 # Frustum of a Regular Pyramid Formula

The 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 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

$$\begin{array}{l}B_{1}\end{array}$$
and
$$\begin{array}{l}B_{2}\end{array}$$
.

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

$$\begin{array}{l}P_{1} and P_{2}\end{array}$$
= Perimeter of Bases
h=Height
$$\begin{array}{l}B_{1} and B_{2}\end{array}$$
= 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 respectively and height is 9 cm.

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

Volume formula,

V =

$$\begin{array}{l}\frac{h ( B_{1}+B_{2}+\sqrt{B_{1}B_{2}})}{3}\end{array}$$

$$\begin{array}{l}\begin{array}{l}\quad V=\frac{9(9+10+\sqrt{9 \times 10})}{3} \\ V=\frac{9(19+\sqrt{90})}{3} \\ V=57+9 \sqrt{10} \\ V=57+28.458 \\ V=85.458 \mathrm{cm}^{3}\end{array}\end{array}$$