3. The base of a wedge is a square, a side of which is 15 inches; the edge is 24 inches, and the height of the wedge is 24 inches: find the volume.

4. The base of a prism is an equilateral triangle, each side of which is 4 inches: find the volume of the solid obtained by cutting off a piece of this prism, so that the sum of the three parallel edges is 15 inches.

5. The base of a prism is a rectangle which measures 7 inches by 8; find the volume of the solid obtained by cutting off a piece of this prism, so that the sum of the four parallel edges is 42 inches.

6. The edge of a wedge is 21 inches; the length of the base is 27 inches; the area of a section of the wedge made by a plane perpendicular to the edge is 160 square inches: find the volume.

7. The edge of a wedge is 25 inches; the length of the base is 22 inches; a section of the wedge made by a plane perpendicular to the edge is an equilateral triangle, each side of which is 10 inches: find the volume.

8. The edge of a wedge is 15 inches; the length of the base is 24 inches, and the breadth 7 inches; the height of the wedge is 22 inches; the wedge is divided into a pyramid and a prism by a plane through one end of the edge parallel to the triangular face at the other end; find the volume of each part.

9. The edge of a wedge is 2 feet 3 inches; the length of the base is 2 feet 9 inches, and the breadth 8 inches; the height of the wedge is 14 inches; the wedge is divided into two pieces by a plane which passes through a point in the edge, distant 18 inches from one end, and which is parallel to the triangular face at that end: find the volume of each piece.

10. The edge of a wedge is 36 inches; the length of the base is 27 inches, and the breadth 5 inches; the height of the wedge is 12 inches. The wedge is divided by a plane, so that the sum of the three parallel edges in one part is 42 inches: find the volume of each part.

XXVIII. PRISMOID.

283. To find the volume of a Prismoid.

RULE. Add together the areas of the two ends and four times the area of a section parallel to the two ends and midway between them; multiply the sum by the height, and one-sixth of the product will be the volume.

284. Examples.

(1) The area of one end is 4 square feet, of the other end 9 square feet, and of the middle section 6 square feet; the height is 2 feet.

4+24+9=37, x37x2=37-12.

Thus the volume is 123 cubic feet.

(2) The area of one end is 224 square inches, of the other end 216 square inches, and of the middle section 221 square inches: the height is 18 inches.

224+884+216=1324,

1

6

× 1324 × 18=3972.

Thus the volume is 3972 cubic inches.

285. The demonstration of the Rule in Art. 283 depends on the fact that a prismoid can be divided into pyramids and wedges, some having their bases in one end of the prismoid, and some in the other, and all having the same height as the prismoid.

286. Each side of the middle section is equal to half the sum of the corresponding sides of the ends. Thus if the ends are rectangles of known dimensions, the area of the middle section can be easily found: for then four times the area of the middle section is equal to the area

of a rectangle, having for each of its dimensions the sum of the corresponding dimensions of the ends.

Each angle of the middle section is equal to the corresponding angle at the ends.

287. If the ends of a prismoid are similar figures similarly situated, the prismoid is a frustum of a pyramid, and therefore the volume might be found by the Rule of Art. 268. By comparing the two Rules, we infer that in this case, four times the area of the middle section is equal to the sum of the areas of the ends added to twice the square root of the product of these areas.

288. It has been proposed to extend the meaning of the term prismoid so as to apply to cases in which the ends are not rectilineal figures. Accordingly, the following definition may be given: A prismoid has for its ends any two parallel plane figures, and has its other boundary straight. By having the other boundary straight is meant that a straight line may be placed on the boundary at any point, so as to coincide with the surface from end to end. This definition will include a frustum of a cone, or either of the pieces obtained by cutting a frustum of a cone by a plane which meets both ends.

The Rule in Art. 283 holds for this extended meaning of the term prismoid.

289. The Rule for finding the volume of a prismoid holds for many other solids; but it would not be possible to define these solids in an elementary manner: the advanced student may consult the author's Integral Calculus, 3rd edition, Art. 192.

290. We will now solve some exercises.

(1) The ends of a prismoid are trapezoids with four parallel edges; the parallel sides of one end are 100 feet and 32 feet respectively, and the distance between them is 28 feet; the corresponding dimensions of the other end are 80 feet, 30 feet, and 26 feet; the distance between the ends is 112 feet: find the volume.

The middle section is a trapezoid; one of the two parallel sides is half of 100+80 feet, that is, 90 feet; and the other is half of 32+30 feet, that is, 31 feet: the distance between these two parallel sides is half the sum of the corresponding distances for the two ends, that is, half of 28+26 feet, that is, 27 feet.

The area of one end in square feet=

132

x 28=1848;

2

the area of the other end in square feet=

2

Thus the volume is 183157 cubic feet.

(2) The edge of a wedge is 21 inches; the length of the base is 15 inches, and the breadth 9 inches; the height of the wedge is 6 inches; the wedge is divided into three parts of equal heights by planes parallel to the base: find the volume of each part.

The parts are two prismoids and a wedge; the height of each part is 2 inches.

The first prismoid has one end a rectangle which measures 15 inches by 9; it will be found that the other end is a rectangle which measures in the corresponding manner 17 inches by 6. The volume, by Art. 283, is 239 cubic inches.

The second prismoid has one end a rectangle which measures 17 inches by 6; it will be found that the other end is a rectangle which measures in the corresponding manner 19 inches by 3. The volume, by Art. 283, is

161 cubic inches.

The edge of the wedge is 21 inches; the length of the base is 19 inches, and the breadth 3 inches. The volume, by Art. 272, is 59 cubic inches.

The sum of the three volumes in cubic inches is 239 +161+59, that is, 459: it will be found that this is equal to the volume of the original wedge, as of course it should be.

EXAMPLES. XXVIII.

1. Find the number of cubic feet which must be removed to form a prismoidal cavity; the depth is 12 feet, and the top and the bottom are rectangles, the corresponding dimensions of which are 400 feet by 180, and 350 feet by 150.

2. Find the number of cubic feet which must be removed to form a prismoidal cavity; the depth is 12 feet, and the top and the bottom are rectangles, the corresponding dimensions of which are 400 feet by 180, and 150 feet by 350.

3. Find the volume of a coal waggon the depth of which is 47 inches; the top and the bottom are rectangles, the corresponding dimensions of which are 81 inches by 54, and 42 inches by 30.

4. Find the number of gallons of water required to fill a canal the depth of which is 4 feet, and the top and the bottom of which are rectangles, the corresponding dimensions of which are 250 feet by 16, and 240 feet by 14.

5. Find the number of cubic feet which must be removed to form a railway cutting in the form of a prismoidal cavity; the ends are trapezoids with four parallel edges; the parallel sides of one end are 144 feet and 36 feet, and the distance between them 36 feet; the corresponding dimensions of the other end are 108 feet, 36 feet, and 24 feet: the distance between the ends is 137 feet.

6. The ends of a prismoid are rectangles, the corresponding dimensions of which are 12 feet by 10, and 8 feet