Class 10 Maths Chapter 12 Notes: Surface Areas & Volumes Made Simple

Chapter 12 · CBSE Class X

📦

Solid Figure (3D Geometry Foundation)

Surface Area Volume Mensuration CBSE Class X

📘 Definition

💡 Concept

📌 Note

🎨 SVG Diagram

Cube, Cylinder and Cone (Basic 3D Solids)

🔢 Formula

\[ \begin{aligned}\text{Surface Area} &= \text{Sum of areas of all faces}\\\\ \text{Volume} &= \text{Space occupied by the solid} \end{aligned} \]

✏️ Example

Identify whether a football and a dice are solid figures.

Any object with 3 dimensions and occupying space is a solid.

Check for dimensions → Check space occupancy → Identify shape.

Football → Spherical shape → Occupies space → Solid figure
Dice → Cubical shape → Occupies space → Solid figure

📐 Derivation

Conceptual Development

Solid geometry extends plane geometry by introducing the third dimension. A square becomes a cube, a rectangle becomes a cuboid, and a circle generates cylinder, cone, and sphere through rotation.

Example: When a rectangle rotates about one of its sides, it generates a cylinder.

⚡ Exam Tip

⚠️ Warning

📋 Case Study

CBSE Case Study / HOTS

A water tank is designed using a cylinder topped with a hemisphere. Identify how many solid figures are involved and justify.

Answer Insight:

Two solids: Cylinder + Hemisphere
This is a composite solid, a key CBSE HOTS concept.

🌟 Importance

📦

Surface Area of Solids (CSA & TSA)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

The surface area of a solid is the total area covered by all its external surfaces. It represents the boundary that separates the solid from its surroundings.

Type of Surface Area

Curved Surface Area (CSA)

Area of only the curved portion of a solid, excluding bases or flat faces.
Total Surface Area (TSA)

Sum of curved surface area and areas of all plane faces (bases, top, etc.).

🔢 Formula

Cylinder \[\begin{aligned}\text{CSA} &= 2\pi rh, \\ \\\text{TSA} &= 2\pi r(h + r) \end{aligned}\]
Cone \[\begin{aligned}\text{CSA} &= \pi rl, \\\\ \text{TSA} &= \pi r(l + r) \end{aligned}\]
Sphere \[\begin{aligned}\text{Surface Area} = 4\pi r^2 \end{aligned}\]
Hemisphere \[\begin{aligned}\text{CSA} &= 2\pi r^2, \\\\ \text{TSA} &= 3\pi r^2 \end{aligned}\]
Cuboid \[\begin{aligned}\text{TSA} = 2(lb + bh + hl) \end{aligned}\]
Cube \[\begin{aligned}\text{TSA} = 6a^2 \end{aligned}\]

🎨 SVG Diagram

Conceptual Visualization (Cylinder CSA vs TSA)

Cylinder: Curved Surface (lateral) + Circular Bases

📐 Derivation

Derivation Insight (Cylinder CSA)

When a cylinder is cut along its height and unfolded, its curved surface forms a rectangle.

Length = circumference of base = \(2\pi r\)
Breadth = height = \(h\)

Therefore, \[ \text{CSA of cylinder} = 2\pi r \times h = 2\pi rh \]

✏️ Example

Find the total surface area of a cylinder of radius 3 cm and height 7 cm.

TSA = curved surface + top + base

Identify r, h → Apply TSA formula → Substitute values

\[ \begin{aligned} \text{TSA} &= 2\pi r(h + r)\\ &= 2 \times \pi \times 3 \times (7 + 3)\\ &= 60\pi \text{ cm}^2 \end{aligned} \]

TSA = \(60\pi \text{ cm}^2\)

⚡ Exam Tip

Exam Tips (High Scoring Strategy)

⚠️ Warning

Common Mistakes

📋 Case Study

CBSE Case Study / HOTS

A cylindrical tank is open from the top. Which surface area will be used to calculate the cost of painting?

Answer Insight:

Only curved surface + base → Not full TSA
Required Area: \[ 2\pi rh + \pi r^2 \]

🌟 Importance

Why This Topic is Crucial

📦

Volume of Solids (Capacity & Space Measurement)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

💡 Concept

Conceptual Understanding

🔢 Formula

Standard Volume Formulas (CBSE Core)

Cuboid: \[ V = l \times b \times h \]
Cube: \[ V = a^3 \]
Cylinder: \[ V = \pi r^2 h \]
Cone: \[ V = \frac{1}{3}\pi r^2 h \]
Sphere: \[ V = \frac{4}{3}\pi r^3 \]
Hemisphere: \[ V = \frac{2}{3}\pi r^3 \]

🎨 SVG Diagram

Geometric Visualization (Cylinder Volume)

Volume = Base Area × Height

🗒️ Derivation

Derivation Insight (Cylinder Volume)

A cylinder can be visualized as a stack of circular discs.

Area of base = \( \pi r^2 \)
Height = \( h \)

Therefore, \[ V = \text{Base Area} \times \text{Height} = \pi r^2 h \]

✏️ Example

Find the volume of a cylinder of radius 3 cm and height 7 cm.

Volume = base area × height

Identify r, h → Apply formula → Substitute values

\[ \begin{aligned} V &= \pi r^2 h \\&= \pi \times 3^2 \times 7 \\&= 63\pi \text{ cm}^3 \end{aligned} \]

🧠 Remember

Units & Conversions

⚡ Exam Tip

Exam Tips (CBSE Strategy)

⚠️ Warning

Common Mistakes

📋 Case Study

CBSE Case Study / HOTS

A cylindrical container is filled with water and poured into a conical container of same base radius and height. What fraction of the cone is filled?

Answer Insight:

Cylinder volume = \( \pi r^2 h \)
Cone volume = \( \frac{1}{3}\pi r^2 h \)
Therefore, cylinder fills 3 cones

🌟 Importance

Importance for Board Exams

📦

Right Circular Solids (Cylinder & Cone)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

A right circular solid is a three-dimensional object whose base is circular and whose axis (line joining centers) is perpendicular to the base.

Common examples include right circular cylinder and right circular cone.

Types of Right Circular Solids

Cylinder: Two parallel circular bases connected by a curved surface.
Cone: One circular base tapering smoothly to a vertex.

💡 Concept

Key Characteristics

🎨 SVG Diagram

Geometrical Visualization

Right Circular Cylinder and Cone with Perpendicular Axis

🔢 Formula

Cylinder Volume: \[ V = \pi r^2 h \]
Cone Volume: \[ V = \frac{1}{3}\pi r^2 h \]
Cone Slant Height: \[ l = \sqrt{r^2 + h^2} \]

📐 Derivation

Conceptual Derivation (Cone vs Cylinder)

A cone can be visualized as a pyramid with circular base. Experimentally, it is observed:

3 identical cones fill exactly 1 cylinder (same base & height)

Hence, \[ V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} \]

✏️ Example

A cone and cylinder have same base radius and height. Compare their volumes.

Use volume formulas

Write both formulas → Take ratio

\[ \begin{aligned} \frac{V_{\text{cone}}}{V_{\text{cylinder}}} &= \frac{\frac{1}{3}\pi r^2 h}{\pi r^2 h}\\ &= \frac{1}{3} \end{aligned} \]

⚡ Exam Tip

Exam Tips

⚠️ Warning

Common Mistakes

📋 Case Study

CBSE Case Study / HOTS

A cone is placed inside a cylinder of same radius and height. What percentage of cylinder volume remains empty?

Answer Insight:

Cone = 1/3 cylinder
Empty space = 2/3
Percentage empty = 66.67%

🌟 Importance

Importance for Board Exams

📦

Cuboid (Rectangular Prism)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

🔎 Key Fact

Key Properties

Faces = 6, Edges = 12, Vertices = 8
All angles are \(90^\circ\)
Opposite faces are equal and parallel
Space diagonal: \[ d = \sqrt{l^2 + b^2 + h^2} \]

🎨 SVG Diagram

Dimensions of a Cuboid: Length (l), Breadth (b), Height (h)

🔢 Formula

Lateral Surface Area (LSA): \[ \boxed{2h(l + b)} \]
Total Surface Area (TSA): \[ \boxed{2(lb + bh + hl)} \]
Volume: \[ \boxed{l \times b \times h} \]

📐 Derivation

Derivation Insight

Total Surface Area:

Top & Bottom → \(2(lb)\)
Front & Back → \(2(bh)\)
Left & Right → \(2(hl)\)

Adding all: \[ TSA = 2(lb + bh + hl) \]

Volume:

A cuboid can be seen as layers of rectangles: \[ V = \text{Base Area} \times \text{Height} = (l \times b) \times h \]

✏️ Example

Find the TSA and volume of a cuboid with \(l=5\) cm, \(b=3\) cm, \(h=2\) cm.

Substitute values in formulas

\[ \begin{aligned} TSA &= 2(lb + bh + hl)\\ &= 2(15 + 6 + 10) \\&= 62 \, \mathrm{cm^2} \end{aligned} \]

\[ \begin{aligned} V &= lbh \\&= 5 \times 3 \times 2 \\&= 30 \, \mathrm{cm^3} \end{aligned} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A cuboidal box is painted on all sides and then cut into smaller cubes. How many cubes will have exactly two faces painted?

Answer Insight:

Edge cubes excluding corners → \(4(n-2)\)
Concept of painted surfaces is frequently asked in CBSE.

🌟 Importance

📦

Cube (Special Case of Cuboid)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

🔎 Key Fact

Key Properties

Faces = 6 (all squares)
Edges = 12 (all equal)
Vertices = 8
Face diagonal: \[ d_f = \sqrt{2}a \]
Space diagonal: \[ d = \sqrt{3}a \]

🎨 SVG Diagram

All edges of cube are equal (a)

🔢 Formula

Lateral Surface Area (LSA): \[ \boxed{\bbox[5pt]{4a^2}} \]
Total Surface Area (TSA): \[ \boxed{\bbox[5pt]{6a^2}} \]
Volume: \[ \boxed{\bbox[5pt]{a^3}} \]

📐 Derivation

Derivation Insight

Total Surface Area:

A cube has 6 square faces, each of area \(a^2\)

\[ TSA = 6 \times a^2 = 6a^2 \]

Lateral Surface Area:

4 side faces (excluding top and bottom) \[ LSA = 4a^2 \]

Volume:

\[ V = a \times a \times a = a^3 \]

✏️ Example

Find the TSA and volume of a cube of side 4 cm.

\[ \begin{aligned} TSA &= 6a^2 \\&= 6 \times 16 \\&= 96 \, \mathrm{cm^2} \end{aligned} \]

\[ \begin{aligned} V &= a^3 \\&= 4^3 \\&= 64 \, \mathrm{cm^3} \end{aligned} \]

📌 Note

Cube vs Cuboid (Quick Insight)

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A cube is painted on all faces and cut into \(n^3\) smaller cubes. How many cubes will have exactly three faces painted?

Answer Insight:

Only corner cubes → Always 8
Independent of n (if \(n \ge 2\))

🌟 Importance

📦

Right Circular Cylinder

Trigonometry Heights and Distances CBSE Class X

📘 Definition

💡 Concept

🎨 SVG Diagram

Right Circular Cylinder with radius r and height h

🔢 Formula

Curved Surface Area (CSA): \[ \boxed{\bbox[5pt]{2\pi r h}} \]
Total Surface Area (TSA): \[ \boxed{\bbox[5pt]{2\pi r(r + h)}} \]
Volume: \[ \boxed{\bbox[5pt]{\pi r^2 h}} \]

📐 Derivation

Derivation Insight

Curved Surface Area:

When the cylinder is cut along its height and opened, it forms a rectangle:

Length = circumference = \(2\pi r\)
Breadth = height = \(h\)

\[ \begin{aligned} CSA &= 2\pi r \times h \\&= 2\pi rh \end{aligned} \]

Total Surface Area:

\[ \begin{aligned} TSA &= CSA + 2(\text{area of base}) \\&= 2\pi rh + 2\pi r^2 \\&= 2\pi r(r + h) \end{aligned} \]

Volume:

\[ \begin{aligned} V &= \text{Base Area} \times \text{Height} \\&= \pi r^2 h \end{aligned} \]

✏️ Example

Find the CSA and volume of a cylinder with radius 3 cm and height 7 cm.

\[ \begin{aligned} CSA &= 2\pi rh \\&= 2 \times \pi \times 3 \times 7 \\&= 42\pi \, \mathrm{cm^2}\\\\ V &= \pi r^2 h \\&= \pi \times 9 \times 7 \\&= 63\pi \, \mathrm{cm^3} \end{aligned} \]

\beginaligned CSA &= 2π rh \\&= 2 × π × 3 × 7 \\&= 42π \, \mathrmcm^2\\\\ V &= π r^2 h \\&= π × 9 × 7 \\&= 63π \, \mathrmcm^3 \endaligned

📌 Note

Real-Life Applications

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A cylindrical tank is filled with water and poured into a cuboidal tank. How will you compare their volumes?

Answer Insight:

Equate volumes: \[ \pi r^2 h = lbh \]
Used in conversion-based CBSE questions.

🌟 Importance

📦

Right Circular Cone

Trigonometry Heights and Distances CBSE Class X

📘 Definition

🔎 Key Fact

💡 Concept

🎨 SVG Diagram

Right Circular Cone showing r, h, and l

🔢 Formula

Curved Surface Area (CSA): \[ \boxed{\bbox[5pt]{\pi r l}} \]
Total Surface Area (TSA): \[ \boxed{\bbox[5pt]{\pi r(r + l)}} \]
Volume: \[ \boxed{\bbox[5pt]{\frac{1}{3}\pi r^2 h}} \]

📐 Derivation

Curved Surface Area:

When the cone is cut and opened, it forms a sector of a circle.

Radius of sector = \(l\)
Arc length = circumference of base = \(2\pi r\)

Hence, \[ CSA = \pi r l \]

Volume Relation:

A cone has one-third the volume of a cylinder with same base and height: \[ V = \frac{1}{3}\pi r^2 h \]

✏️ Example

Find the slant height and volume of a cone with \(r=3\) cm and \(h=4\) cm.

\[ \begin{aligned} l &= \sqrt{r^2 + h^2} \\&= \sqrt{9 + 16} \\&= 5 \, cm\\\\ V &= \frac{1}{3}\pi r^2 h \\&= \frac{1}{3}\pi \times 9 \times 4 \\&= 12\pi \, cm^3 \end{aligned} \]

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A cone is melted and recast into a sphere of same radius. Compare their volumes.

Answer Insight:

Use volume conservation
\[ \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi R^3 \]
Common in CBSE conversion problems

🌟 Importance

📦

Sphere (Perfectly Symmetric Solid)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

🔎 Key Fact

🎨 SVG Diagram

Sphere with centre O and radius r

🔢 Formula

Surface Area: \[ \boxed{\bbox[5pt]{4\pi r^2}} \]
Volume: \[ \boxed{\bbox[5pt]{\frac{4}{3}\pi r^3}} \]

📐 Derivation

Volume:

A sphere’s volume is derived using integration or by comparing with a cylinder and cone:

Volume of sphere = \( \frac{4}{3}\pi r^3 \)

Surface Area:

Surface area of a sphere is equal to four times the area of its great circle:

\[ \begin{aligned} \text{Surface Area} &= 4 \times (\pi r^2) \\&= 4\pi r^2 \end{aligned} \]

✏️ Example

Find the surface area and volume of a sphere of radius 7 cm.

\[ \begin{aligned} \text{Surface Area} &= 4\pi r^2 \\&= 4 \times \pi \times 49 \\&= 196\pi \, \mathrm{cm^2}\\\\ \text{Volume} &= \frac{4}{3}\pi r^3 \\&= \frac{4}{3}\pi \times 343 \\&= \frac{1372}{3}\pi \, \mathrm{cm^3} \end{aligned} \]

\beginaligned \textSurface Area &= 4π r^2 \\&= 4 × π × 49 \\&= 196π \, \mathrmcm^2\\\\ \textVolume &= \frac43π r^3 \\&= \frac43π × 343 \\&= \frac13723π \, \mathrmcm^3 \endaligned

📌 Note

Real-Life Applications

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A solid sphere is melted and recast into smaller identical spheres. How is the number of smaller spheres determined?

Answer Insight:

Volume remains conserved
\[ n = \frac{R^3}{r^3} \]
Very common CBSE concept

🌟 Importance

📦

Hemisphere (Half of a Sphere)

Trigonometry Heights and Distances CBSE Class X

📘 Definition

💡 Concept

🎨 SVG Diagram

Hemisphere with base and curved surface

🔢 Formula

Curved Surface Area (CSA): \[ \boxed{\bbox[5pt]{2\pi r^2}} \]
Total Surface Area (TSA): \[ \boxed{\bbox[5pt]{3\pi r^2}} \]
Volume: \[ \boxed{\bbox[5pt]{\frac{2}{3}\pi r^3}} \]

📐 Derivation

Curved Surface Area:

Sphere surface area = \(4\pi r^2\)
Hemisphere is half:

\[ CSA = \frac{1}{2} \times 4\pi r^2 = 2\pi r^2 \]

Total Surface Area:

Add base area: \[ TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2 \]

Volume:

Sphere volume = \( \frac{4}{3}\pi r^3 \)
Half of it: \[ V = \frac{2}{3}\pi r^3 \]

✏️ Example

Find the TSA and volume of a hemisphere of radius 7 cm.

\[ \begin{aligned} TSA &= 3\pi r^2 \\&= 3 \times \pi \times 49 \\&= 147\pi \, \mathrm{cm^2}\\\\ V &= \frac{2}{3}\pi r^3 \\&= \frac{2}{3}\pi \times 343 \\&= \frac{686}{3}\pi \, \mathrm{cm^3} \end{aligned} \]

\beginaligned TSA &= 3π r^2 \\&= 3 × π × 49 \\&= 147π \, \mathrmcm^2\\\\ V &= \frac23π r^3 \\&= \frac23π × 343 \\&= \frac6863π \, \mathrmcm^3 \endaligned

⚡ Exam Tip

⚠️ Warning

Common Mistakes

📋 Case Study

A solid hemisphere is melted and recast into a cylinder of same radius. Find the height of the cylinder.

Answer Insight:

Volume conserved
\[ \frac{2}{3}\pi r^3 = \pi r^2 h \Rightarrow h = \frac{2r}{3} \]

🌟 Importance

📦

Important Aspects & Key Observations

Trigonometry Heights and Distances CBSE Class X

🔎 Key Fact

Units of Measurement

Surface Area → square units: \(cm^2,\ m^2\)
Volume → cubic units: \(cm^3,\ m^3\)
Capacity: \[ \boxed{1\ cm^3 = 1\ mL} \]
\[ 1\ L = 1000\ cm^3 = 1\ dm^3 \]

Exam Insight: Always convert units before substitution. Most calculation errors occur due to unit mismatch.

💡 Concept

Slant Height Concept

In cones, slant height is essential for surface area calculations:

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

Used only in CSA and TSA of cone
Not used in volume calculation

Exam Trap: Students often substitute \(h\) instead of \(l\) in CSA.

📌 Note

Conversion of Solids (Volume Conservation Principle)<

In problems involving melting, reshaping, or recasting:

Volume before transformation = Volume after transformation

No loss or gain of material (ideal condition)
Only shape changes, volume remains constant

Example: \[ \text{Volume of sphere} = \text{Volume of cylinder} \]

CBSE Focus: Very frequently asked in case-study questions.

🔎 Key Fact

Composite Solids

Many real-life objects are formed by combining basic solids such as cylinders, cones, and hemispheres.

Surface Area: Add only exposed surfaces (exclude hidden/joined parts)
Volume: Add volumes of all individual solids

Exam Insight: Hidden surfaces are the most common source of mistakes.

🗺️ Roadmap

Problem Solving Strategy

Step 1: Identify the solid (cube, cone, cylinder, etc.)
Step 2: Write given dimensions clearly
Step 3: Convert units if required
Step 4: Select correct formula (CSA, TSA, or Volume)
Step 5: Substitute carefully and simplify

⚠️ Warning

Common Mistakes (High-Alert)

📋 Case Study

A metal sphere is melted and recast into smaller identical cones. How will you find the number of cones formed?

Answer Framework:

Apply volume conservation
\[ n = \frac{\text{Volume of sphere}}{\text{Volume of cone}} \]
Substitute formulas carefully

🌟 Importance

📦

Example=1

Trigonometry Heights and Distances CBSE Class X

❓ Question

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

💡 Concept

🧩 Solution

Given

Diameter = 3.5 cm → Radius \( r = 1.75 \) cm
Total height = 5 cm

Step by step Solution

CSA of Hemisphere
\[\begin{aligned} CSA_{hs} &= 2\pi r^2\\ &= 2 \times \frac{22}{7} \times (1.75)^2\\ &= 2 \times \frac{22}{7} \times 3.0625\\ &= 19.25 \, cm^2 \end{aligned} \]
Height of cone: \[ \begin{aligned} h &= 5 - 1.75 \\&= 3.25 \, \mathrm{cm} \end{aligned} \]
Slant height: \[ \begin{aligned} l &= \sqrt{r^2 + h^2}\\ &= \sqrt{(1.75)^2 + (3.25)^2}\\ &= \sqrt{3.0625 + 10.5625}\\ &= \sqrt{13.625}\\ &\approx 3.69 \, \mathrm{cm} \end{aligned} \]
CSA of Cone
\[ \begin{aligned} CSA_{c} &= \pi r l\\ &= \frac{22}{7} \times 1.75 \times 3.69\\ &= 22 \times 0.25 \times 3.69\\ &= 20.30 \, \mathrm{cm^2} \ (\text{approx}) \end{aligned} \]
Total Surface Area
\[ \begin{aligned} TSA &= CSA_{hs} + CSA_{c}\\ &= 19.25 + 20.30\\ &= 39.55 \, \mathrm{cm^2}\\ &\approx 39.6 \, \mathrm{cm^2} \end{aligned} \]

Area to be coloured \( \approx 39.6 \, \mathrm{cm^2} \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistake

📦

Example-2

Trigonometry Heights and Distances CBSE Class X

❓ Question

The decorative block shown in Fig. 12.7 is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take \( \pi = \frac{22}{7} \))

💡 Concept

🧩 Solution

Given

Edge of cube \(a = 5\) cm
Diameter of hemisphere = 4.2 cm → Radius \(r = 2.1\) cm

Step by step Solution

Surface Area of Cube
\[ \begin{aligned} TSA_{cube} &= 6a^2 \\&= 6 \times 5^2 \\&= 6 \times 25 \\&= 150 \, \mathrm{cm^2} \end{aligned} \]
Curved Surface Area of Hemisphere
\[ \begin{aligned} CSA_{h} &= 2\pi r^2\\ &= 2 \times \frac{22}{7} \times (2.1)^2\\ &= 2 \times \frac{22}{7} \times 4.41\\ &= 27.72 \, \mathrm{cm^2} \end{aligned} \]
Area of Circular Base (Hidden)
\[ \begin{aligned} \pi r^2 &= \frac{22}{7} \times (2.1)^2\\ &= 13.86 \, \mathrm{cm^2} \end{aligned} \]
Total Surface Area
\[ \begin{aligned} TSA &= 150 + 27.72 - 13.86\\ &= 163.86 \, \mathrm{cm^2} \end{aligned} \]

Total Surface Area \( = 163.86 \, \mathrm{cm^2} \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistakes

📦

example-3

Trigonometry Heights and Distances CBSE Class X

❓ Question

A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the rocket is 26 cm, and the height of the conical part is 6 cm. The diameter of the cone is 5 cm and that of the cylinder is 3 cm. If the conical part is painted orange and the cylindrical part yellow, find the area painted with each colour. (Take \( \pi = 3.14 \))

💡 Concept

🧩 Solution

Given

Total height = 26 cm
Height of cone = 6 cm → Cylinder height \(= 20\) cm
Radius of cone \(= 2.5\) cm
Radius of cylinder \(= 1.5\) cm

Step by step Solution

Orange Colour (Cone Part)
Slant height: \[ \begin{aligned} l &= \sqrt{r^2 + h^2}\\ &= \sqrt{(2.5)^2 + 6^2}\\ &= \sqrt{6.25 + 36}\\ &= \sqrt{42.25} = 6.5 \, \mathrm{cm} \end{aligned} \]
Curved Surface Area: \[ \begin{aligned} CSA &= \pi r l \\&= 3.14 \times 2.5 \times 6.5 \\&= 51.025 \, \mathrm{cm^2} \end{aligned} \]
Base of cone: \[ \begin{aligned} \pi r^2 &= 3.14 \times (2.5)^2 \\&= 19.625 \, \mathrm{cm^2} \end{aligned} \]
Covered area (top of cylinder): \[ \begin{aligned} \pi (1.5)^2 &= 3.14 \times 2.25 \\&= 7.065 \, \mathrm{cm^2} \end{aligned} \]
\[ \begin{aligned} \text{Orange Area} &= 51.025 + 19.625 - 7.065\\ &= 63.585 \, \mathrm{cm^2} \approx 63.6 \, \mathrm{cm^2} \end{aligned} \]
Yellow Colour (Cylinder Part)
Curved Surface Area: \[ \begin{aligned} CSA &= 2\pi r h \\&= 2 \times 3.14 \times 1.5 \times 20 \\&= 188.4 \, \mathrm{cm^2} \end{aligned} \]
Bottom base: \[ \pi r^2 = 7.065 \, \mathrm{cm^2} \]
\[ \begin{aligned} \text{Yellow Area} &= 188.4 + 7.065 \\&= 195.465 \, \mathrm{cm^2}\\ &\approx 195.5 \, \mathrm{cm^2} \end{aligned} \]

Orange Area \( \approx 63.6 \, cm^2 \)
Yellow Area \( \approx 195.5 \, cm^2 \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistakes

📦

Example-4

Trigonometry Heights and Distances CBSE Class X

❓ Question

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take \( \pi = \frac{22}{7} \))

💡 Concept

🧩 Solution

Given

Height of cylinder \(h = 1.45\, m = 145\, cm\)
Radius \(r = 30\, cm\)

Step by step Solution

Curved Surface Area of Cylinder
\[ \begin{aligned} \text{CSA }&= 2\pi rh \\&= 2 \times \frac{22}{7} \times 30 \times 145\\ &= \frac{44}{7} \times 4350 \\&= 27342.86 \, \mathrm{cm^2} \end{aligned} \]
Curved Surface Area of Hemisphere
\[ \begin{aligned} \text{CSA }&= 2\pi r^2 \\&= 2 \times \frac{22}{7} \times 30^2\\ \\&= \frac{44}{7} \times 900 \\&= 5657.14 \, \mathrm{cm^2} \end{aligned} \]
Area of Bottom Base
\[\begin{aligned}\pi r^2 &= \frac{22}{7} \times 900 \\&= 2828.57 \, \mathrm{cm^2}\end{aligned}\]
Total Surface Area
\[ \begin{aligned} \text{TSA }&= 27342.86 + 5657.14 + 2828.57\\ &= 35828.57 \, \mathrm{cm^2}\\ &\approx 3.58 \, \mathrm{m^2} \end{aligned} \]

Total Surface Area \( \approx 3.58 \, m^2 \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistakes

📦

Example-5

Trigonometry Heights and Distances CBSE Class X

❓ Question

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder. The base of the shed is \(7 \times 15\) m and the height of the cuboidal portion is 8 m. Find the volume of air in the shed.
Further, if machinery occupies 300 m\(^3\) and 20 workers occupy \(0.08\) m\(^3\) each, find the actual air available. (Take \( \pi = \frac{22}{7} \))

💡 Concept

🧩 Solution

Given

Length \(l = 15\) m, Breadth \(b = 7\) m
Height of cuboid \(= 8\) m
Radius of cylinder \(r = 3.5\) m
Length of cylinder \(= 15\) m

Step by Step Solution

Volume of Cuboid
\[\begin{aligned}V_1 &= l \times b \times h \\&= 15 \times 7 \times 8 \\&= 840 \, \mathrm{m^3}\end{aligned}\]
Volume of Half Cylinder
\[ \begin{aligned} V_2 &= \frac{1}{2} \pi r^2 h\\ &= \frac{1}{2} \times \frac{22}{7} \times (3.5)^2 \times 15 \\ &= \frac{1}{2} \times \frac{22}{7} \times 12.25 \times 15\\ &= \frac{1}{2} \times 22 \times 1.75 \times 15\\ &= \frac{1}{2} \times 577.5\\ &= 288.75 \, \mathrm{m^3} \end{aligned} \]
Total Volume of Shed
\[ \begin{aligned} V &= 840 + 288.75 \\&= 1128.75 \, \mathrm{m^3} \end{aligned} \]
Actual Air Available
Volume occupied by machinery: \[ = 300 \, m^3 \]
Volume occupied by workers: \[ \begin{aligned} &= 20 \times 0.08 \\&= 1.6 \, \mathrm{m^3} \end{aligned} \]
Total occupied volume: \[ \begin{aligned} &= 300 + 1.6 \\&= 301.6 \, \mathrm{m^3} \end{aligned} \]
Available air: \[ \begin{aligned} &= 1128.75 - 301.6 \\&= 827.15 \, \mathrm{m^3} \end{aligned} \]

Air available in shed \( = 827.15 \, m^3 \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistakes

📦

Example-6

Trigonometry Heights and Distances CBSE Class X

❓ Question

A juice seller was serving his customers using glasses as shown. The inner diameter of the cylindrical glass is 5 cm, but the bottom has a hemispherical raised portion. The height of the glass is 10 cm. Find the apparent capacity and the actual capacity of the glass. (Take \( \pi = \frac{22}{7} \))

💡 Concept

🧩 Solution

Given

Diameter = 5 cm → Radius \(r = 2.5\) cm
Height \(h = 10\) cm

Step by step Solution

Apparent Capacity
\[ \begin{aligned} V &= \pi r^2 h \\ &= \frac{22}{7} \times (2.5)^2 \times 10\\ &= \frac{22}{7} \times 6.25 \times 10\\ &= \frac{22}{7} \times 62.5\\ &= 196.43 \, \mathrm{cm^3} \end{aligned} \]
Apparent Capacity \( \approx 196.43 \, \mathrm{cm^3} \)
Volume of Hemisphere
\[ \begin{aligned} V_h &= \frac{2}{3}\pi r^3\\ &= \frac{2}{3} \times \frac{22}{7} \times (2.5)^3\\ &= \frac{2}{3} \times \frac{22}{7} \times 15.625\\ &= \frac{2}{3} \times 49.107\\ &= 32.74 \, \mathrm{cm^3} \ (\text{approx}) \begin{aligned} \]
Actual Capacity
\[ &= 196.43 - 32.74 \\&= 163.69 \, \mathrm{cm^3} \]

Actual Capacity \( \approx 163.7 \, cm^3 \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistakes

📦

Example-7

Trigonometry Heights and Distances CBSE Class X

❓ Question

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Find the volume of the toy. Also, if a right circular cylinder circumscribes the toy, find the difference of their volumes. (Take \( \pi = 3.14 \))

💡 Concept

🧩 Solution

Given

Radius \(r = 2\) cm
Height of cone \(h = 2\) cm

Step by step Solution

Volume of Cone
\[ \begin{aligned} V_c &= \frac{1}{3}\pi r^2 h\\ &= \frac{1}{3} \times 3.14 \times 2^2 \times 2\\ &= \frac{1}{3} \times 3.14 \times 8\\ &= 8.373 \, \mathrm{cm^3} \end{aligned} \]
Volume of Hemisphere
\[ \begin{aligned} V_h &= \frac{2}{3}\pi r^3\\ &= \frac{2}{3} \times 3.14 \times 2^3\\ &= \frac{2}{3} \times 3.14 \times 8\\ &= 16.746 \; mathrm{cm^3} \end{aligned} \]
Total Volume of Toy
\[ \begin{aligned} V &= 8.373 + 16.746 \\&= 25.119 \, \mathrm{cm^3} \\&\approx 25.12 \, \mathrm{cm^3} \end{aligned} \]
Volume of Circumscribing Cylinder
Height of cylinder = height of cone + radius of hemisphere \[ = 2 + 2 = 4 \, \mathrm{cm} \]
\[ \begin{aligned} V_{cyl} &= \pi r^2 h \\&= 3.14 \times 2^2 \times 4 \\&= 3.14 \times 4 \times 4 \\&= 50.24 \, \mathrm{cm^3} \end{aligned} \]
Difference in Volumes
\[ \begin{aligned} &= 50.24 - 25.12 \\&= 25.12 \, \mathrm{cm^3} \end{aligned} \]

Difference of volumes \( = 25.12 \, cm^3 \)

📌 Note

Exam Insight

⚠️ Warning

Common Mistakes

📐 NCERT Class X · Chapter 12

Surface Areas & Volumes

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Complete Formula Reference

All surface area & volume formulas — Chapter 12

Use π = 22/7 or 3.14159 as instructed. l = slant height, r = radius, h = height, a = side length

📦 Cuboid

LSA2h(l + b)

TSA2(lb + bh + hl)

Voll × b × h

Diag√(l² + b² + h²)

🎲 Cube

LSA4a²

TSA6a²

Vola³

Diaga√3

🥫 Right Circular Cylinder

CSA2πrh

TSA2πr(r + h)

Volπr²h

CSA = Curved Surface Area (excludes two circular ends)

🍦 Right Circular Cone

CSAπrl

TSAπr(r + l)

Vol(1/3)πr²h

l√(r² + h²)

l is slant height; always find l first if not given

🌐 Sphere

SA4πr²

Vol(4/3)πr³

A sphere has only one surface (no flat face). SA = Total SA.

🍵 Hemisphere

CSA2πr²

TSA3πr²

Vol(2/3)πr³

TSA = CSA + circular base (πr²). CSA = half of sphere's SA.

🪣 Frustum of a Cone ✦

CSAπ(r₁ + r₂)l

TSAπ[r₁² + r₂² + (r₁+r₂)l]

Vol(πh/3)(r₁² + r₂² + r₁r₂)

l√[h² + (r₁ − r₂)²]

r₁ = larger radius, r₂ = smaller radius. Frustum = cone with top cut off. ✦ Key topic in NCERT Ch.12

🔗 Combined Solids

SASum of exposed surfaces

VolSum of individual volumes

SA of combined solid ≠ sum of individual SAs.
Subtract the joined/hidden faces.
Vol of combined solid = sum of individual volumes.

🔢

Key Relationships & Conversions

Used when converting one solid to another

🔁 Conversion Principle

RuleVolume stays same

When a solid is melted/recast into another shape, the volume is conserved. Set V₁ = V₂ to find unknowns.

📏 Unit Conversions

Vol1 m³ = 10⁶ cm³

Vol1 L = 1000 cm³

Area1 m² = 10⁴ cm²

🧮 π Values

Exact22/7 ≈ 3.1428…

Dec3.14159…

Use 22/7 when radius involves multiples of 7. Use 3.14 otherwise.

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📐 Surface Areas & Volumes — Chapter 12

🥫 Cylinder

🍦 Cone

🌐 Sphere

📦 Cuboid

🪣 Frustum

🍵 Hemisphere

🔗 Combined

🔁 Conversion

🎲 Cube

Click any shape above to see its key formulas and concept summary.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025

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Class 10 Maths Chapter 12 Notes: Surface Areas & Volumes Made Simple

Class 10 Maths Chapter 12 Notes: Surface Areas & Volumes Made Simple — Complete Notes & Solutions · academia-aeternum.com

The study of Surface Areas and Volumes forms a crucial bridge between abstract geometry and its tangible real-world applications. This chapter equips learners with the mathematical tools required to measure, compare, and optimise the space occupied by three-dimensional objects. By extending earlier concepts of mensuration, students move from flat, two-dimensional figures to solid shapes encountered in everyday life such as containers, pipes, spheres, cones, and composite solids. This chapter…

🎓 Class 10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE

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Surface Areas & Volumes

Chapter Snapshot

Why This Chapter Matters for Exams

Key Concept Highlights

Important Formula Capsules

What You Will Learn

Navigate to Chapter Resources

🏆 Exam Strategy & Preparation Tips

Types of Solid Figures (CBSE Scope)

Conceptual Development

Common Mistakes

CBSE Case Study / HOTS

Why This Topic Matters (Board Perspective)

Type of Surface Area

Conceptual Visualization (Cylinder CSA vs TSA)

Derivation Insight (Cylinder CSA)

Exam Tips (High Scoring Strategy)

Common Mistakes

CBSE Case Study / HOTS

Why This Topic is Crucial

Conceptual Understanding

Standard Volume Formulas (CBSE Core)

Geometric Visualization (Cylinder Volume)

Derivation Insight (Cylinder Volume)

Units & Conversions

Exam Tips (CBSE Strategy)

Common Mistakes

CBSE Case Study / HOTS

Importance for Board Exams

Types of Right Circular Solids

Key Characteristics

Geometrical Visualization

Conceptual Derivation (Cone vs Cylinder)

Exam Tips

Common Mistakes

CBSE Case Study / HOTS

Importance for Board Exams

Key Properties

Derivation Insight

Common Mistakes

Key Properties

Derivation Insight

Cube vs Cuboid (Quick Insight)

Common Mistakes

Derivation Insight

Real-Life Applications

Common Mistakes

Common Mistakes

Real-Life Applications

Common Mistakes

Common Mistakes

Units of Measurement

Slant Height Concept

Conversion of Solids (Volume Conservation Principle)<

Composite Solids

Problem Solving Strategy

Common Mistakes (High-Alert)

Given

CSA of Hemisphere

CSA of Cone

Total Surface Area

Exam Insight

Common Mistake

Given

Surface Area of Cube

Curved Surface Area of Hemisphere

Area of Circular Base (Hidden)

Total Surface Area

Exam Insight

Common Mistakes

Given

Orange Colour (Cone Part)

Yellow Colour (Cylinder Part)

Exam Insight

Common Mistakes

Given

Curved Surface Area of Cylinder

Curved Surface Area of Hemisphere

Area of Bottom Base

Total Surface Area