area = 141.43 + 78.57 = 220 mm²

3.3 Hemisphere Carved from a Cylinder / Cube

In some problems, a hemisphere is scooped out (carved) from a solid. For example, a wooden block with a hemispherical pit on top. Here, instead of adding the CSA of the hemisphere, you still add it (the carved surface is exposed inside), and you subtract the base circle of the hemisphere from the top face of the cube or cylinder.

TSA (Cube with hemispherical cavity on top):

= TSA of Cube − Base circle of hemisphere + CSA of hemisphere

= 6a² − πr² + 2πr²

= 6a² + πr²

4. Volume of Combinations of Solids

Unlike surface areas, volumes of combined solids ARE simply added together. When two solids are joined, the total volume is the sum of the individual volumes — there is no subtraction of joined faces because volume is a capacity measure, not a surface measure. However, if a shape is carved out, you subtract that volume.

Golden Rule for Volume of Combinations

Volume of Combination = Sum of individual volumes

Volume when carved out = Volume of outer solid − Volume of carved solid

Key insight: Always check whether the shapes are ADDED or one is REMOVED.

4.1 Volume — Cone on Hemisphere

Volume = Volume of Cone + Volume of Hemisphere

= (1/3)πr²h + (2/3)πr³

Note: The radius r is the same for both the cone base and the hemisphere.

Worked Example 4.1 — Buoy / Float (Cone on Hemisphere)

Problem: A buoy is in the shape of a cone mounted on a hemisphere, both having a radius

of 7 cm. If the total height of the buoy is 19 cm, find its volume. (π = 22/7)

Height of hemisphere = radius = 7 cm

Height of cone = Total height − radius = 19 − 7 = 12 cm

Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 49 × 12 = 616 cm³

Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × 343 = 718.67 cm³

Total Volume = 616 + 718.67 ≈ 1334.67 cm³

4.2 Volume — Hemisphere Removed from Cylinder

Volume = Volume of Cylinder − Volume of Hemisphere

= πr²h − (2/3)πr³

Worked Example 4.2 — Hollow Cylinder with Hemispherical Depression

Problem: A cylindrical vessel of radius 3.5 cm and height 10 cm has a hemispherical

depression at the bottom. Find the inner surface area and volume. (π = 22/7)

Inner CSA of cylinder = 2πrh = 2 × (22/7) × 3.5 × 10 = 220 cm²

CSA of hemisphere = 2πr² = 2 × (22/7) × 12.25 = 77 cm²

Area of annular base (top rim) = 0 [open vessel — no top]

Inner surface area = 220 + 77 = 297 cm²

Volume = πr²h − (2/3)πr³

= (22/7)×12.25×10 − (2/3)×(22/7)×42.875

= 385 − 89.83 = 295.17 cm³

5. Conversion of Solid from One Shape to Another

This is a very frequently tested concept. The idea is simple: when a solid is melted and recast into another shape, the VOLUME remains the same. No material is added or lost in the conversion — only the shape changes. Similarly, when a liquid is poured from one vessel into another, the volume of liquid stays constant.

The Conversion Principle

Volume of original solid = Volume of new solid(s)

If one large solid is recast into n smaller identical solids:

Volume of large solid = n × Volume of small solid

∴ n = Volume of large solid ÷ Volume of small solid

Common scenarios in CBSE board papers:

• Sphere/cone melted and recast into smaller spheres

• Cylinder of metal converted into wires or rods

• Water from a cylindrical tank poured into conical containers

5.1 Sphere Melted into Smaller Spheres

Number of small spheres (n) = Volume of large sphere ÷ Volume of small sphere

= (4/3)πR³ ÷ (4/3)πr³

= R³ / r³

where R = radius of large sphere, r = radius of small sphere

Worked Example 5.1 — Sphere into Smaller Spheres

Problem: A solid metallic sphere of radius 12 cm is melted and recast into 64 smaller spheres.

Find the radius of each small sphere.

Volume of large sphere = (4/3)π × 12³ = (4/3)π × 1728

Volume of one small sphere = (4/3)πr³

Since: 64 × (4/3)πr³ = (4/3)π × 1728

∴ 64r³ = 1728

∴ r³ = 1728/64 = 27

∴ r = 3 cm

Answer: Radius of each small sphere = 3 cm

5.2 Cylinder Melted into Cones

Number of cones (n) = Volume of Cylinder ÷ Volume of Cone

= πR²H ÷ (1/3)πr²h

= 3R²H / r²h

where R, H = radius and height of cylinder; r, h = radius and height of cone

Worked Example 5.2 — Cylinder into Cones

Problem: A cylinder of radius 6 cm and height 8 cm is melted and recast into cones each of

radius 2 cm and height 3 cm. How many cones are formed?

Volume of cylinder = π × 36 × 8 = 288π cm³

Volume of one cone = (1/3) × π × 4 × 3 = 4π cm³

Number of cones = 288π / 4π = 72

Answer: 72 cones are formed.

5.3 Water Flow and Filling Problems

Another common type: a pipe or channel fills a tank. Here, the volume of water that flows out of the pipe in a given time equals the volume of the tank being filled.

Volume of water flowing per second = CSA of pipe × speed of water

= πr² × v

Time to fill tank = Volume of tank ÷ Volume flow per second

Worked Example 5.3 — Pipe Filling a Tank

Problem: A pipe with internal radius 1 cm carries water at a speed of 80 cm/s.

How long will it take to fill a cylindrical tank of radius 40 cm and height 72 cm?

Volume of tank = π × 40² × 72 = 115200π cm³

Volume flowing per second = π × 1² × 80 = 80π cm³/s

Time = 115200π / 80π = 1440 seconds = 24 minutes

6. Frustum of a Cone

A frustum is the portion of a cone that remains after the top part is cut off by a plane parallel to the base. The cut produces a smaller circle at the top and retains the original base circle at the bottom. Common real-life examples of frustums include buckets, lamp shades, flower pots, and drinking glasses.

Key measurements of a frustum: Let R = radius of the larger base, r = radius of the smaller base, h = vertical height, and l = slant height of the frustum.

Slant Height l = √[h² + (R − r)²]

Curved Surface Area (CSA) = π(R + r)l

Total Surface Area (TSA) = π(R + r)l + πR² + πr²

= π [ (R + r)l + R² + r² ]

Volume = (πh/3)(R² + Rr + r²)

where R = larger radius, r = smaller radius, h = height, l = slant height

Memory Aid — Frustum Formulas

CSA looks like a trapezium area: π × (sum of parallel radii) × slant height

TSA = CSA + both circular ends (one large base + one small top)

Volume formula has three terms inside: R², Rr, r² (think: R-Rr-r, like a middle term)

Slant height is always: l = √[h² + (R−r)²] — not h² + R² or h² + r²

6.1 Frustum — Worked Examples

Worked Example 6.1 — Bucket (Frustum of Cone)

Problem: A bucket in the shape of a frustum has radii 20 cm (bottom) and 10 cm (top) and

height 30 cm. Find: (i) volume of water it can hold, (ii) cost of metal sheet at ₹15 per 100 cm².

(Use π = 3.14)

Step 1: Slant height

l = √[h² + (R−r)²] = √[900 + (20−10)²] = √[900 + 100] = √1000 = 31.62 cm

Step 2: Volume

V = (π × 30/3)(20² + 20×10 + 10²)

= (3.14 × 10)(400 + 200 + 100)

= 31.4 × 700 = 21,980 cm³

≈ 21.98 litres

Step 3: CSA of frustum (open bucket — no top)

CSA = π(R+r)l = 3.14 × 30 × 31.62 = 2980.71 cm²

Base = πR² = 3.14 × 400 = 1256 cm²

Total metal sheet = 2980.71 + 1256 = 4236.71 cm²

Step 4: Cost = (4236.71 / 100) × 15 = ₹635.51

Worked Example 6.2 — Frustum into Sphere

Problem: A metallic frustum with R = 6 cm, r = 4 cm, h = 14 cm is melted and recast into a

sphere. Find the radius of the sphere. (Use π = 22/7)

Volume of frustum = (π × 14/3)(6² + 6×4 + 4²)

= (22/7 × 14/3)(36 + 24 + 16)

= (44/3) × 76 = 1114.67 cm³

Volume of sphere = (4/3)πR³ = 1114.67

∴ R³ = 1114.67 × 3 / (4 × 22/7) = 1114.67 × 21/88 = 266

∴ R = ∛266 ≈ 6.43 cm

7. Common Mistake Analysis

This section lists the most common errors students make in this chapter, along with the correct approach. Understanding these mistakes before the exam can save you valuable marks.

Common Mistake	Why It Is Wrong	Correct Approach
Adding all TSAs of individual solids	Counts joined surfaces twice	Add only exposed surface areas; subtract joined faces
Using h instead of l in cone CSA	CSA = πrl, not πrh	Always compute l = √(r²+h²) first if l is not given
Forgetting base circle of hemisphere in TSA	3πr² not 2πr²	TSA of hemisphere = CSA + flat base = 2πr² + πr² = 3πr²
Subtracting volumes in combinations	Volumes add, not subtract (unless carved)	Volume of combination = sum of volumes
Wrong frustum slant height formula	Using l = √(h²+R²) instead of √(h²+(R−r)²)	Always use l = √[h² + (R−r)²]
Not converting units	Mixing cm and m gives wrong answers	Convert all measurements to the same unit before solving

8. Complete Formula Summary

Use this table as a quick-reference sheet before your exam. All formulas from this chapter are consolidated here.

Solid / Concept	Formula	Key Notes
Cylinder — CSA	2πrh	No circles included
Cylinder — TSA	2πr(r+h)	Includes both bases
Cylinder — Volume	πr²h
Cone — Slant l	l = √(r²+h²)	Compute before CSA
Cone — CSA	πrl	l is slant height
Cone — TSA	πr(r+l)	Includes base circle
Cone — Volume	(1/3)πr²h	1/3 of cylinder
Sphere — SA	4πr²	No flat surface
Sphere — Volume	(4/3)πr³
Hemisphere — CSA	2πr²	Curved part only
Hemisphere — TSA	3πr²	CSA + base circle
Hemisphere — Vol	(2/3)πr³	Half of sphere
Frustum — l	√[h²+(R−r)²]	R>r always
Frustum — CSA	π(R+r)l
Frustum — TSA	π[(R+r)l + R² + r²]	Both bases included
Frustum — Volume	(πh/3)(R²+Rr+r²)	Three-term bracket

9. Key Points to Remember

• Surface area of combination = sum of EXPOSED surfaces only (exclude hidden joined faces).

• Volume of combination = sum of all individual volumes (nothing excluded unless carved out).

• Conversion problems: volume before = volume after melting/recasting.

• Slant height (l) must be calculated using l = √(r² + h²) if only h is given.

• Frustum slant height: l = √[h² + (R−r)²] — most commonly confused formula in this chapter.

• For sphere-related conversions: number of small spheres = (R/r)³.

• Hemisphere TSA = 3πr² (includes flat circular base).

• Use π = 22/7 for multiples of 7; π = 3.14 otherwise.

• Always draw a clear labelled diagram — it helps identify which surfaces are exposed.

• 1 litre = 1000 cm³ — useful when converting volume to capacity in water problems.

10. Practice Questions

These practice questions are modelled after actual CBSE board exam patterns. Attempt all three categories. Answers involve applying formulas directly — show all steps for full marks.

10.1 — 1 Mark Questions (VSA)

1. A cone and a cylinder have the same base radius and height. What is the ratio of their volumes?

2. The slant height of a cone is 10 cm and the radius is 6 cm. Find its vertical height.

3. A solid sphere of radius r is melted into 8 identical smaller spheres. Find the radius of each small sphere.

4. What is the curved surface area of a frustum with R = 10 cm, r = 4 cm, and slant height l = 13 cm?

5. A hemisphere of radius 7 cm has its flat face open. Write the formula for the total exposed surface area.

6. Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

10.2 — 3 Mark Questions (SA)

7. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. (Use π = 22/7)

8. A metallic sphere of radius 6 cm is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire. (Use π = 3.14)

9. A vessel is in the form of an inverted cone open at the top. Its height is 8 cm and the radius of the top is 5 cm. It is filled with water up to the brim. When lead shots, each a sphere of radius 0.5 cm, are dropped, one-quarter of water flows out. Find the number of lead shots dropped.

10. A solid is in the shape of a cone standing on a hemisphere with both radii equal to 1 cm and height of cone 1 cm. Find the volume of the solid in terms of π.

11. Water is flowing at the rate of 2.52 km/h through a cylindrical pipe of diameter 14 cm into a rectangular tank which is 10 m long and 4 m wide. Determine the time in which the level in the tank will rise by 3 m. (Use π = 22/7)

10.3 — 5 Mark Questions (LA)

12. A tent is in the shape of a cylinder surmounted by a conical top. The cylindrical portion has height 2.1 m and diameter 4 m. The conical part has a slant height of 2.8 m. Find: (i) the area of the canvas needed to make the tent; (ii) the cost of canvas at ₹500 per m². (Use π = 22/7)

13. A bucket is in the shape of a frustum of a cone with radii of its two circular ends as 40 cm and 20 cm and a height of 45 cm. Find its capacity and the cost of the tin sheet used for making it, if the cost of the tin sheet is ₹1.50 per dm². (Take π = 22/7)

14. A hemispherical tank full of water is emptied by a pipe at the rate of 3 and 4/7 litres per second. How much time will it take to empty half the tank if it is 3 m in diameter? (Use π = 22/7; 1 litre = 0.001 m³)

15. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base radius is 3.5 cm, find the total surface area of the article. (Use π = 22/7)

16. From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find the volume and total surface area of the remaining solid. (Take π = 3.14)

11. Comparison — Surface Area vs Volume of Combinations

Aspect	Surface Area	Volume
What it measures	Outer boundary / exposed face area	Space/capacity enclosed inside
For joined solids	Add only exposed surfaces; subtract joined face areas	Simply add all volumes
For carved-out solids	Add TSA of outer + CSA of carved; subtract base of cavity	Subtract carved volume from outer volume
Unit	cm², m², mm²	cm³, m³, mm³, litres
Real-life use	Amount of material / paint needed	Capacity of a container / liquid
Frustum CSA formula	π(R+r)l	Not applicable — use Volume formula
Conversion problems	Not applicable	Volume before = Volume after