In the third case, is greater than L, and denotes the altitude of the prism; the volume of each part is equal to the difference of the prism and pyramid, and is of the same form as before. Hence, the following RULE.-Add twice the length of the back to the length of the edge; multiply the sum by the breadth of the back, und 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? 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 Through the edges L and l', let a plane be passed, and it will m M B 504 cu. ft. 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, V = {Bh(l + 2L) + jbh(L + 21); which may be written under the form, V = дh[(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 + BI + bĽ + bl. But BL is the area of the lower base, or lower section, 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 sec tion; 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 excrcise 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. 2. What is the volume of a stick whose ends are 30 inches by 27, and 24 length being 24 feet ? Ans. 3700 cu. ft. of hewn timber, inches by 18, its 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 208). To find the diedral angle between the faces of a regular polyedron. will this polyedral 125. Let the vertex of any polyedral angle be taken 38 the centre of a sphere whose radius is 1: then sphere, by its intersections with the faces of the 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. E 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 4Q be equal to BQ. If we denote the number of sides by n AQ the angle APQ will be equal to 360° P B 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 14, To find the volume of a regular polyedron. 126. If planes be passed through the centre of the poly edron 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 poly. edron, 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. |