Volume of Parallelepiped Formula
Last Updated :
23 Jul, 2025
A parallelepiped is defined as a three-dimensional shape formed by six parallelograms. It consists of six faces, eight vertices, and twelve edges. The face diagonals of a parallelepiped are two diagonals on each face. It has 12 face diagonals in total. The body or space diagonal of a parallelepiped is the diagonal connecting the vertices that are not on the same face. It can be interpreted as a prism with a parallelogram-shaped base. One of its every two faces is a mirror image of the other.
Volume of a Parallelepiped Formula
The volume of a parallelepiped is defined as the space filled by it in a three-dimensional plane. Knowing the base area and height of the parallelepiped is enough to calculate its volume. It is equal to the product of its base area and height.
V = B × h
where,
V is the volume,
B is the base area,
h is the height.
Find the Volume of a Parallelepiped?
To calculate the volume of a parallelepiped using the formula, it is essential to determine two measurements:
- S, representing the area of the base.
- H, denoting the height of the parallelepiped.
Solved Example for Rectangular Parallelepiped Formula
Problem 1. Calculate the volume of a parallelepiped if its base area is 20 m2 and height is 4 m.
Solution:
We have,
B = 20
h = 4
Using the formula we get,
V = B × h
= 20 (4)
= 80 m3
Problem 2. Calculate the volume of a parallelepiped if its base area is 15 m2 and height is 3 m.
Solution:
We have,
B = 15
h = 3
Using the formula we get,
V = B × h
= 15 (3)
= 45 m3
Problem 3. Calculate the volume of a parallelepiped if its base area is 23 m2 and height is 6 m.
Solution:
We have,
B = 23
h = 6
Using the formula we get,
V = B × h
= 23 (6)
= 138 m3
Problem 4. Calculate the base area of a parallelepiped if its volume is 100 m3 and height is 5 m.
Solution:
We have,
V = 100
h = 5
Using the formula we get,
V = B × h
=> B = V/h
= 100/5
= 20 m2
Problem 5. Calculate the base area of a parallelepiped if its volume is 350 m3 and height is 7 m.
Solution:
We have,
V = 350
h = 7
Using the formula we get,
V = B × h
=> B = V/h
= 350/7
= 50 m2
Problem 6. Calculate the height of a parallelepiped if its volume is 3375 m3 and the base area is 225 m2.
Solution:
We have,
V = 3375
B = 225
Using the formula we get,
V = B × h
=> h = V/B
= 3375/225
= 15 m
Problem 7. Calculate the height of a parallelepiped if its volume is 600 m3 and the base area is 120 m2.
Solution:
We have,
V = 600
B = 120
Using the formula we get,
V = B × h
=> h = V/B
= 600/120
= 5 m
Problem 8: Calculate the volume of a parallelepiped with a base area of 12 m² and a height of 7 m.
Solution:
The volume of a parallelepiped is given by the formula:
V = B x h
where:
B is the base area.
h is the height.
GIven. B = 12m2 , h = 7m
Substituting the values into the formula:
V = 12 x 7 = 84m3
The volume of the parallelepiped is 84m3.
Problem 9: The volume of a parallelepiped is 96 m³, and its height is 8 m. What is the base area?
Solution:
We know the volume formula:
V = B x h
To find the base area B, rearrange the formula:
B = V/h
GIven : V = 96m3 , h = 8m
Substituting the values:
B = 96/8 = 12m2
The base area of the parallelepiped is 12m2 .
Problem 10: If the base area of a parallelepiped is 20 m² and its volume is 160 m³, find the height.
Solution:
Using the volume formula:
V = B x h
To find the base area B, rearrange the formula:
h = V/B
GIven : V = 160m3 , B = 20m2
Substituting the values:
h = 160/20 = 8m
The base area of the parallelepiped is 8m.
Unsolved Pratice Problems
Problem 1. Calculate the volume of a parallelpiped if its base area is 30m2 and Height is 5m.
Problem 2. If the volume of a parallelpiped is 200m3 and its height is 10m, What is its base area.
Problem 3. Find the height of a parallelpiped if its volume is 450m3 and the base area is 90 m2
Problem 4. A parallelpiped has dimensions such that its base area is 22m2 and its height is 10m .Compute its volume.
Problem 5. If the base area of a parallelpiped is 35m2 and its volume is 455m3, find its height.
Conclusion
We can gain valuable insights on geometric truths and how they are put into practice, as with the right dimension- mapping shown here of a parallelepiped. An interesting and useful geometric figure, see its application in architecture to engineering physics. It shows an intuitive way to understand shape of a 3D object, i.e. volume (Area by Height). By learning how to calculate volume, base area and as well as altitude will help the reader appreciate geometrical concepts of space such as they are realized in practical contexts.Combine the using slicing idea of calculus and volumeIn order to understand that space (), take a solid of infinite thin leafs, each leaf height delta h, dividing into such a thin how many intermediate height.
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