Class 10 Mathematics – Surface Areas and Volumes | Interactive Visual Revision Notes

📝 Section 1 — Classroom Notes

🌎 Real-World Connection: A factory making tin cans must know how much metal sheet to cut (surface area) and how much liquid each can holds (volume). A decorator covering a dome with tiles needs the curved surface area. An ice cream cone combines a cone and a hemisphere. These everyday objects show why surface area and volume calculations matter in manufacturing, construction, and packaging.

📏 Topic 1: What Are Combined Solids?

A combined solid joins two or more shapes.
We combine basic shapes like cones and cylinders.
A water tank may be a cylinder + hemisphere.
A sharpened pencil is a cylinder + cone.
We break the shape into known parts.
Then we use formulas for each part.

📏 Topic 2: Surface Area of Combined Solids

Add the visible surfaces of each part.
Do not count surfaces hidden inside.
Where two parts join, that area is hidden.
Example: cone on a cylinder top.
We add cone CSA + cylinder CSA + one base.
The top circle of the cylinder is hidden.
Always draw the shape and mark hidden parts.

📏 Topic 3: Volume of Combined Solids

Volume = total space inside the solid.
Add the individual volumes of each part.
No surface is hidden for volume work.
Volume of combined = V₁ + V₂ + V₃ ...
Sometimes we subtract a carved-out part.
A hollow sphere = outer volume − inner volume.

📏 Topic 4: Cone

A cone has a circular base and a pointed top.
r = radius, h = height, l = slant height.
Slant height: l = √(r² + h²)
Curved Surface Area = πrl
Total Surface Area = πrl + πr²
Volume = 1/3 × πr²h
A cone is one-third of a cylinder with same base.

CSA of Cone = πrl | Volume = ⅓ πr²h

📏 Topic 5: Cylinder

A cylinder has two equal circular ends.
r = radius of the base, h = height.
Curved Surface Area = 2πrh
Total Surface Area = 2πrh + 2πr²
This simplifies to 2πr(h + r).
Volume = πr²h
Think of it as a stack of circles.

CSA of Cylinder = 2πrh | Volume = πr²h

📏 Topic 6: Sphere

A sphere is perfectly round in 3D.
Every point on it is the same distance from the centre.
r = radius of the sphere.
Surface Area = 4πr²
Volume = 4/3 × πr³
A ball, a globe, and a marble are spheres.

Surface Area = 4πr² | Volume = ⅔ πr³

📏 Topic 7: Hemisphere

A hemisphere is half of a sphere.
Cut a sphere through its centre = two hemispheres.
Curved Surface Area = 2πr²
Total Surface Area = 2πr² + πr² = 3πr²
The flat circle on top adds πr².
Volume = 2/3 × πr³
A bowl shape is a good example.

CSA = 2πr² | TSA = 3πr² | Volume = ⅔ πr³ ÷ 2

📏 Topic 8: What Is a Frustum?

A frustum is a cone with its top sliced off.
The cut is made parallel to the base.
It has two circular ends of different sizes.
R = radius of the bigger end (bottom).
r = radius of the smaller end (top).
h = height, l = slant height of the frustum.
A bucket shape is a common frustum.

📏 Topic 9: Surface Area of a Frustum

Slant height: l = √[h² + (R − r)²]
Curved Surface Area = π(R + r)l
Total Surface Area = CSA + both circular ends.
TSA = π(R + r)l + πR² + πr²
Remember to add both circles for total.
Use the slant height formula first if not given.

CSA of Frustum = π(R + r)l

📏 Topic 10: Volume of a Frustum

Volume = πh/3 × (R² + r² + Rr)
This formula has three terms inside.
R² = square of the big radius.
r² = square of the small radius.
Rr = product of both radii.
Multiply everything by πh/3.
Always use consistent units for R, r, and h.

Volume of Frustum = πh/3 × (R² + r² + Rr)

🎨 Section 2 — Easy Illustrations

Picture 1: Cone with Labels

A cone has a circular base, a pointed apex, and slant height connecting them.

Picture 2: Cylinder with Labels

A cylinder has two equal circular ends connected by a curved surface.

Picture 3: Sphere and Hemisphere

A sphere is a full ball. A hemisphere is exactly half of a sphere.

Picture 4: Frustum of a Cone

A frustum has two parallel circular faces of different sizes, connected by a sloping surface.

Picture 5: Combined Solid — Cylinder with Hemisphere

The top circle of the cylinder is hidden inside the hemisphere. Do not count it twice.

✏️ Section 3 — Worked Examples

Example 1: Cone on a Hemisphere — Total Surface Area

Question: A toy is shaped like a cone sitting on top of a hemisphere. The cone has height 9 cm and the common radius is 3.5 cm. Find the total surface area.

1 Cone radius r = 3.5 cm, height h = 9 cm.

2 Find the slant height of the cone.

3 l = √(h² + r²) = √(81 + 12.25) = √93.25

4 l ≈ 9.66 cm.

5 CSA of cone = πrl = 22/7 × 3.5 × 9.66

6 CSA of cone = 11 × 9.66 = 106.26 cm².

7 CSA of hemisphere = 2πr² = 2 × 22/7 × 12.25

8 CSA of hemisphere = 2 × 38.5 = 77 cm².

9 Total = 106.26 + 77 = 183.26 cm².

Total Surface Area ≈ 183.26 cm²

No base circle is exposed — the cone sits on the hemisphere.

Example 2: Volume of a Cylindrical Tank in Litres

Question: A cylindrical water tank has radius 7 m and height 10 m. Find the volume in litres.

1 r = 7 m, h = 10 m.

2 Volume = πr²h = 22/7 × 7² × 10

3 = 22/7 × 49 × 10

4 = 22 × 7 × 10 = 1540 m³.

5 1 m³ = 1000 litres.

6 Volume = 1540 × 1000 = 15,40,000 litres.

Volume = 1540 m³ = 15,40,000 litres

Always convert m³ to litres by multiplying by 1000.

Example 3: Volume of a Frustum

Question: A frustum has big radius R = 14 cm, small radius r = 7 cm, and height h = 24 cm. Find the volume.

1 Use: Volume = πh/3 × (R² + r² + Rr).

2 R² = 14² = 196.

3 r² = 7² = 49.

4 Rr = 14 × 7 = 98.

5 Sum = 196 + 49 + 98 = 343.

6 V = 22/7 × 24/3 × 343

7 V = 22/7 × 8 × 343

8 V = 22 × 8 × 49 = 8624 cm³.

Volume of Frustum = 8624 cm³

343 ÷ 7 = 49, so the 7 in the denominator cancels neatly.

Shape	Curved / Surface Area	Volume
Cone	CSA = πrl, TSA = πr(l + r)	⅓ πr²h
Cylinder	CSA = 2πrh, TSA = 2πr(h + r)	πr²h
Sphere	SA = 4πr²	⅔ πr³
Hemisphere	CSA = 2πr², TSA = 3πr²	⅔ πr³ ÷ 2
Frustum	CSA = π(R + r)l	πh/3 (R²+r²+Rr)
Cone slant	l = √(r² + h²)	Use for CSA
Frustum slant	l = √[h² + (R−r)²]	Use for CSA
Combined solid	Add visible surfaces only	Add all volumes
1 m³	= 1000 litres	Conversion
1 cm³	= 1 mL = 0.001 litre	Conversion
Hidden surface	Do NOT add to total SA	Draw and check
π value	Use 22/7 or 3.14	As told in question

📄 Section 5 — Take-Home Handout

📝 A. Fill in the Blanks

1. The curved surface area of a cone is _________.
2. A frustum is a cone with its _________ cut off.
3. The volume of a sphere is _________ πr³.
4. 1 m³ equals _________ litres.
5. The total surface area of a hemisphere is _________.

✓ B. True or False

1. A cone has two circular faces. _______
2. The volume of a hemisphere is 2/3 πr³. _______
3. To find TSA of a combined solid, add all surfaces including hidden ones. _______
4. A frustum has two parallel circular ends of different sizes. _______
5. The slant height of a cone is always shorter than its height. _______

✏️ C. Solve These

1. A cylinder has radius 5 cm and height 12 cm. Find its total surface area.

Show your work here

2. A solid sphere of radius 6 cm is melted and recast into a cone of radius 6 cm. Find the height of the cone.

Show your work here

🔗 D. Match the Columns

Column A	Column B
1. Cone volume	a. 4πr²
2. Sphere surface area	b. πr²h
3. Cylinder volume	c. π(R + r)l
4. Frustum CSA	d. ⅓ πr²h

🎨 E. Draw and Label

Draw a frustum and label R, r, h, and l clearly.

Draw your frustum here

✅ Detailed Solutions

Solution to Q1: Total Surface Area of a Cylinder

Question: Find the total surface area of a cylinder with radius 5 cm and height 12 cm. (Use π = 22/7.)

1 Formula: TSA of cylinder = 2πr(r + h).

2 r = 5 cm, h = 12 cm, r + h = 17 cm.

3 TSA = 2 × (22/7) × 5 × 17 = 2 × (22/7) × 85.

4 = (2 × 22 × 85) ÷ 7 = 3740/7 ≈ 534.29 cm².

∴ Total Surface Area = 3740/7 ≈ 534.29 cm²

Solution to Q2: Height of Cone from Melted Sphere

Question: A sphere of radius 6 cm is melted and recast into a cone of base radius 6 cm. Find the height of the cone.

1 Volume of sphere = Volume of cone (material is conserved).

2 (4/3)πr³ = (1/3)πR²h. Here r = R = 6 cm.

3 (4/3) × π × 6³ = (1/3) × π × 6² × h.

4 (4/3) × 216 = (1/3) × 36 × h ⇒ 288 = 12h.

5 h = 288 ÷ 12 = 24 cm.

∴ Height of the cone = 24 cm

🃏 Section 6 — Revision Flashcards

👉

📏

What is the curved surface area of a cone?

πrl

where l = √(r² + h²)

📦

What is the volume of a cylinder?

πr²h

Base area times height.

🌎

What is the volume of a sphere?

4/3 πr³

Surface area = 4πr²

🍴

What is a frustum?

A cone with its top sliced off parallel to the base. It has two circular ends of different sizes.

📐

What is the volume of a frustum?

πh/3 × (R² + r² + Rr)

Three terms inside the bracket.

💡

How do you find the TSA of a combined solid?

Add only the visible surfaces. Do not count surfaces that are hidden where shapes join.

⚫

What is the TSA of a hemisphere?

3πr²

CSA (2πr²) + flat circle (πr²).

📏

How many times the volume of a cone equals a cylinder of same base and height?

A cone is exactly one-third the volume of the cylinder with the same base and height.

3 cones = 1 cylinder

📏 Class 10 Mathematics — Surface Areas and Volumes Interactive Visual Revision Notes Aligned with the Class 10 Mathematics syllabus. Independently created study material. Not affiliated with or endorsed by NCERT.

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