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Frustum of a Regular Pyramid Formula

Frustum is a pyramid that is the result of chopping off the top of a regular pyramid. That is the reason why it is called a truncated pyramid.

The distance between the the base and the top of the pyramid is the height and is denoted by h. Similarly, it has a slant height that is denoted by “s” and two bases (top and bottom), whose area is defined by $B_{1}$ and $B_{2}$.

We need to find the lateral surface area and the volume of Frustum of regular pyramid formula.

\[\large V=\frac{h(B_{1}+B_{2}+\sqrt{B_{1}B_{2}})}{3}\]

\[\large S=\frac{s(P_{1}+P_{2})}{2}\]

Here,
S = Lateral Surface Area
$P_{1} and P_{2}$ = Perimeter of Bases
h=Height
$B_{1} and B_{2}= Area of bases
s = Slant height
V=Volume

Solved Examples

Question: Find the volume of a frustum of a regular pyramid whose area of bases are 9 cm2, 10 cm2 and height is 9 cm ?

Solution:
Given
B1 = 9 cm2
B= 10 cm2
h = 9 cm

Volume formula,

V = $\frac{h ( B_{1}+B_{2}+\sqrt{B_{1}B_{2}})}{3}$ 

V = $\frac{9(10 + 12 +\sqrt{10\times 12})}{3}$

V = 98.86 cm3

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