Set of pyramidal right -gonal frustums | |
Faces: | isosceles trapezoids, regular -gons |
Symmetry: | |
Dual: | asymmetric bipyramid |
Properties: | convex |
Net: | Net of right trigonal frustum.png |
Net Caption: | Example: net of right trigonal frustum |
In geometry, a la|'''frustum'''|italic=no|morsel; (: frusta or frustums) is the portion of a solid (normally a pyramid or a cone) that lies between two parallel planes cutting the solid. In the case of a pyramid, the base faces are polygonal and the side faces are trapezoidal. A right frustum is a right pyramid or a right cone truncated perpendicularly to its axis;[1] otherwise, it is an oblique frustum. In a truncated cone or truncated pyramid, the truncation plane is necessarily parallel to the cone's base, as in a frustum.If all its edges are forced to become of the same length, then a frustum becomes a prism (possibly oblique or/and with irregular bases).
A frustum's axis is that of the original cone or pyramid. A frustum is circular if it has circular bases; it is right if the axis is perpendicular to both bases, and oblique otherwise.
The height of a frustum is the perpendicular distance between the planes of the two bases.
Cones and pyramids can be viewed as degenerate cases of frusta, where one of the cutting planes passes through the apex (so that the corresponding base reduces to a point). The pyramidal frusta are a subclass of prismatoids.
Two frusta with two congruent bases joined at these congruent bases make a bifrustum.
The formula for the volume of a pyramidal square frustum was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty :
V=
h | |
3 |
\left(a2+ab+b2\right),
The Egyptians knew the correct formula for the volume of such a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus.
The volume of a conical or pyramidal frustum is the volume of the solid before slicing its "apex" off, minus the volume of this "apex":
V=
h1B1-h2B2 | |
3 |
,
Considering that
B1 | ||||||
|
=
B2 | ||||||
|
=
\sqrt{B1B2 | |
\alpha
V=
| |||||||||||||||||||
3 |
=\alpha
| ||||||||||||||||
3 |
.
V=(h1-
h | |||||||||||||||||||
|
,
Distributing
\alpha
B1+\sqrt{B1B2 | |
+ |
B2}{3};
V=
h | |
3 |
\left(B1+\sqrt{B1B2}+B2\right).
In particular:
V=
\pih | |
3 |
2 | |
\left(r | |
1 |
+r1r2+
2\right), | |
r | |
2 |
where and are the base and top radii.
V=
nh | |
12 |
2 | |
\left(a | |
1 |
+a1a2+
| ||||
a | ||||
2 |
,
where and are the base and top side lengths.
s