# 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 cm^{2}, 10 cm^{2} and height is 9 cm ?

**Solution:
**Given

B

_{1}= 9 cm

^{2 }B

_{2 }= 10 cm

^{2 }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 cm^{3}