# 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

#### Practise This Question

The vertices of a  ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that
ADAB=AEAC=14. Calculate the ratio of the area of the ΔADE and ΔABC