Volume of Cone
A cone is a three-dimensional solid that has a circular base and tapers smoothly to a point called the apex or vertex. Common examples include ice cream cones, party hats, and traffic cones.
The volume of a cone is exactly one-third the volume of a cylinder with the same base radius and height. This relationship is fundamental to understanding mensuration in Class 9 Mathematics.
This topic is part of the NCERT Class 9 chapter on Surface Areas and Volumes. Mastering cone volume calculations is essential for solving problems involving funnels, conical tanks, heaps of sand, and combined solids.
What is Volume of Cone?
Definition: The volume of a cone is the amount of space enclosed within the conical surface and its circular base.
For a cone with base radius r and height h:
Volume = (1/3)πr²h
Where:
- r = radius of the circular base
- h = perpendicular height (from base to apex)
- π = 22/7 or 3.14159...
- The result is in cubic units (cm³, m³, etc.)
Important:
- The height h is the perpendicular distance from the base to the apex, NOT the slant height.
- The slant height l is related by: l = √(r² + h²).
- Volume of cone = (1/3) × Volume of cylinder with the same base and height.
Volume of Cone Formula
Key Formulas:
1. Volume of a cone:
V = (1/3)πr²h
2. Relationship with cylinder:
- Volume of cone = (1/3) × Volume of cylinder (same r and h)
- Three cones fill one cylinder exactly.
3. Slant height:
l = √(r² + h²)
4. Height from slant height:
- h = √(l² − r²)
5. Finding radius from volume:
- r² = 3V / (πh)
- r = √[3V / (πh)]
Units:
- If r and h are in cm, volume is in cm³.
- 1 litre = 1000 cm³
Derivation and Proof
Derivation of Volume of Cone (Experimental Method):
Step 1: Take a cone and a cylinder
- Prepare a hollow cone and a hollow cylinder with the same base radius r and the same height h.
Step 2: Fill the cone with water
- Fill the cone completely with water or sand.
Step 3: Pour into the cylinder
- Pour the contents of the cone into the cylinder.
- The cylinder is NOT fully filled after one pour.
Step 4: Repeat
- Fill the cone again and pour into the cylinder — second time.
- Fill the cone a third time and pour into the cylinder.
- After three pours, the cylinder is exactly full.
Step 5: Establish the formula
- 3 × Volume of cone = Volume of cylinder
- 3 × Volume of cone = πr²h
- Volume of cone = (1/3)πr²h
This experiment verifies that the volume of a cone is exactly one-third the volume of a cylinder with the same dimensions.
Types and Properties
Types of cone problems:
1. Right Circular Cone
- The apex is directly above the centre of the base.
- The axis is perpendicular to the base.
- V = (1/3)πr²h
2. Finding volume when slant height is given
- First find h using: h = √(l² − r²)
- Then apply V = (1/3)πr²h
3. Conical heap problems
- Sand or grain poured from a height forms a conical heap.
- Given: radius of heap and height (or slant height)
- Find volume to determine the quantity of material.
- A cone is melted and recast into another shape (sphere, cylinder, smaller cones).
- Total volume is conserved.
5. Water flow problems
- A conical vessel is filled with water.
- Calculate the volume of water, or the time to fill/empty.
- A cone with the top portion cut off by a plane parallel to the base.
- V = (1/3)πh(R² + r² + Rr), where R and r are the two radii.
Solved Examples
Example 1: Example 1: Volume given radius and height
Problem: Find the volume of a cone with radius 7 cm and height 12 cm. (Use π = 22/7)
Solution:
Given:
- r = 7 cm, h = 12 cm
Using V = (1/3)πr²h:
- V = (1/3) × (22/7) × 7² × 12
- V = (1/3) × (22/7) × 49 × 12
- V = (1/3) × 22 × 7 × 12
- V = (1/3) × 1848
- V = 616 cm³
Answer: Volume = 616 cm³.
Example 2: Example 2: Volume given slant height
Problem: Find the volume of a cone with radius 6 cm and slant height 10 cm. (Use π = 22/7)
Solution:
Given:
- r = 6 cm, l = 10 cm
Finding height:
- h = √(l² − r²) = √(100 − 36) = √64 = 8 cm
Using V = (1/3)πr²h:
- V = (1/3) × (22/7) × 36 × 8
- V = (1/3) × (22/7) × 288
- V = (1/3) × 905.14
- V = 301.71 cm³ (approx.)
Answer: Volume ≈ 301.71 cm³.
Example 3: Example 3: Volume of conical heap
Problem: Sand is poured to form a conical heap of radius 3.5 m and height 4.2 m. Find the volume of sand. (Use π = 22/7)
Solution:
Given:
- r = 3.5 m, h = 4.2 m
Using V = (1/3)πr²h:
- V = (1/3) × (22/7) × (3.5)² × 4.2
- V = (1/3) × (22/7) × 12.25 × 4.2
- V = (1/3) × (22/7) × 51.45
- V = (1/3) × 161.7
- V = 53.9 m³
Answer: Volume of sand = 53.9 m³.
Example 4: Example 4: Comparing cone and cylinder
Problem: A cylinder and a cone have the same base radius 5 cm and same height 9 cm. Find the ratio of their volumes.
Solution:
Volume of cylinder:
- V_cyl = πr²h = π × 25 × 9 = 225π
Volume of cone:
- V_cone = (1/3)πr²h = (1/3) × 225π = 75π
Ratio:
- V_cyl : V_cone = 225π : 75π = 3 : 1
Answer: The cylinder’s volume is 3 times the cone’s volume. Ratio = 3 : 1.
Example 5: Example 5: Finding height from volume
Problem: The volume of a cone is 462 cm³ and its radius is 7 cm. Find the height. (Use π = 22/7)
Solution:
Given:
- V = 462 cm³, r = 7 cm
Using V = (1/3)πr²h:
- 462 = (1/3) × (22/7) × 49 × h
- 462 = (1/3) × 22 × 7 × h
- 462 = (154/3) × h
- h = 462 × 3 / 154 = 1386/154 = 9 cm
Answer: Height = 9 cm.
Example 6: Example 6: Cone melted into sphere
Problem: A cone of radius 6 cm and height 7 cm is melted and recast into a sphere. Find the radius of the sphere.
Solution:
Volume of cone:
- V_cone = (1/3)π × 36 × 7 = 84π cm³
Volume of sphere:
- V_sphere = (4/3)πR³
Equating volumes:
- (4/3)πR³ = 84π
- R³ = 84 × 3/4 = 63
- R = ³√63 = 3.98 cm (approx.)
Answer: Radius of the sphere ≈ 3.98 cm.
Example 7: Example 7: Finding radius from volume and height
Problem: The volume of a cone is 1232 cm³ and its height is 24 cm. Find the radius. (Use π = 22/7)
Solution:
Given:
- V = 1232 cm³, h = 24 cm
Using V = (1/3)πr²h:
- 1232 = (1/3) × (22/7) × r² × 24
- 1232 = (22 × 24) / (3 × 7) × r²
- 1232 = (528/21) × r²
- r² = 1232 × 21 / 528 = 25872/528 = 49
- r = 7 cm
Answer: Radius = 7 cm.
Example 8: Example 8: Water flow problem
Problem: A conical vessel of radius 12 cm and height 16 cm is completely filled with water. This water is poured into a cylindrical vessel of radius 8 cm. Find the height of water in the cylinder.
Solution:
Volume of cone:
- V = (1/3)π(12)²(16) = (1/3)π(144)(16) = 768π cm³
Volume of water in cylinder:
- π(8)²h = 768π
- 64h = 768
- h = 12 cm
Answer: Height of water in the cylinder = 12 cm.
Example 9: Example 9: Capacity in litres
Problem: A conical tent has a radius of 7 m and a height of 24 m. Find the volume of air inside the tent. (Use π = 22/7)
Solution:
Given:
- r = 7 m, h = 24 m
Using V = (1/3)πr²h:
- V = (1/3) × (22/7) × 49 × 24
- V = (1/3) × 22 × 7 × 24
- V = (1/3) × 3696
- V = 1232 m³
Answer: Volume of air = 1232 m³.
Example 10: Example 10: Effect of doubling dimensions
Problem: If both the radius and height of a cone are doubled, how does the volume change?
Solution:
Original volume:
- V = (1/3)πr²h
New dimensions:
- New radius = 2r, New height = 2h
New volume:
- V' = (1/3)π(2r)²(2h)
- V' = (1/3)π(4r²)(2h)
- V' = (1/3)π(8r²h)
- V' = 8 × (1/3)πr²h = 8V
Answer: The volume becomes 8 times the original.
Real-World Applications
Applications of Volume of Cone:
- Ice cream industry: Calculating the quantity of ice cream a cone can hold.
- Construction: Estimating sand, gravel, or cement in conical heaps at construction sites.
- Agriculture: Measuring grain stored in conical heaps on farms.
- Civil engineering: Designing conical tanks, funnels, and hoppers for water treatment and storage.
- Party supplies: Determining material for party hats and conical containers.
- Road safety: Manufacturing traffic cones with specific volume and height requirements.
Key Points to Remember
- Volume of cone = (1/3)πr²h, where r is the base radius and h is the perpendicular height.
- Volume of cone = (1/3) × Volume of cylinder with the same base and height.
- Three cones of the same dimensions fill one cylinder exactly.
- Slant height l = √(r² + h²). Do NOT use slant height in the volume formula.
- Height h = √(l² − r²) when slant height and radius are given.
- Doubling both radius and height increases volume 8 times.
- In melting and recasting problems, the total volume is conserved.
- Volume is always in cubic units.
- A cone is a special case of a pyramid with a circular base.
- The formula applies only to a right circular cone (axis perpendicular to the base).
Practice Problems
- Find the volume of a cone with radius 3.5 cm and height 12 cm. (Use π = 22/7)
- A cone has a slant height of 13 cm and base radius 5 cm. Find its volume.
- The volume of a cone is 154 cm³ and its height is 6 cm. Find the radius. (Use π = 22/7)
- A conical vessel has base diameter 14 cm and height 18 cm. How many litres of water can it hold?
- A cone of height 15 cm and radius 6 cm is melted and recast into smaller cones of height 5 cm and radius 2 cm. How many smaller cones are formed?
- Sand is poured to form a conical heap of diameter 7 m and height 3 m. Find the volume of sand.
- A conical tent has a radius of 10.5 m and height 8 m. Find the volume of air inside.
- If the radius of a cone is halved and the height is tripled, what is the effect on the volume?
Frequently Asked Questions
Q1. What is the formula for the volume of a cone?
Volume of a cone = (1/3)πr²h, where r is the base radius and h is the perpendicular height.
Q2. Why is the volume of a cone one-third of a cylinder?
Experimentally, if you fill a cone with water and pour it into a cylinder of the same base and height, it takes exactly 3 cone-fills to completely fill the cylinder. Hence, cone volume = (1/3) × cylinder volume.
Q3. What is the difference between height and slant height of a cone?
The height (h) is the perpendicular distance from the base to the apex. The slant height (l) is the distance along the surface from the base edge to the apex. They are related by l = √(r² + h²).
Q4. Can the slant height be used directly in the volume formula?
No. The volume formula uses the perpendicular height h, not the slant height l. If only slant height is given, first calculate h = √(l² − r²), then use V = (1/3)πr²h.
Q5. How do you find the height of a cone if the volume and radius are given?
Rearrange: h = 3V / (πr²). Substitute the known values and simplify.
Q6. What happens to the volume if the radius is doubled?
The volume becomes 4 times the original, since volume depends on r². New volume = (1/3)π(2r)²h = 4 × (1/3)πr²h.
Q7. Is volume of a cone in the CBSE Class 9 syllabus?
Yes. Volume of cone is part of CBSE Class 9 Mathematics, Chapter: Surface Areas and Volumes.
Q8. What is a frustum of a cone?
A frustum is formed when a cone is cut by a plane parallel to the base. It has two circular faces. Volume = (1/3)πh(R² + r² + Rr), where R and r are the radii of the two ends.
Related Topics
- Surface Area of Cone
- Volume of Cylinder
- Volume of Sphere
- Slant Height of Cone
- Surface Area of Sphere
- Surface Area of Hemisphere
- Volume of Hemisphere
- Combination of Solids
- Conversion of Solids
- Frustum of a Cone
- Surface Area of Combined Solids
- Volume Word Problems
- Volume of Frustum of Cone
- Curved Surface Area of Frustum
