Chapter 12 · CBSE Class X
📘 Definition
💡 Concept
📌 Note
🎨 SVG Diagram
🔢 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
📘 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)
📐 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
📘 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)
📄 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
📘 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
🔢 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
📘 Definition
🔎 Key Fact
Key Properties
🎨 SVG Diagram
🔢 Formula
📐 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
📘 Definition
🔎 Key Fact
Key Properties
🎨 SVG Diagram
🔢 Formula
📐 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
📘 Definition
💡 Concept
🎨 SVG Diagram
🔢 Formula
📐 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.
\[ \begin{aligned}
CSA &= 2\pi rh \\&= 2 \times \pi \times 3 \times 7 \\&= 42\pi \, \mathrm{cm^2}\\\\
V &= \pi r^2 h \\&= \pi \times 9 \times 7 \\&= 63\pi \, \mathrm{cm^3}
\end{aligned} \]
📌 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
📘 Definition
🔎 Key Fact
💡 Concept
🎨 SVG Diagram
🔢 Formula
📐 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
📘 Definition
🔎 Key Fact
🎨 SVG Diagram
🔢 Formula
📐 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.
\[ \begin{aligned}
\text{Surface Area} &= 4\pi r^2 \\&= 4 \times \pi \times 49 \\&= 196\pi \, \mathrm{cm^2}\\\\
\text{Volume} &= \frac{4}{3}\pi r^3 \\&= \frac{4}{3}\pi \times 343 \\&= \frac{1372}{3}\pi \, \mathrm{cm^3}
\end{aligned} \]
📌 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
📘 Definition
💡 Concept
🎨 SVG Diagram
🔢 Formula
📐 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.
\[ \begin{aligned}
TSA &= 3\pi r^2 \\&= 3 \times \pi \times 49 \\&= 147\pi \, \mathrm{cm^2}\\\\
V &= \frac{2}{3}\pi r^3 \\&= \frac{2}{3}\pi \times 343 \\&= \frac{686}{3}\pi \, \mathrm{cm^3}
\end{aligned} \]
⚡ 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
🔎 Key Fact
Units of Measurement
💡 Concept
Slant Height Concept
📌 Note
Conversion of Solids (Volume Conservation Principle)<
🔎 Key Fact
Composite Solids
🗺️ 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
❓ 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
❓ 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
❓ 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
❓ 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
❓ 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 \( \pi = \frac{22}{7} \))
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
❓ 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
❓ 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 \( = 25.12 \, cm^3 \)
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📐 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.
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
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Surface Areas And Volumes | Mathematics Class 10 | Academia Aeternum
Surface Areas And Volumes | Mathematics Class 10 | Academia Aeternum — 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.
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