THE PRISMATOID FORMULA.

731. DEF. A polyhedron is called a prismatoid if it has for bases two polygons in parallel planes, and for lateral faces triangles or trapezoids with one side common with one base and the opposite vertex or side common with the other base.

732. DEF. The altitude of a prismatoid is the perpendicular distance between the planes of its bases.

The mid-section of a prismatoid is the section made by a plane parallel to its bases and midway between them.

The mid-section bisects the altitude and all the lateral edges.

PROPOSITION XLIV. THEOREM.

733. The volume of a prismatoid is equal to the product of one sixth of its altitude into the sum of its bases and four times its mid-section.

Let V denote the volume, B and b the bases, M the mid-section,

and H the altitude, of a given prismatoid.

To prove that V = } H (B + b + 4 M).

Proof. If any lateral face is a trapezoid, divide it into two triangles by a diagonal.

Take any point P in the mid-section and join P to the vertices of the polyhedron and of the mid-section.

Divide the prismatoid into pyramids which have their vertices at P, and for their respective bases the lower base B, the upper base b, and the lateral faces of the prismatoid.

The lateral pyramid P-ADE is composed of three pyramids P-AFC, P-FCE, and P-FDE.

Now P-AFC may be regarded as having vertex A and base PFC, and P-FCE, as having vertex E and base PFC.

Hence, the volume of P-AFC is equal to

and

the volume of P-FCE is equal to

H × PFC,

H × PFC. § 651 The pyramid P-FDE is equivalent to twice P-FCE.

For they have the same vertex P, and the base FDE is twice the base FCE, since the ▲ FDE has its base DE twice the base FC of the AFCE (§ 405), and these triangles have the same altitude.

Hence, the volume of P-FDE is equal to & H × PFC. Therefore, the volume of P-ADE, which is composed of P-AFC, P-FCE, and P-FDE, is equal to & H× PFC.

In like manner, the volume of each lateral pyramid is equal to H × the area of that part of the mid-section which is included within it; and, therefore, the total volume of all these lateral pyramids is equal to H × M.

The volume of the pyramid with base B is and the volume of the pyramid with base bis

Therefore,

V = { H (B + b + 4 M).

H×B,

H× b.

$ 651

Q. E.D.

734. The prismatoid formula may be used for finding the volumes of all the solids of Elementary Geometry:

A A D

B

e

The formula for the volume of the frustum of a pyramid is V = } H (B + b + √ B x b).

But 3 H (B+ b + √ B × b) = } H (B + b + 4 M).

$ 657 (1)

For if e and e' are corresponding sides of B and b, then (ee) is the corresponding side of M.

§ 190

VB

and

√M

(e + e')

2 √B + √b

1

=

√M

2√M = √B+ √b.

4 MB + b + 2 √ B x b.

If we put this value of 4 M in (1), the two members become identically equal.

735. If the base b becomes zero, we have a pyramid, and the prismatoid formula becomes VHx B.

$ 652

736. If the bases B and b are equal, we have a prism, and the prismatoid formula becomes V = H × B.

NOTE.

§ 628

The Prismatoid Formula is taken by permission from the work on Mensuration by Dr. George Bruce Halsted, Professor of Mathematics in the University of Texas.

Ex. 701. Show that the prismatoid formula can be used for finding the volume of the frustum of a cone, for finding the volume of a cone, for finding the volume of a cylinder.

FRUSTUMS OF PYRAMIDS AND OF CONES.

Ex. 702. How many square feet of tin are required to make a funnel, if the diameters of the top and bottom are 28 inches and 14 inches, respectively, and the height is 24 inches?

Ex. 703. Find the expense, at 60 cents a square foot, of polishing the curved surface of a marble column in the shape of the frustum of a right circular cone whose slant height is 12 feet, and the radii of whose bases are 3 feet 6 inches and 2 feet 4 inches, respectively.

Ex. 704. The slant height of the frustum of a regular pyramid is 20 feet; the sides of its square bases 40 feet and 16 feet. Find the volume.

Ex. 705. If the bases of the frustum of a pyramid are regular hexagons whose sides are 1 foot and 2 feet, respectively, and the volume of the frustum is 12 cubic feet, find the altitude.

Ex. 706. The frustum of a right circular cone 14 feet high has a volume of 924 cubic feet. Find the radii of its bases if their sum is 9 feet.

Ex. 707. From a right circular cone whose slant height is 30 feet, and the circumference of whose base is 10 feet, there is cut off by a plane parallel to the base a cone whose slant height is 6 feet. Find the lateral area and the volume of the frustum.

Ex. 708. Find the difference between the volume of the frustum of a pyramid whose bases are squares, 8 feet and 6 feet, respectively, on a side and the volume of a prism of the same altitude whose base is a section of the frustum parallel to its bases and equidistant from them.

Ex. 709. A Dutch windmill in the shape of the frustum of a right cone is 12 meters high. The outer diameters at the bottom and the top are 16 meters and 12 meters, the inner diameters 12 meters and 10 meters. How many cubic meters of stone were required to build it?

Ex. 710. The chimney of a factory has the shape of a frustum of a regular pyramid. Its height is 180 feet, and its upper and lower bases are squares whose sides are 10 feet and 16 feet, respectively. The flue is throughout a square whose side is 7 feet.

material does the chimney contain?

How many cubic feet of

Ex. 711. Find the volume V of the frustum of a cone of revolution, having given the slant height L, the height H, and the lateral area S.

EQUIVALENT SOLIDS.

Ex. 712. A cube each edge of which is 12 inches is transformed into a right prism whose base is a rectangle 16 inches long and 12 inches wide. Find the height of the prism, and the difference between its total area and the total area of the cube.

Ex. 713. The dimensions of a rectangular parallelopiped are a, b, c. Find (i) the height of an equivalent right circular cylinder, having a for the radius of its base; (ii) the height of an equivalent right circular cone having a for the radius of its base.

Ex. 714. A regular pyramid 12 feet high is transformed into a regular prism with an equivalent base. Find the height of the prism.

Ex. 715. The diameter of a cylinder is 14 feet, and its height is 8 feet. Find the height of an equivalent right prism, the base of which is a square with a side 4 feet long.

Ex. 716. If one edge of a cube is a, what is the height H of an equivalent right circular cylinder whose radius is R ?

Ex. 717. The heights of two equivalent right circular cylinders are in the ratio 4: 9. If the diameter of the first is 6 feet, what is the diameter of the other?

Ex. 718. A right circular cylinder 6 feet in diameter is equivalent to a right circular cone 7 feet in diameter. If the height of the cone is 8 feet, what is the height of the cylinder ?

Ex. 719. The frustum of a regular pyramid 6 feet high has for bases squares 5 feet and 8 feet on a side. Find the height of an equivalent regular pyramid whose base is a square 12 feet on a side.

Ex. 720. The frustum of a cone of revolution is 5 feet high, and the diameters of its bases are 2 feet and 3 feet, respectively. Find the height of an equivalent right circular cylinder whose base is equal in area to the section of the frustum made by a plane parallel to its bases and equidistant from the bases.

Ex. 721. Find the edge of a cube equivalent to a regular tetrahedron whose edge measures 3 inches.

Ex. 722. Find the edge of a cube equivalent to a regular octahedron whose edge measures 3 inches.