There are three cases;

1st, When the length of the edge is equal to the length of the back;

2d. When it is less; and

3d, When it is greater.

In the first case, the wedge is equal in volume to a right prism, whose base is the triangle ADG, and altitude GH or AB: hence, its volume is equal to ADG multiplied by AB.

In the second case, through H, a point of the edge, pass a plane HCB perpendicular to the back, and intersecting it in the line BC parallel to AD. This plane will divide the wedge into two parts, one of which is represented by the figure.

N

P

M

G

Through G, draw the plane GNM parallel to HCB, and it will divide the part of the wedge represented by the figure into the right triangular prism GNM-B, and the quadrangular pyramid ADNM-G. Draw GP perpendicular to NM: it will also be perpendicular to the back of the wedge (B. VI., P. XVII.), and hence, will be equal to the altitude of the wedge.

Denote AB by L, the breadth AD by b, the edge GH by 1, the altitude by h, and the volume by V; then,

We can find a similar expression for the remaining part of the wedge, and by adding, the factor within the parenthesis becomes the entire length of the edge plus twice the length of the back.

RULE. Add twice the length of the back to the length of the edge; multiply the sum by the breadth of the back, and that result by one sixth of the altitude; the final product will be the volume required.

Examples.

1. If the back of a wedge is 40 by 20 feet, the edge 35 feet, and the altitude 10 feet, what is the volume?

Ans. 3833.33 cu. ft.

2. What is the volume of a wedge, whose back is 18 feet by 9, edge 20 feet, and altitude 6 feet?

Ans. 504 cu. ft.

To find the volume of a prismoid.

122. A PRISMOID is a frustum of a wedge. Let L and B denote the length and breadth of the lower base, I and b the length and breadth of the upper base, M and m the length and breadth of the section equidistant from the bases, and h the altitude of the prismoid.

m

M

Through the edges L and l', let a plane be passed, and it will divide the prismoid into two wedges, having for bases the bases of the prismoid, and for edges the lines L and l'.

The volume of the prismoid, denoted by V, will be equal to the sum of the volumes of the two wedges; hence,

or,

V =

Bh (l + 2L) + fbh (L + 21);

which may be written under the form,

V =

th [(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, 4Mm = (L + 1) (B + b) = BL + Bl + bL + bl.

Substituting in (A), we have

V = th (BL + bl + 4Mm).

1

But BL is the area of the lower base, or lower section, bl 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 ex tensive 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 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 one sixth 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.

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

stick of hewn timber, and 24 inches by 18,

Ans. 102 cu. ft.

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 219).

To find the diedral angle contained between two consecutive faces of a regular polyedron.

125. As in the figure, let the vertex, O, of a polyedral angle of a

tetraedron be taken as the centre of a sphere whose radius is 1: then will the three faces of this polyedral angle, by their intersections with the surface of the sphere, determine the spherical

triangle FAB. The plane angles FOA, FOB, and AOB, being equal to each other, the arcs FA, FB, and AB, which measure these angles, are also equal to each other, and the spherical triangle FAB is equilateral. The angle FAB of the triangle is equal to the diedral angle of the planes FOA and AOB, that is, to the diedral angle between the faces of the tetraedron.

In like manner, if the vertex of a polyedral angle of any one of the regular polyedrons be taken as the centre of a sphere whose radius is 1, the faces of this polyedral angle will, by their intersections with the surface of the sphere, determine a regular spherical polygon; the number of sides of this spherical polygon will be equal to the