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340 LearnersLast updated on December 11, 2025

The volume of a truncated cone is the amount of space it occupies.
It is calculated using the formula: Volume = (1/3)πh(R² + r² + Rr) Where ‘R’ is the radius of the larger base, ‘r’ is the radius of the smaller base, and ‘h’ is the height of the truncated cone.
A truncated cone is a 3-dimensional shape with two circular ends of different sizes. To calculate its volume, we add the areas of the two circular ends and their geometric mean, and then multiply by the height and π divided by 3.
The formula for the volume of a truncated cone is given as follows: Volume = (1/3)πh(R² + r² + Rr)
To derive the volume of a truncated cone, we start with the concept of volume for a cone.
A truncated cone is formed by cutting a smaller cone off the top of a larger cone.
The volume of a cone is: Volume = (1/3)πr²h
For a truncated cone: Volume = (1/3)πh(R² + r² + Rr)
This formula is derived by subtracting the volume of the smaller cone from that of the larger cone.
The volume of a truncated cone is expressed in cubic units, for example, cubic centimeters (cm³) or cubic meters (m³).
To find the volume, follow these steps: Write down the formula: Volume = (1/3)πh(R² + r² + Rr) Where R is the radius of the larger base, r is the radius of the smaller base, and h is the height.
Substitute the values of R, r, and h into the formula to find the volume. Ensure all measurements are in the same units.


Remember the formula: Volume = (1/3)πh(R² + r² + Rr) Break it down: Understanding the formula helps in visualizing the problem. The volume is the space inside the truncated cone.
Simplify the numbers: For ease of calculation, ensure the units are consistent, and simplify the expression (R² + r² + Rr) before multiplying by πh/3.
Check for errors in measurement: Ensure you have the correct radii and height, as mistakes in these values will lead to incorrect volume.
Learning the volume of a truncated cone can lead to some common mistakes. Let’s look at these mistakes and learn how to avoid them.
A truncated cone has a height of 5 cm, a larger base radius of 4 cm, and a smaller base radius of 2 cm. What is its volume?
The volume of the truncated cone is approximately 125.67 cm³.
To find the volume of a truncated cone, use the formula: V = (1/3)πh(R² + r² + Rr)
Here, R = 4 cm, r = 2 cm, and h = 5 cm,
so: V = (1/3)π * 5 * (4² + 2² + 4*2) V = (1/3)π * 5 * (16 + 4 + 8) V = (1/3)π * 5 * 28 V ≈ 125.67 cm³
A truncated cone has a height of 10 m, a larger base radius of 6 m, and a smaller base radius of 3 m. Find its volume.
The volume of the truncated cone is approximately 942.48 m³.
To find the volume of a truncated cone, use the formula: V = (1/3)πh(R² + r² + Rr)
Substitute the values: R = 6 m, r = 3 m, h = 10 m:
V = (1/3)π * 10 * (6² + 3² + 6*3) V = (1/3)π * 10 * (36 + 9 + 18) V = (1/3)π * 10 * 63 V ≈ 942.48 m³
The volume of a truncated cone is 250 cm³. If the height is 5 cm and the larger base radius is 5 cm, what is the smaller base radius?
The smaller base radius of the truncated cone is approximately 2.29 cm.
Using the formula for volume: V = (1/3)πh(R² + r² + Rr)
Rearrange to solve for r: 250 = (1/3)π * 5 * (5² + r² + 5r) 250 = (5/3)π * (25 + r² + 5r)
Rearrange and solve for r using algebraic methods, yielding r ≈ 2.29 cm.
A truncated cone has a height of 7 inches, a larger base radius of 8 inches, and a smaller base radius of 3 inches. Find its volume.
The volume of the truncated cone is approximately 861.9 inches³.
Using the formula for volume: V = (1/3)πh(R² + r² + Rr)
Substitute the values: R = 8 inches, r = 3 inches, h = 7 inches:
V = (1/3)π * 7 * (8² + 3² + 8*3)
V = (1/3)π * 7 * (64 + 9 + 24)
V = (1/3)π * 7 * 97
V ≈ 861.9 inches³
You have a lampshade shaped like a truncated cone with a height of 12 cm, a larger base radius of 10 cm, and a smaller base radius of 5 cm. How much space (in cubic centimeters) does it occupy?
The lampshade occupies approximately 2,617.99 cm³.
Using the formula for volume: V = (1/3)πh(R² + r² + Rr)
Substitute the values: R = 10 cm, r = 5 cm, h = 12 cm:
V = (1/3)π * 12 * (10² + 5² + 10*5)
V = (1/3)π * 12 * (100 + 25 + 50)
V = (1/3)π * 12 * 175 V ≈ 2,617.99 cm³

Nishtha is a passionate Maths teacher with 3 years of teaching experience. She focuses on making mathematics simple, engaging, and stress-free through student-centric, interactive, and exam-oriented teaching. With strong fundamentals, regular practice, an
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