SURFACE AREAS AND VOLUMES — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 11  ·  Q–
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Class 9 Mathematics Exercise-11.3 NCERT Solutions Olympiad Board Exam
Chapter 11

SURFACE AREAS AND VOLUMES

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

9 Questions
20–30 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
📘 Concept & Theory Concept Used ›

A right circular cone is a three-dimensional solid having a circular base and a pointed vertex. The volume of a cone tells us how much space is enclosed inside it.

The formula for the volume of a cone is:

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

where:

  • \(\small V\) = Volume of the cone
  • \(\small r\) = Radius of the circular base
  • \(\small h\) = Height of the cone
  • \(\small \pi = \frac{22}{7}\) or \(\small 3.14\)

The factor \(\small \frac{1}{3}\) shows that the volume of a cone is one-third the volume of a cylinder having the same base radius and height.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the formula for the volume of a cone.

  2. Substitute the given values of radius and height.

  3. Simplify step-by-step carefully.

  4. Write the final answer with correct cubic units.

📊 Graph / Figure Graph / Figure ›
h r Right Circular Cone
Cone showing radius and height
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Formula for the volume of a right circular cone:\[\small V = \frac{1}{3}\pi r^2 h\]

    where \(\small r\) is the radius and \(\small h\) is the height of the cone.

  2. (i) Radius \(\small = 6\) cm, Height \(\small = 7\) cm
  3. Substituting the values into the formula: \[\small \require{cancel} \begin{aligned} V &= \frac{1}{3}\times \pi \times r^2 \times h \\ &= \frac{1}{3}\times \frac{22}{7}\times (6)^2 \times 7 \\ &= \frac{1}{3}\times \frac{22}{\cancel{7}}\times 36 \times \cancel{7}\\ &= \frac{1}{3}\times 22 \times 36\\ &= \frac{1}{\cancel{3}}\times \cancelto{264}{792}\\ &= 264 \end{aligned} \]
  4. Therefore,\[\small \boxed{V = 264\ \text{cm}^3}\]
  5. (ii) Radius \(\small = 3.5\) cm, Height \(\small = 12\) cm
  6. Substituting the values into the formula: \[\small \begin{aligned} V &= \frac{1}{3}\times \pi \times r^2 \times h \\ &= \frac{1}{3}\times \frac{22}{7}\times (3.5)^2 \times 12\\ &= \frac{1}{\cancel{3}}\times \frac{22}{7}\times 12.25 \times \cancelto{4}{12}\\ &= \frac{1}{\cancel{3}}\times \frac{22}{\cancel{7}}\times \frac{\cancelto{7}{49}}{\cancel{4}}\times \cancelto{\cancel{4}}{12}\quad (12.25 =\frac{49}{4})\\ &=154 \end{aligned} \]
  7. Therefore,\[\small \boxed{V = 154\ \text{cm}^3}\]
💡 Answer Final Answer ›
Answer: \(\small (i)\;V = 264\ \text{cm}^3\) \(\small (ii)\; V = 154\ \text{cm}^3\)
🎯 Exam Significance Exam Significance ›
  • Questions based on the volume of cones are frequently asked in CBSE and state board examinations.
  • Competitive examinations such as NTSE, Olympiads, SSC, Polytechnic entrance exams, and foundation-level JEE preparation often include direct formula-based problems from mensuration.
  • Students must learn correct substitution of values and proper unit writing because step marking is important in board examinations.
  • Understanding the relation between cone and cylinder volumes helps in solving higher-level aptitude and geometry problems.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone is calculated using: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. Always square the radius before multiplication.

  3. Volume is always written in cubic units such as \(\small \text{cm}^3\).

  4. Simplification should be done carefully step-by-step to avoid calculation mistakes.

  5. A cone has one-third the volume of a cylinder with the same radius and height.
↑ Top
1 / 9  ·  11%
Q2 →
Q2
NUMERIC3 marks
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
📘 Concept & Theory Concept Used ›

The capacity of a conical vessel is equal to the volume of the cone. Since liquids occupy three-dimensional space, the capacity is measured in cubic units or litres.

The volume of a cone is given by:

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

where:

  • \(\small V\) = Volume or capacity of the cone
  • \(\small r\) = Radius of the base
  • \(\small h\) = Vertical height
  • \(\small \pi = \frac{22}{7}\) or \(\small 3.14\)

In a right circular cone, the radius, height, and slant height form a right triangle. Therefore, the Pythagoras theorem is used:

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

where \(\small l\) is the slant height.

Also,

\[\small 1000\ \text{cm}^3 = 1\ \text{litre} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Use the Pythagoras theorem to find the missing dimension of the cone.

  2. Apply the cone volume formula.

  3. Simplify step-by-step carefully.

  4. Convert cubic centimetres into litres.

  5. Write the final answer with proper units.

📊 Graph / Figure Graph / Figure ›
h r l Conical Vessel Dimensions
Cone showing radius, height and slant height
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  12 steps
  1. (i) Radius \(\small = 7\) cm, Slant Height \(\small = 25\) cm
  2. First, find the height of the cone using: \[\small l^2 = r^2 + h^2\]
  3. Substituting the values: \[\small \begin{aligned} h^2 &= l^2 - r^2 \\ &= 25^2 - 7^2 \\ &= 625 - 49 \\ &= 576 \Rightarrow h&=\sqrt{576}\\ &h=24\ \text{cm} \end{aligned} \]
  4. Now, volume of the cone: \[\small \require{cancel} \begin{aligned} V &= \frac{1}{3}\pi r^2 h \\ &= \frac{1}{3}\times \frac{22}{7}\times (7)^2 \times 24 \\ &= \frac{1}{\cancel{3}}\times \frac{22}{\cancel{7}}\times \cancelto{7}{49} \times \cancelto{8}{24}\\ &=22\times 7\times8\\ &= 1232\ \text{cm}^3 \end{aligned} \]
  5. Convert into litres: \[\small \begin{aligned} 1000\ \text{cm}^3 &= 1\ \text{litre} \\ 1232\ \text{cm}^3 &= \frac{1232}{1000} \\ &= 1.232\ \text{litres} \end{aligned} \]
  6. Therefore, \[\small \boxed{1.232\ \text{litres}}\]
  7. (ii) Height \(\small = 12\) cm, Slant Height \(\small = 13\) cm
  8. Let the radius of the cone be \(\small r\).
  9. Using Pythagoras Theorem: \[\small l^2 = r^2 + h^2\]
  10. Substituting the values: \[\small \begin{aligned} r^2 &= l^2 - h^2 \\ &= 13^2 - 12^2 \\ &= 169 - 144 \\ &= 25 \end{aligned} \]
  11. Therefore, \[\small \begin{aligned} r &= \sqrt{25} \\ &= 5\ \text{cm} \end{aligned} \]
  12. Now calculate the volume: \[\small \require{cancel} \begin{aligned} V &= \frac{1}{3}\pi r^2 h \\ &= \frac{1}{3}\times \frac{22}{7}\times (5)^2 \times 12 \\ &= \frac{1}{\cancel{3}}\times \frac{22}{7}\times 25 \times \cancelto{4}{12}\\ &=\frac{22}{7}\times 100\\ &\approx 314 \end{aligned} \]
  13. Converting into litres: \[\small \begin{aligned} V&\approx 314\times \frac{1}{1000}\\ &\approx 0.314\ \text{litres} \end{aligned} \]
  14. Therefore,\[\small \boxed{V=0.314\ \text{litres}}\]
🎯 Exam Significance Exam Significance ›
  • Questions involving cone volume and capacity are frequently asked in Class 9 board examinations.
  • Competitive examinations such as NTSE, Olympiads, SSC, and Polytechnic entrance tests often include mensuration-based numerical problems.
  • This question strengthens understanding of the relation between slant height, radius, and height using the Pythagoras theorem.
  • Unit conversion between cubic centimetres and litres is very important for practical application-based questions.
  • Stepwise simplification improves accuracy and helps students secure full marks in subjective examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. Relation among slant height, radius, and height: \[\small l^2 = r^2 + h^2 \]

  3. Capacity of a vessel is numerically equal to its volume.

  4. Always convert cubic centimetres into litres using: \[\small 1000\ \text{cm}^3 = 1\ \text{litre} \]

  5. Proper step-by-step simplification reduces calculation mistakes.
← Q1
2 / 9  ·  22%
Q3 →
Q3
NUMERIC3 marks
The height of a cone is 15 cm. If its volume is \(\small 1570\ \text{cm}^3\), find the radius of the base. (Use \(\small \pi = 3.14\))
📘 Concept & Theory Theory and Concept ›

A cone is a three-dimensional solid having a circular base and a pointed vertex. The volume of a cone represents the space enclosed inside it.

The formula for the volume of a cone is:

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

where:

  • \(\small V\) = Volume of the cone
  • \(\small r\) = Radius of the base
  • \(\small h\) = Height of the cone
  • \(\small \pi = 3.14\)

In this question, the volume and height are given, so we use the formula to find the unknown radius.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the formula for the volume of a cone.

  2. Substitute the given values of volume and height.

  3. Simplify the equation carefully to find \(\small r^2\).

  4. Take square root to obtain the radius.

  5. Write the answer with correct units.

📊 Graph / Figure Graph / Figure ›
h = 15 cm r = ? Volume of a Cone
Cone showing height and radius
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. Given
  2. \[\small h = 15\ \text{cm}\] and \[\small V = 1570\ \text{cm}^3\]
  3. Step-by-step Solution
  4. Let the radius of the cone be \(\small r\) cm.
  5. Formula for the volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h\]
  6. Substituting the given values: \[\small \begin{aligned} 1570 &= \frac{1}{3}\times 3.14 \times r^2 \times 15\\ &= \frac{1}{3}\times 47.1 \times r^2\\ &= 15.7r^2\\ \Rightarrow r^2 &= \frac{1570}{15.7}\\ &= 100\\ \Rightarrow r&=\sqrt{100}\\ r&=10\text{ cm} \end{aligned} \]
🎯 Exam Significance Exam Significance ›
  • Questions based on finding unknown dimensions from volume formulas are commonly asked in board examinations.
  • This problem strengthens algebraic manipulation skills along with mensuration concepts.
  • Competitive exams such as NTSE, Olympiads, SSC, and foundation-level engineering entrance preparation frequently include formula-based geometry questions.
  • Students learn how to rearrange formulas correctly to find unknown quantities.
  • Proper stepwise presentation is important for securing full marks in descriptive mathematics papers.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone is: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. When volume and height are known, radius can be found by rearranging the formula.

  3. Always simplify calculations step-by-step to avoid mistakes.

  4. Radius is obtained after taking the square root of \(\small r^2\).

  5. Volume is expressed in cubic units such as \(\small \text{cm}^3\).
← Q2
3 / 9  ·  33%
Q4 →
Q4
NUMERIC3 marks
If the volume of a right circular cone of height 9 cm is \(\small 48\pi\ \text{cm}^3\), find the diameter of its base.
📘 Concept & Theory Theory and Concept Used ›

A right circular cone is a three-dimensional solid having a circular base and a pointed vertex. The volume of a cone gives the amount of space occupied by it.

The formula for the volume of a cone is:

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

where:

  • \(\small V\) = Volume of the cone
  • \(\small r\) = Radius of the base
  • \(\small h\) = Height of the cone

The diameter of the base is related to the radius by:

\[\small \text{Diameter} = 2r \]

In this question, the volume and height are given, so first the radius is calculated using the volume formula. Then the diameter is obtained.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the formula for the volume of a cone.

  2. Substitute the given values of volume and height.

  3. Simplify the equation to find \(\small r^2\).

  4. Find the radius by taking square root.

  5. Use the relation \(\small D = 2r\) to calculate the diameter.

📊 Graph / Figure Graph / Figure ›
h = 9 cm r Diameter = 2r Right Circular Cone
Cone showing height, radius and diameter
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Given
  2. \[\small V = 48\pi\ \text{cm}^3\] and \[\small h = 9\ \text{cm}\]
  3. Step-by-step Solution
  4. Let the radius of the cone be \(\small r\) cm.
  5. Formula for the volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h\]
  6. Substituting the given values: \[\small \require{cancel} \begin{aligned} 48\pi &= \frac{1}{\cancel{3}}\times \pi \times r^2 \times \cancelto{3}{9}\\ &=3\pi r^2\\ \Rightarrow r^2&=\frac{48\pi}{3\pi}\\ r^2&=16\\ \Rightarrow r&=\sqrt{16}\\ r=4 \end{aligned} \]
  7. Diameter of the base: \[\small \begin{aligned} D &= 2r \\ &= 2 \times 4 \\ &= 8\ \text{cm} \end{aligned} \]
  8. Therefore, the diameter of the base of the cone is:\[\small \boxed{8\ \text{cm}}\]
🎯 Exam Significance Exam Significance ›
  • Questions involving unknown dimensions of cones are very common in school examinations.
  • This problem develops the ability to rearrange mensuration formulas correctly.
  • Competitive exams such as NTSE, Olympiads, SSC, and Polytechnic entrance tests frequently include similar formula-based numerical questions.
  • Students learn the importance of cancelling common terms carefully during simplification.
  • Such problems improve conceptual understanding of the relation between radius and diameter.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. Diameter is twice the radius: \[\small D = 2r \]

  3. When \(\small \pi\) appears on both sides of an equation, it can be cancelled during simplification.

  4. Radius is obtained after taking the square root of \(\small r^2\).

  5. Stepwise calculations help avoid algebraic mistakes.
← Q3
4 / 9  ·  44%
Q5 →
Q5
NUMERIC3 marks
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
📘 Concept & Theory Theory and Concept Used ›

A conical pit is shaped like a right circular cone. Its capacity means the amount of material or liquid it can hold, which is equal to its volume.

The volume of a cone is calculated using:

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

where:

  • \(\small V\) = Volume of the cone
  • \(\small r\) = Radius of the circular base
  • \(\small h\) = Height or depth of the cone

Since the diameter is given, the radius is obtained by dividing the diameter by \(\small 2\):

\[\small r = \frac{d}{2} \]

Also,

\[\small 1\ \text{m}^3 = 1\ \text{kilolitre} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the radius using the given diameter.

  2. Apply the formula for the volume of a cone.

  3. Simplify the calculation carefully step-by-step.

  4. Convert cubic metres into kilolitres.

  5. Write the final answer with proper units.

📊 Graph / Figure Graph / Figure ›
h = 12 m r = 1.75 m Conical Pit Capacity
Conical pit showing radius and depth
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  8 steps
  1. Given
  2. Diameter of the conical pit:\[\small d = 3.5\ \text{m}\]
  3. Depth of the pit:\[\small h = 12\ \text{m}\]
  4. Radius of the pit:\[\small \begin{aligned} r &= \frac{d}{2} \\ &= \frac{3.5}{2} \\ &= 1.75\ \text{m} \end{aligned} \]
  5. Step-by-step Solution
  6. Formula for the volume of a cone:\[\small V = \frac{1}{3}\pi r^2 h\]
  7. Substituting the given values: \[\small \begin{aligned} V &= \frac{1}{3}\times \frac{22}{7}\times (1.75)^2 \times 12\\ &= \frac{1}{3}\times \frac{22}{7}\times 3.0625 \times 12\\ &= 4\times \frac{22}{7}\times 3.0625 \\ &= \frac{88}{7}\times 3.0625 \\ &= 12.57142857 \times 3.0625 \\ &= 38.5\ \text{m}^3 \end{aligned} \]
  8. Since:\[\small 1\ \text{m}^3 = 1\ \text{kilolitre}\]
  9. Therefore, \[\small 38.5\ \text{m}^3 = 38.5\ \text{kilolitres}\]
  10. Hence, the capacity of the conical pit is: \[\small \boxed{38.5\ \text{kilolitres}}\]
🎯 Exam Significance Exam Significance ›
  • Questions based on practical applications of cone volume are frequently asked in CBSE and state board examinations.
  • This problem helps students understand the real-life use of mensuration formulas in storage and excavation problems.
  • Competitive examinations such as NTSE, Olympiads, SSC, and Polytechnic entrance tests often include capacity and volume-based questions.
  • Students learn the importance of unit conversion between cubic metres and kilolitres.
  • Careful simplification and proper substitution are essential for accurate answers in exams.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. Radius is half of the diameter: \[\small r = \frac{d}{2} \]

  3. Capacity of a vessel or pit is numerically equal to its volume.

  4. Conversion relation: \[\small 1\ \text{m}^3 = 1\ \text{kilolitre} \]

  5. Step-by-step calculations reduce chances of numerical mistakes.
← Q4
5 / 9  ·  56%
Q6 →
Q6
NUMERIC3 marks
The volume of a right circular cone is \(\small 9856\ \text{cm}^3\). If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
📘 Concept & Theory Theory and Concept Used ›

A right circular cone is a three-dimensional solid having a circular base and a pointed vertex. Different measurements related to a cone include volume, slant height, and curved surface area.

Formula for the volume of a cone:

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

Formula for slant height:

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

Formula for curved surface area:

\[\small \text{CSA} = \pi r l \]

where:

  • \(\small V\) = Volume of the cone
  • \(\small r\) = Radius of the base
  • \(\small h\) = Height of the cone
  • \(\small l\) = Slant height of the cone
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the radius using the given diameter.

  2. Use the volume formula to calculate the height.

  3. Apply the Pythagoras theorem to find the slant height.

  4. Use the curved surface area formula.

  5. Write all answers with proper units.

📊 Graph / Figure Graph / Figure ›
h r = 14 cm l Cone: Height, Slant Height & CSA
Cone showing radius, height and slant height
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  10 steps
  1. Given

  2. \[\small V = 9856\ \text{cm}^3 \]

    Diameter of the base:

    \[\small d = 28\ \text{cm} \]

    Radius of the base:

    \[\small \begin{aligned} r &= \frac{d}{2} \\ &= \frac{28}{2} \\ &= 14\ \text{cm} \end{aligned} \]

  3. (i) Finding the Height of the Cone
  4. Formula for volume:\[\small V = \frac{1}{3}\pi r^2 h\]
  5. Substituting the known values: \[\small \begin{aligned} 9856 &= \frac{1}{3}\times \frac{22}{7}\times (14)^2 \times h\\ &= \frac{1}{3}\times \frac{22}{7}\times 196 \times h\\ &= \frac{1}{3}\times 22 \times 28 \times h\\ &= \frac{616}{3}h\\ \Rightarrow 29568 &= 616h\\ \Rightarrow h&=\frac{29568}{616h}\\ &= 48\ \text{cm} \end{aligned} \]
  6. Therefore,\[\small \boxed{h = 48\ \text{cm}}\]
  7. (ii) Finding the Slant Height
  8. Using the relation:\[\small l^2 = r^2 + h^2\]
  9. Substituting the values: \[\small \begin{aligned} l^2 &= (14)^2 + (48)^2 \\ &= 196 + 2304 \\ &= 2500\\ \Rightarrow l&=\sqrt{2500}\\ l&=50 \end{aligned} \]
  10. Therefore,\[\small \boxed{l = 50\ \text{cm}}\]
  11. (iii) Finding the Curved Surface Area
  12. Formula for curved surface area:\[\small \text{CSA} = \pi r l\]
  13. Substituting the values: \[\small \begin{aligned} \text{CSA} &= \frac{22}{7}\times 14 \times 50\\ &= 22 \times 2 \times 50\\ &= 44 \times 50\\ &= 2200\ \text{cm}^2 \end{aligned} \]
  14. Therefore,\[\small \boxed{\text{CSA} = 2200\ \text{cm}^2}\]
🎯 Exam Significance Exam Significance ›
  • This is a complete mensuration problem involving multiple concepts of cones in a single question.
  • Such questions are commonly asked in CBSE board examinations and state-level school examinations.
  • Competitive examinations like NTSE, Olympiads, SSC, and Polytechnic entrance tests frequently include multi-step geometry and mensuration problems.
  • Students learn how different cone formulas are interconnected.
  • Stepwise derivation and careful simplification help in scoring full marks in descriptive examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. Slant height relation: \[\small l^2 = r^2 + h^2 \]

  3. Curved surface area: \[\small \text{CSA} = \pi r l \]

  4. Radius is half of the diameter.

  5. Proper substitution and systematic calculations reduce mistakes in exams.
← Q5
6 / 9  ·  67%
Q7 →
Q7
NUMERIC3 marks
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
📘 Concept & Theory Theory and Concept Used ›

When a right triangle is revolved about one of its perpendicular sides, a right circular cone is formed.

In this question:

  • The side about which the triangle rotates becomes the height of the cone.
  • The other perpendicular side becomes the radius of the cone.
  • The hypotenuse becomes the slant height of the cone.

Formula for the volume of a cone:

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

where:

  • \(\small V\) = Volume of the cone
  • \(\small r\) = Radius of the base
  • \(\small h\) = Height of the cone
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the dimensions of the cone formed after revolution.

  2. Determine the radius and height from the triangle.

  3. Apply the formula for the volume of a cone.

  4. Simplify step-by-step carefully.

  5. Write the final answer with proper units.

📊 Graph / Figure Graph / Figure ›
12 cm 5 cm 13 cm Volume = 1/3 π r² h = 1/3 × π × 5² × 12 = 100π ≈ 314.16 cm³ Revolution of Right Triangle
Right triangle revolved about side 12 cm
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  11 steps
  1. The given right triangle has sides:
  2. \[\small 5\ \text{cm}, \quad 12\ \text{cm}, \quad 13\ \text{cm}\]
  3. Since the triangle is revolved about the side measuring \(\small 12\ \text{cm}\):
  4. Height of the cone: \[\small h = 12\ \text{cm} \]
  5. Radius of the cone: \[\small r = 5\ \text{cm} \]
  6. Slant height: \[\small l = 13\ \text{cm} \]
  7. Formula for the volume of a cone:
  8. \[\small V = \frac{1}{3}\pi r^2 h \]
  9. Substituting the values: \[\small \begin{aligned} V &= \frac{1}{3}\times \pi \times (5)^2 \times 12 \\ &= \frac{1}{3}\times \pi \times 25 \times 12 \\ &= \pi \times 25 \times 4\\ &= 100\pi\ \text{cm}^3\\ &= 100 \times 3.1416\\ &= 314.16\ \text{cm}^3 \end{aligned} \]
  10. Hence, the volume of the solid obtained is:\[\small \boxed{100\pi\ \text{cm}^3}\]
  11. or approximately,\[\small \boxed{314.16\ \text{cm}^3}\]
🎯 Exam Significance Exam Significance ›
  • This question is important because it combines geometry and mensuration concepts.
  • Such application-based problems are frequently asked in CBSE and state board examinations.
  • Competitive examinations such as NTSE, Olympiads, SSC, and Polytechnic entrance tests often include solids formed by revolution.
  • Students learn how two-dimensional figures generate three-dimensional solids.
  • Understanding the relation between triangle dimensions and cone dimensions is important for higher geometry topics.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Revolving a right triangle about one perpendicular side forms a cone.

  2. The axis of rotation becomes the height of the cone.

  3. The other perpendicular side becomes the radius.

  4. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  5. Always write the volume in cubic units such as \(\small \text{cm}^3\).
← Q6
7 / 9  ·  78%
Q8 →
Q8
NUMERIC3 marks
If the triangle ABC in Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
📘 Concept & Theory Theory and Concept Used ›

When a right triangle is revolved about one of its perpendicular sides, it forms a right circular cone.

In this case:

  • The side about which the triangle rotates becomes the height of the cone.
  • The other perpendicular side becomes the radius of the cone.
  • The hypotenuse becomes the slant height of the cone.

Formula for the volume of a cone:

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

Ratio of volumes is found by simplifying the two obtained volumes.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the radius and height of the cone formed after revolution.

  2. Apply the formula for the volume of a cone.

  3. Simplify step-by-step carefully.

  4. Use the volume obtained in Question 7.

  5. Find and simplify the ratio of the two volumes.

📊 Graph / Figure Graph / Figure ›
5 cm (axis) 12 cm (radius) 13 cm Revolution: About 5cm Side
Right triangle revolved about side 5 cm
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  15 steps
  1. The given right triangle has sides:\[\small 5\ \text{cm}, \quad 12\ \text{cm}, \quad 13\ \text{cm}\]
  2. The triangle is revolved about the side measuring \(\small 5\ \text{cm}\).
  3. Therefore: Height of the cone: \[\small h = 5\ \text{cm} \]
  4. Radius of the cone: \[\small r = 12\ \text{cm} \]
  5. Slant height: \[\small l = 13\ \text{cm} \]
  6. Formula for the volume of a cone:\[\small V = \frac{1}{3}\pi r^2 h\]
  7. Substituting the values: \[\small \begin{aligned} V &= \frac{1}{3}\times \pi \times (12)^2 \times 5 \\ &= \frac{1}{3}\times \pi \times 144 \times 5 \\ &= \pi \times 48 \times 5\\ &= 240\pi\ \text{cm}^3 \end{aligned} \]
  8. Therefore, the volume of the solid obtained is:\[\small \boxed{240\pi\ \text{cm}^3}\]
  9. Using:\[\small \pi \approx 3.1416\]
  10. Approximate value: \[\small \begin{aligned} V &= 240 \times 3.1416 \\ &= 753.984\ \text{cm}^3 \end{aligned} \]
  11. Ratio of the Volumes of the Solids in Questions 7 and 8
  12. From Question 7:\[\small V_1 = 100\pi\ \text{cm}^3\]
  13. V_2 = 240\pi\ \text{cm}^3
  14. Therefore, the ratio: \[\small \begin{aligned} V_1 : V_2 &= 100\pi : 240\pi \\ &= 100 : 240 \\ &= 5 : 12 \end{aligned} \]
  15. Hence, the ratio of the volumes is:\[\small \boxed{5 : 12}\]
🎯 Exam Significance Exam Significance ›
  • This question combines concepts of geometry, solids formed by revolution, and ratio of volumes.
  • Such integrated mensuration problems are commonly asked in CBSE and state board examinations.
  • Competitive examinations such as NTSE, Olympiads, SSC, and Polytechnic entrance tests frequently include application-based cone problems.
  • Students learn how changing the axis of revolution changes the dimensions and volume of the cone.
  • Understanding ratios of volumes strengthens algebraic simplification skills.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Revolving a right triangle about one perpendicular side forms a cone.

  2. The axis of rotation becomes the height of the cone.

  3. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  4. Ratios should always be simplified to lowest terms.

  5. The value of \(\small \pi\) cancels automatically while finding ratios.
← Q7
8 / 9  ·  89%
Q9 →
Q9
NUMERIC3 marks
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
📘 Concept & Theory Theory and Concept Used ›

A heap of wheat shaped like a cone is a practical application of mensuration. The quantity of wheat stored is measured using the volume of the cone.

Formula for the volume of a cone:

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

To cover the heap with canvas, only the curved outer surface is covered. Therefore, the required area is the curved surface area (CSA) of the cone.

Formula for curved surface area:

\[\small \text{CSA} = \pi r l \]

where:

  • \(\small r\) = Radius of the base
  • \(\small h\) = Height of the cone
  • \(\small l\) = Slant height of the cone

The slant height is found using:

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

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the radius from the given diameter.

  2. Use the volume formula to calculate the volume of the heap.

  3. Find the slant height using the Pythagoras theorem.

  4. Calculate the curved surface area for the canvas required.

  5. Write the final answers with proper units.

📊 Graph / Figure Graph / Figure ›
h = 3 m r = 5.25 m l Conical Wheat Heap with Canvas
Conical heap showing height, radius and canvas covering
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  11 steps
  1. GIven

  2. Diameter of the heap:

    \[\small d = 10.5\ \text{m} \]

    Height of the heap:

    \[\small h = 3\ \text{m} \]

    Radius of the heap:

    \[\small \begin{aligned} r &= \frac{d}{2} \ &= \frac{10.5}{2} \ &= 5.25\ \text{m} \end{aligned} \]

  3. Step-by-step Solution
  4. Finding the Volume of the Heap
  5. Formula for volume of a cone:\[\small V = \frac{1}{3}\pi r^2 h\]
  6. Substituting the values: \[\small \begin{aligned} V &= \frac{1}{3}\times \frac{22}{7}\times (5.25)^2 \times 3\\ &= \frac{1}{3}\times \frac{22}{7}\times 27.5625 \times 3\\ &= \frac{22}{7}\times 27.5625\\ &= 86.625\ \text{m}^3 \end{aligned} \]
  7. Hence, the volume of the heap is: \[\small \boxed{86.625\ \text{m}^3}\]
  8. Finding the Area of Canvas Required
  9. The canvas covers only the curved outer surface of the heap.
  10. First find the slant height: \[\small l = \sqrt{r^2 + h^2} \]
  11. Substituting the values: \[\small \begin{aligned} l &= \sqrt{(5.25)^2 + (3)^2} \\ &= \sqrt{27.5625 + 9} \\ &= \sqrt{36.5625} \\ &\approx 6.047\ \text{m} \end{aligned} \]
  12. Formula for curved surface area: \[\small \text{CSA} = \pi r l \]
  13. Substituting the values: \[\small \begin{aligned} \text{CSA} &= \frac{22}{7}\times 5.25 \times 6.047\\ &\approx 99.78\ \text{m}^2 \end{aligned} \]
  14. Therefore, the area of canvas required is:\[\small \boxed{99.78\ \text{m}^2}\]
🎯 Exam Significance Exam Significance ›
  • This is a practical application-based mensuration problem frequently asked in board examinations.
  • The question combines volume, slant height, and curved surface area concepts in one problem.
  • Competitive examinations such as NTSE, Olympiads, SSC, and Polytechnic entrance tests often include such real-life geometry applications.
  • Students learn the importance of identifying whether total surface area or curved surface area is required.
  • Proper stepwise calculations improve accuracy and presentation in examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Volume of a cone: \[\small V = \frac{1}{3}\pi r^2 h \]

  2. Curved surface area: \[\small \text{CSA} = \pi r l \]

  3. Slant height: \[\small l = \sqrt{r^2 + h^2} \]

  4. Radius is half of the diameter.

  5. Canvas required for covering a cone equals its curved surface area.
← Q8
9 / 9  ·  100%
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Chapter Complete!

All 9 solutions for SURFACE AREAS AND VOLUMES covered.

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NCERT Class 9 Maths Exercise 11.3 Solutions
NCERT Class 9 Maths Exercise 11.3 Solutions — Complete Notes & Solutions · academia-aeternum.com
Explore expertly solved textbook exercises from NCERT Mathematics Class 9 Chapter 11, “Surface Areas and Volumes.” These clear, stepwise solutions help you master concepts of three-dimensional geometry, applying key formulas for cubes, cuboids, cones, cylinders, spheres, and hemispheres. Perfect for CBSE exam preparation, revision, and building strong problem-solving skills.
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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