which may be written under the form,

V = {h[(BL + bl+ Bl + bL) + BL+ bl].

· (A.)

Because the auxiliary section is midway between the bases,

we have,

2M = L+ 1, and

2m = B + b ;

hence,

4 Mm

(L+ 1) (B + 6) = BL + bl + BL + bl.

But BL is the area of the lower base, or lower section, b is the area of the upper base, or upper section, and Mm is the area of the middle section; hence, the following

RULE.

To find the volume of a prismoid, find the sum of the areas of the extreme sections and four times the middle section; multiply the result by one-sixth of the distance between the extreme sections; the result will be the volume required.

This rule is used in computing volumes of earth-work in railroad cutting and embankment, and is of very extensive application. It may be shown that the same rule holds for every one of the volumes heretofore discussed in this work. Thus, in a pyramid, we may regard the base as one extreme section, and the vertex (whose area is 0), as the other extreme; their sum is equal to the area of the base. The area of a section midway between between them is equal to one-fourth of the base: hence, four times the middle section is equal to the base. Multiplying the sum of these by onesixth of the altitude, gives the same result as that already found. The application of the rule to the case of cylinders, frustums of cones, spheres, &c., is left as an exercise for the student.

EXAMPLES.

1. One of the bases of a rectangular prismoid is 25 feet by 20, the other 15 feet by 10, and the altitude 12 feet; required the volume.

Ans. 3700 cu. ft.

stick

of hewn timber,

inches by 18, its

Ans. 102 cu. ft.

2. What is the volume of a stick whose ends are 30 inches by 27, and 24 length being 24 feet?

MENSURATION OF REGULAR

POLYEDRONS.

123. A REGULAR POLYEDRON is a polyedron bounded by equal regular polygons.

The polyedral angles of any regular polyedron are all equal.

124. There are five regular polyedrons (Book VII., Page 208).

To find the diedral angle between the faces of a regular

polyedron.

125. Let the vertex of any polyedral angle be taken as the centre of a sphere whose radius is 1: then will this sphere, by its intersections with the faces of the polyedral angle, determine a regular spherical polygon whose sides will be equal to the plane angles that bound the polyedral angle, and whose angles are equal to the diedral angles between the faces.

It only remains to deduce a formula for finding one. angle of a regular spherical polygon, when the sides are given.

D

Let ABCDE represent a regular spherical polygon, and let P be the pole of a small circle passing through its vertices. Suppose P to be connected with each of the vertices by arcs of great circles; there will thus be formed as many equal isosceles triangles as the polygon has sides, the vertical angle in each being equal to 360° divided by the number of sides. Through P draw PQ perpendicular to AB: then will AQ

E

C

P

B

be equal to BQ. If we denote the number of sides by n,

the angle APQ will be equal to

360°
2n

180°

or

In the right-angled spherical triangle APQ, we know the base AQ, and the vertical angle APQ; hence, by Napier's rules for circular parts, we have,

sin (90° APQ) =cos (90° PAQ) cos AQ;

or, by reduction, denoting the side AB by s, and the angle PAB, by A,

To find the volume of a regular polyedron.

126. If planes be passed through the centre of the polyedron and each of the edges, they will divide the polyedron into as many equal right pyramids as the polyedron has faces. The common vertex of these pyramids will be at the centre of the polyedron, their bases will be the faces of the polyedron, and their lateral faces will bisect the diedral angles of the polyedron. The volume of each pyramid will be equal to its base into one-third of its altitude, and this multiplied by the number of faces, will be the volume of the polyedron.

It only remains to deduce a formula for finding the distance from the centre to one face of the polyedron.

Conceive a perpendicular to be drawn from the centre of the polyedron to one face; the foot of this perpendicular will be the centre of the face. From the foot of this perpendicular, draw a perpendicular to either side of the face in which it lies, and connect the point thus determined with the centre of the polyedron. There will thus be formed a right-angled triangle, whose base is the apothem of the face, whose angle at the base is half the diedral angle of the polyedron, and whose altitude is the required altitude of the pyramid, or in other words, the radius of the inscribed sphere.

Denoting the perpendicular by P, the base by b, and diedral angle by A, we have Formula (3), Art. 37,

the Trig.,

Pb tan A ;

but is the apothem of one face; if, therefore, we denote the number of sides in that face by n, and the length of each side by s, we shall have (Art. 101, Mens.),

The volumes of all

hence, the volume may be computed.

the regular polyedrons have been computed on the supposition that their edges are each equal to 1, and the results are given in the following

From the principles demonstrated in Book VII., we may write the following

RULE.

To find the volume of any regular polyedron, 'multiply the cube of its edge by the corresponding tabular volume ;. the product will be the volume required.