G' L M H' H D A B I к: projected on a horizontal plane. Also, let E'A'B'F', G'C'D'H represent the vertical sections at the extremities. If we suppose ver D: tical planes to pass through the lines AC, BD, the middle part of the excavation, or that contained between these vertical planes, will be a rectangular prismoid, of which A'B'KI E I's will be one base, and C'D'ML the other base. Its solidity will therefore. be given by Art. 114. The parts upon each side of the middle prismoid are also halves of rectangular prismoids; or, if the two parts are equal, they may be regarded as constituting a second prismoid, one of whose bases is the sum of the triangles A'E'I, B'F/K ; and the other base is the sum of the triangles C'G'L, D'H'M. Therefore the volume of the entire solid is equal to the product of one sixth of its length, by the sum of the areas of the sections at the two extremities, and four times the area of a parallel and equidistant section. Ex. 1. Let ABCDEFG represent the profile of a tract of B' land selected for the line of a rail-way; and suppose it is required, by cutting and embankment, to reduce it from its present hilly surface to one uniform slope from the point A to the point G. The distance AH is 561 feet; the distance DK is 820 feet; HI is 858 feet; KL is 825 feet; ID is 825 feet; LG is 330 feet. The perpendicular BH is 18 feet; the perpendicular KE is 19 feet; CI is 20 feet; " LF is 8 feet. The annexed figure repre- c d sents a cross section, showing the form of the excavation. The base of the cutting is to 66 e a be 50 feet wide, the slope 11 horizontal to 1 perpendicular; that is, where the depth ad is 10 feet, the width of the slope cd at the surface will be 15 feet. Calculation of the portion ABH. Since BH is 18 feet, the length of cd in the cross section will be 27 feet, and cf, the breadth at the top of the section, will be 104 feet. We accordingly find, by Art. 87, the area of the trapezoid forming the cross section at BH equal to 104+50 -X18=1386 foet. 2 For the middle section, the height is 5 feet, cd is 13.5 feet, and cf is 77 feet. The area of the cross section is therefore equal to 77 +50 x9=571.5. 2 561 6. Calculation of the portion BCIH. Since CI is 20 feet, the length of cd is 30 feet, and cf is 110 feet. The area of the section at CI is therefore equal to 110+50 X 20=1600. 2 For the middle section, the height is 19 feet, cd is 28.E foet, and cf is 107 feet. The area of the cross section is therefore equal to 107 +50 -X19=1491.5. 2 858 6 Calculation of the portion CID. The height of the middle section is 10 feet; therefore cf is 80 feet, and the area of the cross section is 80+50 X10=650. 2 825 6 21388.9 cubic yards. ABH=12716.0 cubic yards. CDI=21388.9 The following is a cross section, showing the form of the embankment. The top of the embankment 6 ъ is to be 50 feet wide, the slope 2 to 1; that is, where the height ad is 10 feet, the base cd is to C be 20 feet. d e Calculation of the portion DKE. Since EK is 19 feet, the length of cd is 38 feet, and cf is 126 feet. The area of the cross section at EK is therefore equal to 126+-50 x19=1672. 2 For the middle section, the height is 9.5 feet; cd is therefore 19 feet, and cf is 88 feet. The area of the cross section is therefore 88 +50 X9.5=655.5 2 820 6 Calculation of the portion KEFL. Since LF is 8 feet, cd is 16 feet, and cf is 82 feet. The area of the section at LF is therefore equal to 82+50 X8=528. 2 The height of the middle section is 13.5 feer; therefore cd is 27 feet, and cf is 104 feet. The area of the cross section is therefore equal to 104+50 -x13.5=1039.5. 2 825 6 Calculation of the portion LFG. The height of the middle section is 4 feet; therefore cf is 66 feet, and the area of the cross section is equal to 66 +50 x4=232. 2 330 6 2965.9 cubic yards. DKE=21735.1 cubic yards. LFG= 2965.9 Total embankment, 57079.7 Ex. 2. Compute the amount of excavation of the hill ABCD from the following data : The distance AH is 325 feet; the perpendicular BH is 12 feet; HI is 672 feet; CI is 13 feet. ID is 534 feet. The base of the cutting to be 50 feet wide, and the slope 11 horizontal to 1 perpendicular. Ans., 33969 cubic yards. PROBLEM X. (116.) To find the surface of a regular polyedron. RULE. Multiply the area of one of the faces by the number of F faces; or, Multiply the square of one of the edges by the surface of a similar solid whose edge is unity. Since all the faces of a regular polyedron are equal, it is evident that the area of one of them, multiplied by their number, will give the entire surface. Also, regular solids of the same name are similar, and similar polygons are as the squares of their homologous sides (Geom., Prop. 26, B. IV.). The following table shows the surface and solidity of regular polyedrons whose edge is unity. The surface is obtained by multiplying the area of one of the faces, as given in Art. 92, by the number of faces. Thus the area of an equilateral triangle, whose side is 1, is 0.4330127. Hence the surface of a regular tetraedron =.4330127x4=1.7320508, and so on for the other solids. Names. A Table of the regular Polyedrons whose Edges are unity No. of Faces. Solidity. 8 3.4641016 0.4714045. 20 8.6602540 2.1816950. Ex. 1. What is the surface of a regular octaedron whose edges are each 8 feet? Ans., 221.7025 feet. Ex. 2. What is the surface of a regular dodecaedron whose edge is 12 feet? Ans., 2972.985 feet. PROBLEM XI. (117.) To find the solidity of a regular polyedron. RULE. Multiply the surface by one third of the perpendicular let fall from the center on one of the faces; or, Multiply the cube of one of the edges by the solidity of a similar polyedron, whose edge is unity. Since the faces of a regular polyedron are similar and equal, |