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128 LearnersLast updated on December 11, 2025

The volume of a truncated pyramid is the total space it occupies or the number of cubic units it can hold. A truncated pyramid is a 3D shape with a top and bottom base that are parallel but not of the same size. To find the volume of a truncated pyramid, we use the formula that involves the areas of the bases and the height of the pyramid. In real life, kids relate to the volume of a truncated pyramid by thinking of things like a frustum of a pyramid-shaped glass or a lampshade. In this topic, letโs learn about the volume of a truncated pyramid.
The volume of a truncated pyramid is the amount of space it occupies.
It is calculated using the formula:\( [ \text{Volume} = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) ] where ( h ) \)is the height of the truncated pyramid,\( ( A_1 ) \)is the area of the top base, and\( ( A_2 )\) is the area of the bottom base.
This formula accounts for both bases and the height of the pyramid.
To derive the volume of a truncated pyramid, we use the concept of volume as the total space occupied by a 3D object.
The formula for the volume of a truncated pyramid is derived from the volume of a full pyramid: \([ \text{Volume} = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) ]\)
This formula considers the average of the areas of the bases and the height to determine the volume.
The volume of a truncated pyramid is usually expressed in cubic units, for example, cubic centimeters\( (cm^3), cubic meters (m^3).\)
To find the volume, calculate the areas of the top and bottom bases, determine the height, and apply the formula: \([ \text{Volume} = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) ]\)
Once we know the areas of the bases and the height, substitute these values into the formula to find the volume.


Remember the formula: The formula for the volume of a truncated pyramid is:\( [ \text{Volume} = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) ] \)
Break it down: Calculate the areas of the bases and the height separately, then substitute them into the formula. Ensure the bases are parallel and the height is perpendicular to them.
Simplify the numbers: If the areas and height are simple numbers, it is easier to substitute them directly into the formula.
Check for unit consistency: Ensure all measurements are in the same units before calculating the volume.
Making mistakes while learning the volume of a truncated pyramid is common. Let’s look at some common mistakes and how to avoid them to get a better understanding of the volume of truncated pyramids.
A truncated pyramid has a top base area of 20 cmยฒ, a bottom base area of 50 cmยฒ, and a height of 10 cm. What is its volume?
The volume of the truncated pyramid is 366.03 cm³.
To find the volume of a truncated pyramid, use the formula: \([ V = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) ]\)
Here, \(( A_1 = 20 \, \text{cm}^2 ), ( A_2 = 50 \, \text{cm}^2 ), and ( h = 10 \, \text{cm} ).\)
\(V = \frac{1}{3} \times 10 \times (20 + 50 + \sqrt{20 \times 50}) \)
\(V = \frac{1}{3} \times 10 \times (70 + \sqrt{1000}) \)
\(V = \frac{1}{3} \times 10 \times (70 + 31.62) \)
\( V = 366.03 \, \text{cm}^3 \)
A truncated pyramid has a top base area of 30 mยฒ, a bottom base area of 70 mยฒ, and a height of 8 m. Find its volume.
The volume of the truncated pyramid is 800.34 m³.
To find the volume of a truncated pyramid, use the formula:\( V = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) \)
Substitute \( A_1 = 30 \, \text{m}^2 , A_2 = 70 \, \text{m}^2 , and h = 8 \text{m} \)
\(V = \frac{1}{3} \times 8 \times (30 + 70 + \sqrt{30 \times 70}) \)
\( V = \frac{1}{3} \times 8 \times (100 + \sqrt{2100}) \)
\(V = \frac{1}{3} \times 8 \times (100 + 45.83) \)
\(V = 800.34 \, \text{m}^3 \)
The volume of a truncated pyramid is 500 cmยณ. If the top base area is 15 cmยฒ, the bottom base area is 35 cmยฒ, find the height.
The height of the truncated pyramid is 7.21 cm.
If you know the volume of the truncated pyramid and need to find the height, rearrange the formula: \(V = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) \)
\(500 = \frac{1}{3} \times h \times (15 + 35 + \sqrt{15 \times 35})
\)
\(500 = \frac{1}{3} \times h \times (50 + \sqrt{525}) \)
\(500 = \frac{1}{3} \times h \times (50 + 22.91) \)
\( h = \frac{500 \times 3}{72.91} \)
h = 7.21 cm
A truncated pyramid has a top base area of 25 inchesยฒ, a bottom base area of 60 inchesยฒ, and a height of 5 inches. Find its volume.
The volume of the truncated pyramid is 391.03 inches³.
Using the formula for volume: \(V = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) \)
Substitute \(A_1 = 25 , \text{inches}^2 , A_2 = 60 , \text{inches}^2 , and h = 5 \, \text{inches} \)
\(V = \frac{1}{3} \times 5 \times (25 + 60 + \sqrt{25 \times 60}) \)
\(V = \frac{1}{3} \times 5 \times (85 + \sqrt{1500}) \)
\(V = \frac{1}{3} \times 5 \times (85 + 38.73) \)
\(V = 391.03 \, \text{inches}^3 \)
You have a truncated pyramid-shaped container with a top base area of 12 ftยฒ, a bottom base area of 45 ftยฒ, and a height of 6 ft. How much space (in cubic feet) is available inside the container?
The container has a volume of 378.25 cubic feet.
Using the formula for volume: \(V = \frac{1}{3} \times h \times (A_1 + A_2 + \sqrt{A_1 \times A_2}) \)
Substitute \(A_1 = 12 \text{ft}^2 , A_2 = 45 \, \text{ft}^2 , and h = 6 \text{ft} \)
\(V = \frac{1}{3} \times 6 \times (12 + 45 + \sqrt{12 \times 45})
\)
\(V = \frac{1}{3} \times 6 \times (57 + \sqrt{540})
\)
\(V = \frac{1}{3} \times 6 \times (57 + 23.24)ย
\)
\(V = 378.25 \, \text{ft}^3ย \)

Seyed Ali Fathima S a math expert with nearly 5 years of experience as a math teacher. From an engineer to a math teacher, shows her passion for math and teaching. She is a calculator queen, who loves tables and she turns tables to puzzles and songs.
: She has songs for each table which helps her to remember the tables






