Frustum Of A Regular Pyramid Formula

Frustum of A Regular Pyramid Formula

The frustum is a pyramid formed by slicing the top off of a conventional pyramid. It's termed a truncated pyramid for this reason.
The height of the pyramid is the distance between its base and top, denoted by h. It, too, has a slant height represented by "s," as well as two bases (top and bottom) whose area is described by "a", B₁ and B₂.

We need to find the lateral surface area and the volume of Frustum of the regular pyramid formula.
$V = \frac{h(B_1 + B_2 + \sqrt{B_1{B}_2})}{3}$
Here, S = Lateral Surface Area, P₁ and P₂ = Perimeter of Bases, h = Height, B₁ and B₂ = Area of bases, s = Slant height, V = Volume

Find solved example of Frustum Of A Regular Pyramid Formula

Example: Find out volume of a frustum of a regular pyramid whose area of bases are 9 cm2, 10 cm²respectively and height is 9 cm.

Solution:
Given
B₁ = 9 cm²
B²= 10 cm²
h = 9 cm

Volume Formula,
$V = \frac{h(B_1+B_2+\sqrt{B_1{B}_2})}{3}$
V = [9(9 + 10 + √9 x 10)]/3
V = [9(19 + √90)]/3
V = 57 + 9√10
V = 57 + 28.458
V = 85.458 cm³

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