V=πr²h V=πr²h/3 V=4πr³/3 l = √(r²+h²) (cone slant) CSA cyl = 2πrh CSA cone = πrl SA sphere = 4πr² Hemisphere: CSA=2πr², TSA=3πr², V=⅔πr³
V=
Chapter 11  ·  Class IX Mathematics

Cubes, Cuboids, Cylinders, Cones, and Spheres

Surface Areas and Volumes

Wrap It, Fill It, Measure It — Every 3D Shape Under One Chapter

Chapter Snapshot

10Concepts
16Formulae
12–14%Exam Weight
5–6Avg Q's
ModerateDifficulty

Why This Chapter Matters for Exams

CBSE Class IXNTSEState Boards

Surface Areas and Volumes is one of the highest-weightage chapters in Class IX, contributing 12–14 marks in CBSE Boards. Every paper includes at least two formula-based calculation problems — one on surface area and one on volume. Cylinders and cones are the most frequently tested solids. NTSE tests all standard formulas directly and also includes "find radius given volume" reverse problems.

Key Concept Highlights

Cuboid: Surface Area and Volume
Cube: Surface Area and Volume
Right Circular Cylinder
Cylinder: Curved and Total Surface Area
Cylinder: Volume
Right Circular Cone
Cone: Slant Height, CSA, TSA
Cone: Volume
Sphere: Surface Area and Volume
Hemisphere: CSA, TSA, and Volume

Important Formula Capsules

$\text{Cuboid: LSA}=2h(l+b),\ \text{TSA}=2(lb+bh+hl),\ V=lbh$
$\text{Cube: TSA}=6a^2,\ V=a^3$
$\text{Cylinder: CSA}=2\pi rh,\ \text{TSA}=2\pi r(r+h),\ V=\pi r^2 h$
$\text{Cone: }l=\sqrt{r^2+h^2},\ \text{CSA}=\pi rl,\ \text{TSA}=\pi r(l+r),\ V=\tfrac{1}{3}\pi r^2 h$
$\text{Sphere: SA}=4\pi r^2,\ V=\tfrac{4}{3}\pi r^3$
$\text{Hemisphere: CSA}=2\pi r^2,\ \text{TSA}=3\pi r^2,\ V=\tfrac{2}{3}\pi r^3$

What You Will Learn

Navigate to Chapter Resources

🏆 Exam Strategy & Preparation Tips

Create a formula card — there are 16 formulas to memorise, but they follow a pattern: cone always has the slant height l = √(r²+h²) which links SA and volume. Use π = 22/7 when radius is a multiple of 7, else π = 3.14. For "find dimension given SA or V", rearrange the formula algebraically before substituting. Time investment: 3–4 days.

Chapter 11 · CBSE · Class IX
🧊

Surface Area of a Right Circular Cone

NCERT Class 9 Mathematics Geometry Surface Area Volume Right Circular Cone Sphere Hemisphere Mensuration
📖 Introduction
📘 Definition
🎨 SVG Diagram
Labeled Diagram of a Right Circular Cone
Vertex h l r Base
📌 Note
Important Terms Related to Cone
🗒️ Relation Between Radius, Height and Slant Height
Relation Between Radius, Height and Slant Height

In a right circular cone, radius, height and slant height form a right-angled triangle. Therefore, by Pythagoras Theorem:

\[\small l^2=h^2+r^2 \]

Hence,

\[\small \boxed{l=\sqrt{h^2+r^2}} \]

This formula is extremely important because surface area formulas of a cone depend on slant height.

In board examinations, students often forget to calculate slant height before applying the CSA or TSA formula.
📘 Definition
Curved Surface Area of a Cone
📐 Derivation
Derivation of Curved Surface Area Formula

When the curved surface of a cone is opened, it forms a sector of a circle of radius \(\small l\).

The arc length of the sector equals the circumference of the cone’s base.

\[\small \text{Arc length}=2\pi r \]

Radius of the sector:

\[\small =l \]

Area of the complete circle of radius \(\small l\):

\[\small \pi l^2 \]

Therefore,

\[\small \frac{\text{Area of sector}}{\pi l^2} = \frac{2\pi r}{2\pi l} \] \[\small \frac{\text{Area of sector}}{\pi l^2} = \frac{r}{l} \] \[\small \text{Area of sector} = \pi l^2 \times \frac{r}{l} \] \[\small \boxed{\text{CSA}=\pi rl} \]
🔢 Formula
Total Surface Area of a Cone
🔢 Formula
Important Formula Chart
✏️ Example
Solved Examples
Find the curved surface area of a cone whose radius is \(\small 7\text{ cm}\) and slant height is \(\small 10\text{ cm}\).

Curved Surface Area of Cone:

\[\small \text{CSA}=\pi rl \]
  1. Write the given values.
  2. Apply the CSA formula.
  3. Simplify carefully.
  1. given
    \[\small r=7\text{ cm}, \quad l=10\text{ cm}\]
  2. \(\small \text{CSA}=\pi rl\)
    \[\small \begin{aligned}&=\frac{22}{7}\times 7\times 10\\ &=220\text{ cm}^2\end{aligned}\]
  3. Therefore,
    \[\small \boxed{\text{CSA}=220\text{ cm}^2}\]
A cone has radius \(\small 6\text{ cm}\) and height \(\small 8\text{ cm}\). Find its total surface area.
First calculate slant height using Pythagoras theorem.
  1. Slant Height
    \[\small \begin{aligned} l&=\sqrt{h^2+r^2}\\ &=\sqrt{8^2+6^2}\\ &=\sqrt{64+36}\\ &=\sqrt{100}\\ l&=10\text{ cm} \end{aligned} \]
  2. Total Surface Area:
    \[\small \begin{aligned} \text{TSA}&=\pi r(l+r)\\ &=\frac{22}{7}\times 6\times (10+6)\\ &=\frac{22}{7}\times 96\\ &=301.71\text{ cm}^2 \end{aligned} \]
  3. Therefore,
    \[\small \boxed{\text{TSA}\approx 301.71\text{ cm}^2}\]
📋 Case Study
CBSE Case Study Based Question

A company manufactures party caps in the shape of cones. Each cap has radius \(\small 5\text{ cm}\) and slant height \(\small 12\text{ cm}\). The company wants to know the amount of colored paper required to make one cap.

Think and Analyze

Since the cap is open at the bottom, only the curved surface area is needed.

Solution
\[\small \text{CSA}=\pi rl \] \[\small =\frac{22}{7}\times 5\times 12 \] \[\small =188.57\text{ cm}^2 \]

Therefore, the paper required is:

\[\small \boxed{188.57\text{ cm}^2} \]
⚡ Exam Tip
❌ Common Mistakes
Common Mistakes Students Make
  • Using height instead of slant height in CSA formula.
  • Forgetting to add base area while calculating TSA.
  • Writing incorrect square units.
  • Incorrect use of Pythagoras theorem.
  • Rounding answers too early in calculations.
⚡ Quick Revision
  • A cone has one circular base and one curved surface.
  • Slant height: \[\small l=\sqrt{h^2+r^2} \]
  • Curved Surface Area: \[\small \pi rl \]
  • Total Surface Area: \[\small \pi r(l+r) \]
  • Surface area is measured in square units.
🧊

Example 1

❓ Question
Find the curved surface area of a right circular cone whose slant height is \(\small 10\text{ cm}\) and base radius is \(\small 7\text{ cm}\).
💡 Concept
Concept Used
🎨 SVG Diagram
Cone Visualization
l = 10 cm r = 7 cm Vertex
🧩 Solution
Given

Radius of the cone:

\[\small r=7\text{ cm} \]

Slant height of the cone:

\[\small l=10\text{ cm} \]
  1. 1
    Write the formula for curved surface area.
  2. 2
    Substitute the values of radius and slant height.
  3. 3
    Simplify carefully.
  4. 4
    Write the answer with proper square units.
  1. Formula for curved surface area of a cone:
    \[\small \text{CSA}=\pi rl\]
  2. Substituting the given values:
    \[\small \begin{aligned} \text{CSA} &= \pi \times 7 \times 10 \\ &= \frac{22}{7}\times 7 \times 10 \\ &= 22 \times 10 \\ &= 220 \end{aligned} \]
  3. Therefore,
    \[\small \boxed{\text{Curved Surface Area}=220\text{ cm}^2}\]
⚡ Exam Tip
❌ Common Mistakes
  • Students sometimes use the height instead of the slant height.
  • Some students mistakenly calculate total surface area instead of curved surface area.
  • Forgetting to use square units in the final answer.
🧊

Eample 2

❓ Question

The height of a cone is \(\small 16\text{ cm}\) and its base radius is \(\small 12\text{ cm}\). Find:

  1. Curved Surface Area (CSA)
  2. Total Surface Area (TSA)

Use:

\[\small \pi=3.14 \]
💡 Concept
Concepts Used
🎨 SVG Diagram
Visual Representation
h = 16 cm l = 20 cm r = 12 cm Vertex
🗺️ Roadmap
Solution Roadmap
  1. Calculate the slant height using Pythagoras theorem.
  2. Use slant height to calculate curved surface area.
  3. Add the base area to calculate total surface area.
  4. Write answers with proper square units.
🧩 Solution

Given

  • Height of the cone: \[\small h=16\text{ cm} \]
  • Radius of the base: \[\small r=12\text{ cm} \]
  • Value of \(\small \pi\): \[\small \pi=3.14 \]

Calculation of Slant Height

Using:

\[\small l=\sqrt{h^2+r^2} \]

Substituting the values:

\[\small \begin{aligned} l &=\sqrt{16^2+12^2} \ &=\sqrt{256+144} \ &=\sqrt{400} \ &=20\text{ cm} \end{aligned} \]

Therefore,

\[\small \boxed{l=20\text{ cm}} \]

Curved Surface Area

Formula:

\[\small \text{CSA}=\pi rl \]

Substituting the values:

\[\small \begin{aligned} \text{CSA} &=3.14\times 12\times 20 \ &=753.6 \end{aligned} \]

Therefore,

\[\small \boxed{\text{CSA}=753.6\text{ cm}^2} \]

Total Surface Area

Formula:

\[\small \text{TSA}=\pi r(l+r) \]

Substituting the values:

\[\small \begin{aligned} \text{TSA} &=3.14\times 12\times (20+12) \ &=3.14\times 12\times 32 \ &=1205.76 \end{aligned} \]

Therefore,

\[\small \boxed{\text{TSA}=1205.76\text{ cm}^2} \]
Quantity Value
Curved Surface Area \(\small 753.6\text{ cm}^2\)
Total Surface Area \(\small 1205.76\text{ cm}^2\)
📌 Important Note
⚡ Exam Tip
❌ Common Mistakes
  • Using height instead of slant height directly in the CSA formula.
  • Forgetting to add the area of the circular base in TSA.
  • Arithmetic mistakes while simplifying decimal calculations.
  • Writing answers without units.
🧊

Example 3

❓ Question
A corn cob, shaped somewhat like a cone, has the radius of its broadest end as \(\small 2.1\text{ cm}\) and height as \(\small 20\text{ cm}\). If each \(\small 1\text{ cm}^2\) of the surface of the cob carries an average of 4 grains, find the total number of grains on the entire cob.
🎨 SVG Diagram
Cone Shaped Corn Cob
h = 20 cm r = 2.1 cm CONE SURFACE DENSITY 4 grains / cm²
💡 Concept
Concept Used
🗺️ Roadmap
Solution Roadmap
  1. Calculate the slant height using Pythagoras theorem.
  2. Find the curved surface area of the cone.
  3. Multiply the curved surface area by 4 to get total grains.
🧩 Solution

Given

  • Radius of the cone: \[\small r=2.1\text{ cm} \]
  • Height of the cone: \[\small h=20\text{ cm} \]
  • Number of grains per \(\small 1\text{ cm}^2\): \[\small 4 \]

Calculation of Slant Height

Using:

\[\small l=\sqrt{h^2+r^2} \]

Substituting the given values:

\[\small \begin{aligned} l &=\sqrt{20^2+(2.1)^2} \\ &=\sqrt{400+4.41} \\ &=\sqrt{404.41} \\ &\approx 20.1\text{ cm} \end{aligned} \]

Therefore,

\[\small \boxed{l\approx 20.1\text{ cm}} \]

Curved Surface Area of the Corn Cob

Formula:

\[\small \text{CSA}=\pi rl \]

Substituting the values:

\[\small \begin{aligned} \text{CSA} &\approx 3.14\times 2.1\times 20.1 \\ &\approx 3.14\times 42.21 \\ &\approx 132.54\text{ cm}^2 \end{aligned} \]

Therefore,

\[\small \boxed{\text{CSA}\approx 132.54\text{ cm}^2} \]

Number of Grains on the Corn Cob

Each \(\small 1\text{ cm}^2\) carries 4 grains.

Therefore:

\[\small \begin{aligned} \text{Total number of grains} &=132.54\times 4 \\ &\approx 530.16 \end{aligned} \]

Since the number of grains must be a whole number:

\[\small \boxed{\text{Total grains}\approx 531} \]
The cone-shaped corn cob contains approximately 531 grains.
⚡ Exam Tip
❌ Common Mistakes
  • Using total surface area instead of curved surface area.
  • Forgetting to calculate slant height first.
  • Rounding off too early during calculations.
  • Ignoring units in intermediate steps.
🧊

Surface Area of a Sphere and Hemisphere

📖 Introduction
📘 Definition
🎨 SVG Diagram
Labeled Diagram of a Sphere
O radius (r) diameter Sphere Geometry α-1.0
🔢 Formula
Surface Area of a Sphere
🗒️ Derivation of Surface Area Formula of Sphere
Derivation of Surface Area Formula of Sphere

Archimedes discovered that the surface area of a sphere is equal to four times the area of its greatest circular cross-section.

Area of a circle of radius \(\small r\):

\[\small \pi r^2 \]

Therefore, surface area of sphere:

\[\small 4\times \pi r^2 \] \[\small \boxed{\text{Surface Area}=4\pi r^2} \]
This is one of the most elegant and historically important formulas in geometry.
🤔 Important Features of a Sphere
  • Perfectly symmetrical in all directions.
  • Has only one curved surface.
  • Has no vertices, no edges, and no flat faces.
  • All points on the surface are equidistant from the centre.
  • Surface area increases with the square of the radius.
📘 Definition
What is a Hemisphere?
🎨 Labeled Diagram of a Hemisphere
O r Curved Surface Circular Base
ℹ️ Information
Structure of a Hemisphere

A hemisphere has two distinct surfaces:

  • Curved Surface
  • Flat Circular Base
🔢 Formula
Curved Surface Area of a Hemisphere
🔢 Formula
Important Formula Chart
📊 Difference Between Sphere and Hemisphere
Sphere Hemisphere
Completely round solid Half of a sphere
No flat surface One flat circular surface
Only one surface area formula Separate CSA and TSA formulas
Surface Area \(\small =4\pi r^2\) TSA \(\small =3\pi r^2\)
⚡ Exam Tip
❌ Common Mistakes
  • Using \(\small 2\pi r^2\) for the surface area of a sphere.
  • Forgetting to add base area while finding TSA of hemisphere.
  • Confusing radius and diameter.
  • Forgetting square units in answers.
📋 Case Study

A decorative dome is shaped like a hemisphere of radius \(\small 7\text{ m}\). The outer curved surface is to be painted.

Since only the curved outer part is painted:

\[\small \text{CSA}=2\pi r^2 \] \[\small =2\times \frac{22}{7}\times 7\times 7 \] \[\small =308\text{ m}^2 \]

Therefore, the area to be painted is:

\[\small \boxed{308\text{ m}^2} \]
⚡ Quick Revision
  • Sphere surface area: \[\small 4\pi r^2 \]
  • Hemisphere curved surface area: \[\small 2\pi r^2 \]
  • Hemisphere total surface area: \[\small 3\pi r^2 \]
  • Sphere has no edges and no vertices.
  • Hemisphere has one curved surface and one circular base.
🧊

Example 4

❓ Question
Find the surface area of a sphere of radius \(\small 7\text{ cm}\).
💡 Concept
🎨 Visual Representation of Sphere
O r = 7 cm Sphere
🗺️ Roadmap
Solution Roadmap
  1. Write the formula for the surface area of a sphere.
  2. Substitute the value of radius.
  3. Simplify step-by-step carefully.
  4. Write the answer with square units.
🧩 Solution

Given

Radius of the sphere:

\[\small r=7\text{ cm} \]

Formula:

\[\small A=4\pi r^2 \]

Substituting the given values:

\[\small \begin{aligned} A &=4\times \pi \times 7^2 \\ &=4\times \frac{22}{7}\times 7\times 7 \\ &=4\times 22\times 7 \\ &=88\times 7 \\ &=616 \end{aligned} \]

Therefore,

\[\small \boxed{\text{Surface Area}=616\text{ cm}^2} \]
⚡ Exam Tip
🗒️ Commo Mistake
  • Forgetting to square the radius.
  • Using diameter directly in place of radius.
  • Writing \(\small \pi r^2\) instead of \(\small 4\pi r^2\).
  • Forgetting square units in the final answer.
📋 Case Study

A decorative spherical balloon has radius \(\small 14\text{ cm}\). If the radius becomes twice its original value, how many times will the surface area increase?

Since:

\[\small A=4\pi r^2 \]

surface area depends on the square of radius.

\[\small (2r)^2=4r^2 \]

Therefore, the surface area becomes:

\[\small \boxed{4\text{ times}} \]
⚡ Quick Revision
  • Surface area of sphere: \[\small 4\pi r^2 \]
  • Sphere has only one curved surface.
  • Radius is always squared in the formula.
  • Surface area is measured in square units.
🧊

Example 5

❓ Question

Find:

  1. the Curved Surface Area (CSA)
  2. the Total Surface Area (TSA)

of a hemisphere of radius \(\small 21\text{ cm}\).

💡 Concept
🎨 SVG Diagram
Visual Representation of Hemisphere
Hemisphere Diagram A 3D-styled diagram of a hemisphere showing its curved surface, circular base, and radius. O r = 21 cm Curved Surface Circular Base
🗺️ Roadmap
  1. Use the CSA formula of hemisphere.
  2. Use the TSA formula of hemisphere.
  3. Substitute the radius carefully.
  4. Write answers with square units.
🧩 Solution

Radius of the hemisphere: \[\small r=21\text{ cm} \]

Curved Surface Area of Hemisphere

Formula:

\[\small \text{CSA}=2\pi r^2 \]

Substituting the given values:

\[\small \begin{aligned} \text{CSA} &=2\times \frac{22}{7}\times 21\times 21 \\ &=2\times 22\times 3\times 21 \\ &=44\times 63 \\ &=2772 \end{aligned} \]

Therefore,

\[\small \boxed{\text{CSA}=2772\text{ cm}^2} \]

Total Surface Area of Hemisphere

Formula:

\[\small \text{TSA}=3\pi r^2 \]

Substituting the values:

\[\small \begin{aligned} \text{TSA} &=3\times \frac{22}{7}\times 21\times 21 \\ &=3\times 22\times 3\times 21 \\ &=66\times 63 \\ &=4158 \end{aligned} \]

Therefore,

\[\small \boxed{\text{TSA}=4158\text{ cm}^2} \]
⚡ Exam Tip
❌ Common Mistakes
  • Using \(\small 4\pi r^2\) instead of \(\small 2\pi r^2\) for CSA.
  • Forgetting to include the circular base in TSA.
  • Arithmetic mistakes while simplifying.
  • Writing answers without square units.
📋 Case Study

A dome-shaped building is hemispherical in shape. The curved outer surface is to be painted while the circular base rests on the ground.

Which formula should be used?

Since only the outer curved part is painted:

\[\small \boxed{\text{CSA}=2\pi r^2} \]

The circular base is not exposed, so it is not included.

⚡ Quick Revision
  • Hemisphere is half of a sphere.
  • CSA of hemisphere: \[\small 2\pi r^2 \]
  • TSA of hemisphere: \[\small 3\pi r^2 \]
  • TSA includes the circular base area.
  • Surface areas are written in square units.
🧊

Example 6

❓ Question
The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of \(\small 7\text{ m}\). Find the area available to the motorcyclist for riding.
💡 Concept
🎨 HOLLOW SPHERE STUNT VISUALIZATION
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 d = 7m 90 91 92 93 HOLLOW SPHERE STUNT VISUALIZATION 96 97 98 99 100 101 102 103
🗺️ Roadmap
  1. Convert diameter into radius.
  2. Use the surface area formula of sphere.
  3. Simplify step-by-step carefully.
  4. Write the answer with square units.
🧩 Solution

Given

Diameter of the hollow sphere:

\[\small d=7\text{ m} \]

Radius of the sphere:

\[\small r=\frac{d}{2}=\frac{7}{2}=3.5\text{ m} \]

Formula for the surface area of a sphere:

\[\small \text{Surface Area}=4\pi r^2 \]

Substituting the value of radius:

\[\small \begin{aligned} \text{Surface Area} &=4\times \frac{22}{7}\times 3.5\times 3.5 \\ &=4\times \frac{22}{7}\times 12.25 \\ &=4\times 38.5 \\ &=154 \end{aligned} \]

Therefore,

\[\small \boxed{\text{Surface Area}=154\text{ m}^2} \]
The area available to the motorcyclist for riding inside the hollow sphere is \[\small \boxed{154\text{ m}^2} \]
⚡ Exam Tip
❌ Common Mistakes
  • Using diameter directly in the formula instead of radius.
  • Forgetting to square the radius.
  • Writing incorrect units.
  • Arithmetic mistakes while multiplying decimals.
📋 Case Study

If the diameter of the spherical cage becomes twice as large, how many times will the riding area increase?

Since:

\[\small \text{Surface Area}=4\pi r^2 \]

and radius also doubles,

\[\small (2r)^2=4r^2 \]

the riding area becomes:

\[\small \boxed{4\text{ times}} \]
⚡ Quick Revision
  • Surface area of sphere: \[\small 4\pi r^2 \]
  • Radius is half of diameter.
  • Surface area is measured in square units.
  • Surface area increases with square of radius.
🧊

Example 7

❓ Question
A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is \(\small 17.6\text{ m}\), find the cost of painting it, given that the painting cost is ₹5 per \(\small 100\text{ cm}^2\).
💡 Concept
🎨 SVG Diagram
Visualization of Hemispherical Dome
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Mathematics Challenge A hemispherical dome of a building needs to be painted. Base Circumference: 17.6 m Painting Cost: ₹5 per 100 cm² 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 Circumference = 17.6m 68 69 70 71 CALCULATE TOTAL SURFACE PAINTING COST
🗺️ Roadmap
  1. Find the radius using circumference formula.
  2. Calculate curved surface area of hemisphere.
  3. Convert painting rate into cost per square metre.
  4. Calculate total painting cost.
🧩 Solution

Finding the radius of the hemisphere:

Circumference of the circular base:

\[\small C=2\pi r \]

Substituting the given circumference:

\[\small \begin{aligned} 17.6 &=2\times \frac{22}{7}\times r \\ 17.6 &=\frac{44}{7}\times r \\ r &=\frac{17.6\times 7}{44} \\ r &=2.8\text{ m} \end{aligned} \]

Calculating the curved surface area of the hemisphere:

Curved surface area formula:

\[\small \text{CSA}=2\pi r^2 \]

Substituting the value of radius:

Substituting $r = 2.8$ m:

\[\small \text{CSA}=2\pi (2.8)^2 \]

Calculating:

\[\small \text{CSA}=2\times \frac{22}{7}\times 7.84 \] \[\small \text{CSA}=\frac{44}{7}\times 7.84 \] \[\small \text{CSA}=49.28\text{ m}^2 \]

Given painting cost:

\[\small \text{₹ }5\text{ per }100\text{ cm}^2 \]

Converting \(\small 100\text{ cm}^2\) into square metres:

\[\small 100\text{ cm}^2=\frac{100}{10,000}\text{ m}^2=0.01\text{ m}^2 \]

Therefore, painting cost per square metre:

\[\small \text{₹ }5\text{ per }0.01\text{ m}^2 = \text{₹ }500\text{ per m}^2 \]

Calculating total painting cost:

Total painting cost can be calculated as:

\[\small \text{Total Cost} = \text{Area to be Painted} \times \text{Cost per Unit Area} \]

Substituting the values:

\[\small \begin{aligned} \text{Total Cost} &=49.28\text{ m}^2\times \text{₹ }500\text{ per m}^2 \\ &=\text{₹ }24,640 \end{aligned} \]

Therefore,

\[\small \boxed{\text{Total Painting Cost}=\text{₹ }24,640} \]
💡 Important Mathematical Concept
⚡ Exam Tip
❌ Common Mistakes
  • Using TSA instead of CSA for painting problems.
  • Forgetting to convert \(\small \text{cm}^2\) into \(\small \text{m}^2\).
  • Incorrect simplification while calculating radius.
  • Missing square units in intermediate calculations.
📋 Case Study

If the radius of the hemispherical dome becomes double, how will the painting cost change?

Since:

\[\small \text{CSA}=2\pi r^2 \]

the area depends on \(\small r^2\). Therefore:

\[\small (2r)^2=4r^2 \]

Hence, the painting cost will become:

\[\small \boxed{4\text{ times}} \]
⚡ Quick Revision
  • Circumference of circle: \[\small 2\pi r \]
  • CSA of hemisphere: \[\small 2\pi r^2 \]
  • \(\small 1\text{ m}^2=10,000\text{ cm}^2\)
  • Cost = Area × Rate
🧊

Volume of a Right Circular Cone

📖 Introduction
💡 Concept
What is a Right Circular Cone?
🎨 Right Circular Cone
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 h 31 r 32 Vertex 33 CIRCULAR BASE 34 35
📌 What Does Volume Mean?
🔢 Formula
Volume Formula
📐 Derivation
Derivation of the Formula

Consider a cone and a cylinder having:

  • same radius \(\small r\),
  • same height \(\small h\).

Volume of cylinder:

\[\small \pi r^2 h \]

Experimentally, it is observed that:

Three cones of same dimensions completely fill one cylinder having the same base and height.

Therefore:

\[\small \text{Volume of cone} = \frac{1}{3}\times \text{Volume of cylinder} \] \[\small \boxed{V=\frac{1}{3}\pi r^2 h} \]
🗒️ Cone and Cylinder Relationship
Cone and Cylinder Relationship
Solid Volume Formula
Cylinder \(\small \pi r^2 h\)
Cone \(\small \frac{1}{3}\pi r^2 h\)

Therefore:

\[\small \boxed{ \text{Volume of cone} = \frac{1}{3} \times \text{Volume of cylinder} } \]
👁️ Observation
Important Mathematical Observations
🔢 Formula
Volume Formulas
⚡ Exam Tip
❌ Common Mistakes
  • Forgetting the factor \(\small \frac{1}{3}\).
  • Using slant height instead of perpendicular height.
  • Forgetting to square the radius.
  • Writing square units instead of cubic units.
  • Incorrect simplification of fractions.
📋 Case Study

A juice vendor fills ice cream cones with flavored ice. To know how much ice the cone can hold, which quantity should be calculated?

Since capacity is required:

\[\small \boxed{ \text{Volume of Cone} = \frac{1}{3}\pi r^2 h } \]

Volume helps determine the storage capacity of the cone.

⚡ Quick Revision
  • Volume of cone: \[\small \frac{1}{3}\pi r^2 h \]
  • Cone volume is one-third the volume of a cylinder of same base and height.
  • Volume is measured in cubic units.
  • Radius is squared in the formula.
  • Height used is perpendicular height, not slant height.
🧊

Example 8

❓ Question
The height and the slant height of a cone are \(\small 21\text{ cm}\) and \(\small 28\text{ cm}\) respectively. Find the volume of the cone.
💡 Concept
🎨 SVG Diagram
h = 21 cm l = 28 cm r = ? Height (h) Radius (r) Slant height (l) Surface Area = πr(l + r)
🗺️ Roadmap
solution roadmap
  1. Use Pythagoras theorem to calculate radius.

  2. Find \(\small r^2\).

  3. Substitute values in the cone volume formula.

  4. Simplify carefully and write answer in cubic units.

🧩 Solution
step-by-step solution
🗒️ Given
  • Height of the cone: \[\small h=21\text{ cm} \]
  • Slant height of the cone: \[\small l=28\text{ cm} \]

Find the Radius

To find the radius of the cone, we can use the Pythagorean theorem:

\[\small r^2 + h^2 = l^2 \]

Substituting the known values:

\[\small r^2 + (21)^2 = (28)^2 \]

\[\small r^2 + 441 = 784 \]

\[\small r^2 = 784 - 441 \]

\[\small r^2 = 343 \]

\[\small r = \sqrt{343} = 7\sqrt{7} \text{ cm} \]

Calculate the Volume

Now that we have the radius, we can calculate the volume of the cone using the formula:

\[\small V = \frac{1}{3} \pi r^2 h \]

Substituting the values:

\[\small V = \frac{1}{3} \pi (7\sqrt{7})^2 (21) \]

\[\small V = \frac{1}{3} \pi (49 \times 7) (21) \]

\[\small V = \frac{1}{3} \pi (343) (21) \]

\[\small V = \frac{1}{3} \pi (7203) \]

\[\small V = 2401\pi \text{ cm}^3 \]

⚡ Exam Tip
❌ Common Mistakes
  • Using slant height directly in the volume formula.
  • Forgetting the factor \(\small \frac{1}{3}\).
  • Arithmetic mistakes during simplification.
  • Writing square units instead of cubic units.
📋 Case Study

If both radius and height of a cone are doubled, how many times will the volume increase?

Since:

\[\small V=\frac{1}{3}\pi r^2 h \]

replacing \(\small r\) by \(\small 2r\) and \(\small h\) by \(\small 2h\):

\[\small \begin{aligned} V' &=\frac{1}{3}\pi (2r)^2(2h) \ &=\frac{1}{3}\pi (4r^2)(2h) \ &=8V \end{aligned} \]

Therefore, the volume becomes:

\[\small \boxed{8\text{ times}} \]
⚡ Quick Revision
  • Volume of cone: \[\small \frac{1}{3}\pi r^2 h \]
  • Pythagoras theorem: \[\small l^2=r^2+h^2 \]
  • Use perpendicular height for volume calculations.
  • Volume is measured in cubic units.
🧊

example 9

❓ Question
Monica has a piece of canvas whose area is \(\small 551\text{ m}^2\). She uses it to make a conical tent with base radius \(\small 7\text{ m}\). Assuming that stitching margins and wastage while cutting amount to approximately \(\small 1\text{ m}^2\), find the volume of the tent.
💡 Concept
🎨 SVG Diagram
Conical Tent h = 24 m l = 25 m r = 7 m Height (h) Radius (r) V = ⅓πr²h • S = πrl + πr²
🗺️ Roadmap
Solution roadmap
  1. Find the usable canvas area.
  2. Use curved surface area formula to find slant height.
  3. Use Pythagoras theorem to calculate vertical height.
  4. Calculate the volume of the cone.
🧩 Solution
step-by-step solution
🗒️ Given
  • Total canvas area: \[\small 551\text{ m}^2 \]
  • Wastage and stitching margin: \[\small 1\text{ m}^2 \]
  • Radius of tent: \[\small r=7\text{ m} \]

Effective Canvas Area

Total canvas area:

\[\small 551\text{ m}^2 \]

Wastage and stitching margin:

\[\small 1\text{ m}^2 \]

Therefore:

\[\small \begin{aligned} \text{Usable Canvas Area} &=551-1 \\ &=550\text{ m}^2 \end{aligned} \]

Finding the Slant Height

Since the canvas forms the curved surface:

\[\small \pi rl=550 \]

Substituting:

\[\small r=7\text{ m}, \quad \pi=\frac{22}{7} \] \[\small \begin{aligned} l &=\frac{550}{\pi r} \\ &=\frac{550}{\frac{22}{7}\times 7} \\ &=\frac{550}{22} \\ &=25\text{ m} \end{aligned} \]

Therefore,

\[\small \boxed{l=25\text{ m}} \]

Finding the Vertical Height

Using Pythagoras theorem:

\[\small l^2=r^2+h^2 \]

Rearranging:

\[\small h^2=l^2-r^2 \]

Substituting the values:

\[\small \begin{aligned} h^2 &=25^2-7^2 \\ &=625-49 \\ &=576 \end{aligned} \]

Therefore:

\[\small h=\sqrt{576} \] \[\small \boxed{h=24\text{ m}} \]

Surface Area Right Circular Cone

Formula:

\[\small V=\frac{1}{3}\pi r^2 h \]

Substituting the values:

\[\small \begin{aligned} V &=\frac{1}{3}\times \frac{22}{7}\times 7\times 7\times 24 \\ &=\frac{1}{3}\times 22\times 7\times 24 \\ &=\frac{3696}{3} \\ &=1232 \end{aligned} \]

Therefore,

\[\small \boxed{V=1232\text{ m}^3} \]
⚡ Exam Tip
❌ Common Mistakes
  • Using total surface area instead of curved surface area.
  • Forgetting to subtract wastage area.
  • Using slant height directly in volume formula.
  • Forgetting the factor \(\small \frac{1}{3}\).
  • Writing square units instead of cubic units.
📋 Case Study

If the radius of the tent doubles while height remains the same, how will the volume change?

Since:

\[\small V=\frac{1}{3}\pi r^2 h \]

replacing \(\small r\) by \(\small 2r\):

\[\small (2r)^2=4r^2 \]

Therefore, the volume becomes:

\[\small \boxed{4\text{ times}} \]
⚡ Quick Revision
  • CSA of cone: \[\small \pi rl \]
  • Volume of cone: \[\small \frac{1}{3}\pi r^2 h \]
  • Pythagoras theorem: \[\small l^2=r^2+h^2 \]
  • Volume is measured in cubic units.
  • Tent problems usually use curved surface area only.
🧊

Volume of a Sphere and Hemisphere

📖 Introduction
🤔 Did You Know?
What is Volume?

Volume represents the amount of space occupied by a three-dimensional object.

It tells us:

  • how much liquid a solid can hold,
  • how much material is needed to fill it,
  • or the storage capacity of the object.

Volume is always measured in cubic units:

\[\small \text{cm}^3,\quad \text{m}^3,\quad \text{mm}^3 \]
🎨 SVG Diagram
O r SPHERE
🔢 Formula
Formula for Volume of a Sphere
📐 Derivation

Archimedes discovered that the volume of a sphere is related to the volume of a cylinder surrounding it.

The formula derived is:

\[\small \boxed{V=\frac{4}{3}\pi r^3} \]

Here:

  • radius is cubed because volume measures three-dimensional space,
  • small changes in radius create large changes in volume.
📌 Note
Volume of a Hemisphere
🎨 SVG Diagram
Labeled Diagram of Hemisphere
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 29 30 31 32 33 34 38 39 40 46 47 48 53 54 55 60 61 62 63 64 65 O 66 r 67 HEMISPHERE 68 Circular Base 69
🔢 Formula
Important Formula Chart
🗒️ Important Features
  • Volume depends on \(\small r^3\), so small changes in radius greatly affect volume.
  • If radius doubles, volume becomes eight times.
  • Sphere is the most space-efficient three-dimensional shape.
  • Spherical shapes store maximum volume for minimum surface area.
  • Volume is always measured in cubic units.
🗒️ Comparison Between Sphere and Hemisphere
👁️ Important Mathematical Observation
⚡ Exam Tip
❌ Common Mistakes
  • Forgetting the factor \(\small \frac{4}{3}\).
  • Using \(\small r^2\) instead of \(\small r^3\).
  • Confusing sphere formulas with cylinder formulas.
  • Writing square units instead of cubic units.
  • Using diameter directly instead of radius.
📋 Case Study

A spherical water tank has radius \(\small 5\text{ m}\). To determine how much water it can store, which quantity should be calculated?

Since storage capacity is needed:

\[\small \boxed{ V=\frac{4}{3}\pi r^3 } \]

Volume determines the capacity of the tank.

⚡ Quick Revision
  • Volume of sphere: \[\small \frac{4}{3}\pi r^3 \]
  • Volume of hemisphere: \[\small \frac{2}{3}\pi r^3 \]
  • Radius is cubed in volume formulas.
  • Volume is measured in cubic units.
  • Doubling radius increases volume eight times.
🧊

example 10

❓ Question
Find the volume of a sphere of radius \(\small 11.2\text{ cm}\).
🎨 SVG Diagram
Visualization of Sphere
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 O 45 r = 11.2 cm 46 SPHERE 47
🗺️ Roadmap
Solution roadmap
  1. Write the volume formula of sphere.
  2. Substitute the value of radius.
  3. Simplify step-by-step carefully.
  4. Write answer in cubic units.
🧩 Solution

given

Radius of sphere:\(\small r=11.2\text{ cm}\)

Formula:

\[\small V=\frac{4}{3}\pi r^3 \]

Substituting the values:

\[\small \begin{aligned} V &=\frac{4}{3}\times \frac{22}{7}\times 11.2\times 11.2\times 11.2 \\ &=\frac{4}{3}\times \frac{22}{7}\times 1404.928 \\ &=\frac{4}{3}\times 4415.488 \\ &=5887.317 \end{aligned} \]

Therefore,

\[\small \boxed{V\approx 5887.3\text{ cm}^3} \]

Rounded to the nearest cubic centimetre:

\[\small \boxed{V\approx 5887\text{ cm}^3} \]
👁️ Important Observation
⚡ Exam Tip
❌ Common Mistakes
  • Using \(\small r^2\) instead of \(\small r^3\).
  • Forgetting the factor \(\small \frac{4}{3}\).
  • Writing square units instead of cubic units.
  • Incorrect decimal multiplication.
🧊

Example 11

❓ Question
A shot-putt is a metallic sphere of radius \(\small 4.9\text{ cm}\). If the density of the metal is \(\small 7.8\text{ g/cm}^3\), find the mass of the shot-putt.
💡 Concept
🎨 SVG Diagram
Metallic Sphere O r = 4.9 r
🗺️ Roadmap
Solution Roadmap
  1. Calculate the volume of the sphere.
  2. Multiply volume by density.
  3. Convert grams into kilograms.
🗒️ Step-by-step Solution
Step-by-step Solution

Given

Radius: \[\small r=4.9\text{ cm} \] Density: \[\small 7.8\text{ g/cm}^3 \]

Calculating Volume

Formula:

\[\small V=\frac{4}{3}\pi r^3 \]

Substituting the values:

\[\small \begin{aligned} V &=\frac{4}{3}\times \frac{22}{7}\times 4.9\times 4.9\times 4.9 \\ &=\frac{4}{3}\times \frac{22}{7}\times 117.649 \\ &=\frac{4}{3}\times 369.754 \\ &=493.005 \end{aligned} \]

Therefore,

\[\small \boxed{V\approx 493.0\text{ cm}^3} \]

Calculating Mass

Formula:

\[\small \text{Mass}= \text{Density}\times \text{Volume} \]

Substituting the values:

\[\small \begin{aligned} \text{Mass} &=7.8\times 493.005 \\ &=3845.439 \end{aligned} \]

Therefore,

\[\small \boxed{\text{Mass}\approx 3845.4\text{ g}} \]

Converting into Kilograms

Since:

\[\small 1000\text{ g}=1\text{ kg} \]

Therefore:

\[\small \begin{aligned} \text{Mass} &=\frac{3845.4}{1000} \\ &=3.8454\text{ kg} \end{aligned} \]

Rounded value:

\[\small \boxed{\text{Mass}\approx 3.85\text{ kg}} \]
💡 Important Concept
⚡ Exam Tip
❌ Common Mistakes
  • Forgetting the factor \(\small \frac{4}{3}\).
  • Incorrect cube calculation of radius.
  • Using wrong unit conversion between grams and kilograms.
  • Writing square units instead of cubic units.
🧊

example 12

❓ Question
A hemispherical bowl has a radius of \(\small 3.5\text{ cm}\). Find the volume of water it can contain.
💡 Concept
🎨 SVG Diagram
Hemispherical Bowl Water Capacity Visualization O r = 3.5 cm Water Level
🗺️ Roadmap
Soplution Raodmap
  1. Write the volume formula for hemisphere.
  2. Substitute the radius value.
  3. Simplify step-by-step carefully.
  4. Write the answer in cubic units.
🧩 Solution
Step-by-setep Solution
💡 Important Capacity Concept
👁️ Observation
Important Mathematical Observation
⚡ Exam Tip
❌ Common Mistakes
  • Using sphere formula instead of hemisphere formula.
  • Forgetting the factor \(\small \frac{2}{3}\).
  • Using \(\small r^2\) instead of \(\small r^3\).
  • Writing square units instead of cubic units.
  • Rounding too early during calculations.
⚡ Quick Revision
  • Volume of hemisphere: \[\small \frac{2}{3}\pi r^3 \]
  • Radius is cubed in volume calculations.
  • Volume is measured in cubic units.
  • \(\small 1\text{ cm}^3=1\text{ mL}\)
  • Capacity equals volume for containers.
🧊

Chapter Summary: Surface Areas and Volumes

🗺️ Chapter Overview

In this chapter, we studied important three-dimensional solids such as:

  • Right Circular Cone
  • Sphere
  • Hemisphere

We learned how to calculate:

  • Curved Surface Area (CSA)
  • Total Surface Area (TSA)
  • Volume
Surface area measures the outer covering of a solid, while volume measures the space occupied inside the solid.
🔢 Complete Formula Chart
🧠 Important Concepts to Remember
⚡ Exam Tip
⚡ Quick Final Revision Notes
  • Cone CSA: \[\small \pi rl \]
  • Cone TSA: \[\small \pi rl+\pi r^2 \]
  • Sphere Surface Area: \[\small 4\pi r^2 \]
  • Hemisphere CSA: \[\small 2\pi r^2 \]
  • Hemisphere TSA: \[\small 3\pi r^2 \]
  • Cone Volume: \[\small \frac{1}{3}\pi r^2 h \]
  • Sphere Volume: \[\small \frac{4}{3}\pi r^3 \]
  • Hemisphere Volume: \[\small \frac{2}{3}\pi r^3 \]
· Updated
NCERT · Class IX · Chapter 11

Surface Areas & Volumes

An intelligent interactive engine with step-by-step solutions, formula sheets, concept visualisations, quizzes and games.

Six Core Concepts

Click any card to expand its full derivation, formulas, and intuition builder.

📦
Cuboid & Cube
Rectangular solids formed by three pairs of congruent rectangular faces.
6 Faces · 12 Edges
🥫
Right Circular Cylinder
Curved surface generated by a rectangle rotating about one side.
CSA + 2 circles
🍦
Right Circular Cone
Apex over a circular base. Slant height is the key link.
ℓ = √(r²+h²)
🌐
Sphere & Hemisphere
A perfectly symmetric solid discovered by Archimedes.
4πr² · (4/3)πr³
🏗️
Combination of Solids
Real-world objects are composite. Add volumes; expose only visible faces.
V = V₁ + V₂ + …
🔄
Conversion of Solids
When recast, total volume is conserved.
Volume Conservation

Master Formula Sheet

All formulas for Chapter 11 in one place. Use π = 22/7 unless stated otherwise.

CuboidCubeCylinderConeSphereHemisphere
Key Formulas
Cuboid TSA2(lb + bh + hl)
Cuboid LSA2(l + b) × h
Cuboid Volumel × b × h
Cube TSA6a²
Cube LSA4a²
Cube Volume
Space Diagonal√(l² + b² + h²)
Key Formulas
CSA2πrh
TSA2πr(r + h)
Volumeπr²h
Hollow Volumeπ(R² − r²)h
Hollow TSA2π(R+r)h + 2π(R²−r²)
Key Formulas
Slant Height ℓ√(r² + h²)
CSAπrℓ
TSAπr(r + ℓ)
Volume(1/3)πr²h
Key Formulas
Sphere SA4πr²
Sphere Volume(4/3)πr³
Hemisphere CSA2πr²
Hemisphere TSA3πr²
Hemisphere Volume(2/3)πr³
Spherical Shell Volume(4/3)π(R³ − r³)
Essential Conversions
1 m= 100 cm = 1000 mm
1 m²= 10,000 cm²
1 m³= 10⁶ cm³ = 1,000 litres
1 litre= 1000 cm³ = 1 dm³
π (approx.)22/7 ≈ 3.14159

Solve Any Problem Instantly

Select a solid, choose the quantity to find, enter dimensions, and get a fully narrated step-by-step solution.

Curated Questions & Full Solutions

Original questions organised by concept with increasing difficulty. Click Show Solution for complete step-by-step workings.

Q 1.1Easy

A rectangular tank measures 3 m × 2 m × 1.5 m. It needs painting on all faces. If one litre covers 6 m², how many litres are needed?

1
Identify Dimensions
l = 3 m, b = 2 m, h = 1.5 m
2
TSA = 2(lb + bh + hl)
= 2(3×2 + 2×1.5 + 1.5×3) = 2(6 + 3 + 4.5) = 27 m²
3
Litres of Paint
= 27 ÷ 6 = 4.5 litres
4.5 litres
paint required
Q 1.2Medium

A cuboid has TSA = 148 cm². Its length is 6 cm and breadth is 4 cm. Find its height and volume.

1
Set Up Equation
2(6×4 + 4h + 6h) = 148 → 2(24 + 10h) = 148
2
Solve for h
24 + 10h = 74 → 10h = 50 → h = 5 cm
3
Volume
V = 6 × 4 × 5 = 120 cm³
h = 5 cm, V = 120 cm³
Height and Volume
Q 2.1Easy

A cylindrical well has diameter 4 m and depth 14 m. Find its volume and cost of plastering the curved inner surface at ₹8 per m².

1
Given: r = 2 m, h = 14 m
2
Volume
V = πr²h = (22/7)×4×14 = 176 m³
3
CSA & Cost
CSA = 2πrh = 2×(22/7)×2×14 = 176 m² → Wait: = 88 m² Cost = 88 × 8 = ₹704
176 m³ · ₹704
Volume and Cost
Q 2.2Hard

A metal pipe is 77 cm long with inner diameter 4 cm and outer diameter 4.4 cm. Find the total surface area of the pipe.

1
r = 2 cm, R = 2.2 cm, h = 77 cm
2
Outer CSA = 2πRh
= 2×(22/7)×2.2×77 = 1064.8 cm²
3
Inner CSA = 2πrh
= 2×(22/7)×2×77 = 968 cm²
4
2 Annular Rings = 2π(R²−r²)
= 2×(22/7)×(4.84−4) = 2×(22/7)×0.84 = 5.28 cm²
5
TSA = 1064.8 + 968 + 5.28
= 2038.08 cm²
2038.08 cm²
Total Surface Area of Pipe
Q 3.1Medium

A conical tent has diameter 24 m and slant height 13 m. Find (i) canvas area needed and (ii) cost at ₹70 per m².

1
r = 12 m, ℓ = 13 m (slant given directly)
2
Verify height: h = √(ℓ²−r²)
= √(169−144) = √25 = 5 m
3
Canvas = CSA = πrℓ
= (22/7)×12×13 = 3432/7 ≈ 490.29 m²
4
Cost
= (3432/7)×70 = 3432×10 = ₹34,320
490.3 m² · ₹34,320
Canvas and Cost
Q 4.1Easy

A hemispherical bowl of radius 10.5 cm is to be silver-plated on the inside. Find the area and cost at ₹5 per cm².

1
r = 10.5 cm; Area = CSA (inner surface only)
2
CSA = 2πr²
= 2×(22/7)×10.5² = 2×(22/7)×110.25 = 693 cm²
3
Cost = 693 × 5 = ₹3,465
693 cm² · ₹3,465
Area and Cost
Q 5.1Hard

A toy is a cylinder (r=3.5 cm, h=4 cm) topped with a hemisphere (same radius). Find total volume and total surface area.

1
Shape: Cylinder + Hemisphere, r = 3.5 cm
2
Total Volume
V = πr²h + (2/3)πr³ = πr²(h + 2r/3) = (22/7)×12.25×(4 + 7/3) ≈ 243.83 cm³
3
Total SA = base + cylinder CSA + hemisphere CSA
= πr² + 2πrh + 2πr² = πr(3r + 2h) = (22/7)×3.5×(10.5+8) = 11×18.5 = 203.5 cm²
V ≈ 243.8 cm³ · TSA = 203.5 cm²
Combination Solid
Q 5.2Hard

A sphere of radius 8 cm is melted into small spheres of radius 1 cm. Find (i) number of balls and (ii) ratio of total new SA to original SA.

1
Volume Conservation
n×(4/3)π(1)³ = (4/3)π(8)³ → n = 512
2
Original SA = 4π(8)² = 256π cm²
3
Total new SA = 512 × 4π(1)² = 2048π cm²
4
Ratio = 2048π / 256π = 8 : 1
Surface area increases 8× — equal to R/r ratio!
512 balls · SA ratio 8 : 1
Conversion Result

Tips, Tricks & Common Mistakes

Exam-tested wisdom. Learn not just how to solve — but how to avoid losing marks.

🎯
TSA vs CSA: "Area of canvas" = CSA; "cost of painting a box" = TSA. Identifying which surface is being asked about is half the solution.
📐
Slant Height First: For any cone, if ℓ is not given, compute ℓ = √(r² + h²) before anything else. Never use h directly in CSA = πrℓ.
🔁
Conversion Shortcut: Sphere of radius R → n spheres of radius r: n = (R/r)³. Saves enormous computation in exams.
🧩
Combination SA: Total SA = SA₁ + SA₂ − 2 × (common face area). Subtract twice because both solids lose that face.
📏
Units Must Match: Convert ALL measurements to the same unit before substituting. 1 m = 100 cm → 1 m³ = 10⁶ cm³.
🌐
Hemisphere Flat Face: TSA = 2πr² + πr² = 3πr². Add the flat base for TSA; use 2πr² only for the curved CSA.
3 Cones = 1 Cylinder: Three cones of same r and h fill one cylinder. Use as a sanity check in conversion problems.
💧
Overflow = Object Volume: When an object is submerged in a full vessel, the volume of water that overflows equals the object's volume.
Mistake 01 — Cone
Using height h instead of slant height ℓ in CSA = πrh (wrong formula).
✅ Fix: Always find ℓ = √(r²+h²) first. CSA = πrℓ.
Mistake 02 — Combinations
Adding TSA₁ + TSA₂ for a combined solid, double-counting the joined face.
✅ Fix: SA = CSA₁ + CSA₂ + exposed base(s) only.
Mistake 03 — Diameter vs Radius
Plugging diameter D directly into r in formulas like πr² — giving 4× the correct area.
✅ Fix: Always halve diameter first: r = D/2.
Mistake 04 — Hemisphere TSA
Writing TSA = 2πr² (forgetting the flat base). This gives CSA, not TSA.
✅ Fix: TSA = 2πr² + πr² = 3πr².
Mistake 05 — Unit Mismatch
Mixing cm and m in the same computation (e.g. r=7 cm, h=1.4 m), giving answers off by 100×.
✅ Fix: Convert h to cm (1.4 m = 140 cm) before computing.
Mistake 06 — Conversion Count
Writing n = R/r for sphere recast count instead of n = (R/r)³.
✅ Fix: Equate full volumes: (4/3)πR³ = n·(4/3)πr³ → n = R³/r³.

MCQ Quiz — 12 Questions

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NCERT Class 9 Maths Chapter 11 Surface Areas & Volumes Notes
NCERT Class 9 Maths Chapter 11 Surface Areas & Volumes Notes — Complete Notes & Solutions · academia-aeternum.com
Chapter 11, "Surface Areas and Volumes," guides students into the world of three-dimensional geometry, where mathematical concepts move from flat surfaces to spatial forms. Through this chapter, learners explore how to calculate surface areas and volumes for common solids such as cubes, cuboids, cylinders, cones, spheres, and hemispheres. Real-life applications and thoughtful examples connect classroom learning to everyday situations, helping students understand how geometry is used in…
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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