11. The ends of the frustum of a pyramid are equilateral triangles, the lengths of the sides being 6 feet and 7 feet respectively; the height of the frustum is 4 feet: find the volume.

12. The ends of the frustum of a pyramid are squares, the lengths of the sides being 20 feet and 30 feet respectively; the length of the straight line which joins the middle point of any side of one end with the middle point of the corresponding side of the other end is 13 feet: find the volume.

13. The shaft of Pompey's pillar, which is situated near Alexandria in Egypt, is a single stone of granite; the height is 90 feet, the diameter at one end is 9 feet, and at the other end 7 feet 6 inches: find the volume.

14. The mast of a ship is 50 feet high; the circumference at one end is 60 inches, and at the other 36: find the number of cubic feet of wood.

15. The radii of the ends of a frustum of a right circular cone are 7 feet and 8 feet respectively; and the height is 3 feet: find the volume of the cone from which the frustum was obtained.

16. The radii of the ends of the frustum of a right circular cone are 7 feet and 8 feet respectively; and the height is 3 feet: find the volumes of the two pieces obtained by cutting the frustum by a plane parallel to the ends and midway between them.

17. The radii of the ends of a frustum of a right circular cone are 7 feet and 8 feet respectively; and the height is 3 feet: the frustum is cut into three, each one foot in height, by planes parallel to the ends: find the number of cubic inches in each of the pieces.

18. The ends of the frustum of a pyramid are regular hexagons, the lengths of the sides being 8 feet and 10 feet respectively; the height of the frustum is 6 feet: find the volumes of the two pieces obtained by cutting the frustum by a plane parallel to the ends and midway between them.

XXVII. WEDGE.

272. To find the Volume of a Wedge.

RULE. Add the length of the edge to twice the length of the base; multiply the sum by the width of the base, and the product by the height of the wedge; one-sixth of the result will be the volume of the wedge.

273. Examples.

(1) The edge of a wedge is 12 inches; the length of the base is 16 inches, and the breadth 7 inches; the height of the wedge is 24 inches.

12+16+16=44;

1

× 44 × 7 × 24=1232.

6

Thus the volume of the wedge is 1232 cubic inches.

(2) The edge of a wedge is 5

inches; the length of the base is 3 inches, and the breadth 2 inches; the height of the wedge is 4 inches.

Thus the volume of the wedge is 15 cubic inches.

274. If the edge of a wedge be equal to the length of the base, the wedge is a triangular prism; so that we are thus furnished with another Rule for finding the volume of such a prism. This Rule is in general not identical with that given in Art. 246, because the dimensions which are supposed to be known are not the same in the two cases. If the prism be a right prism, it is easy to see that the two Rules are really identical.

275. If the edge of a wedge be shorter than the length of the base, the wedge can be divided into an oblique triangular prism and a pyramid on a rectangular base, by drawing a plane through one end of the edge parallel to the triangular face at the other end. And, in like manner, if the edge be shorter than the length of the base, the wedge is equal to the excess of a certain triangular prism over a certain pyramid on a rectangular base. It is by this consideration that the Rule of Art. 272 can be shewn to be true.

276. It has been proposed to extend the meaning of the word wedge to the case in which the base instead of being a rectangle is a parallelogram or a trapezoid. The Rule for finding the volume will still hold, provided we understand the length of the base to be half the sum of the parallel sides, and the breadth of the base to be the perpendicular distance between the parallel sides.

277. Suppose a solid has been obtained by cutting a right triangular prism by a plane, inclined to the length of the prism, which does not meet the base D of the prism.

The volume of this solid may be found by the following Rule:

Multiply the area of the base of the prism by one-third of the sum of the parallel edges of the solid.

A

B

278. The solid considered in the preceding Article is a wedge in the enlarged meaning of the word noticed in Art. 276: the Rule for finding the volume may be demonstrated by a method similar to that explained in Art. 275. For if through E, the upper end of the shortest of the three edges, we draw a plane parallel to the base ABC, we divide the solid into a right prism and a pyramid; the

volumes of each of these can be obtained by known Rules, and it will be found that the sum of the two volumes will agree with that assigned by the Rule of Art. 277.

279. Suppose a solid has been obtained by cutting a right prism having a parallelogram for base by a plane, inclined to the length of the prism, which does not meet the base of the prism. The volume of this solid may be found by the following Rule:

Multiply the area of the base of the prism by one-fourth of the sum of the four parallel edges of the solid.

H

D

B

F

280. It is easy to see that AE+CG=BF+DH; for each of these is equal to twice the distance between the point of intersection of AC and BD and the point of intersection of EG and FH; and thus the Rule of the preceding Article might be given in this form :

Multiply the area of the base of the prism by half the sum of two opposite edges out of the four parallel edges.

281. The Rule of Art. 280 follows from that of Art. 277. For if we cut the solid into two pieces by a plane which passes through AE and CG, the Rule of Art. 277 will determine the volume of each piece; and it will be found that the sum of the two volumes will agree with that assigned by the Rule of Art. 280.

282. We will now solve some exercises.

(1) The edge of a wedge is 18 inches; the length of the base is 20 inches; the area of a section of the wedge made by a plane perpendicular to the edge is 150 square inches: find the volume.

The section made by a plane perpendicular to the edge is a triangle; therefore the product of the base of this

T. M.

11

triangle into its height is 2 × 150, that is, 300: this product is the same as that of the breadth of the wedge into the height of the wedge.

Thus the volume is 2900 cubic inches.

The result can be obtained more readily by the Rule of Art. 277.

(2) The edge of a wedge is 16 inches; the length of the base is 24 inches, and the breadth 6 inches; and the height of the wedge is 10 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.

The length of the base of the pyramid is 24-16 inches, that is, 8 inches; hence, by Art. 263, the volume of the pyramid in cubic inches =

=

1

3

×8×6×10=160.

The prism has three parallel edges, each 16 inches long; and, by Art. 274, its volume in cubic inches

1. The edge of a wedge is 2 feet 3 inches; the length of the base 2 feet 3 inches, and the breadth 8 inches; the height of the wedge is 15 inches: find the volume.

2. The edge of a wedge is 9 feet; the length of the base is 6 feet, and the breadth is 3 feet; the height of the wedge is 2 feet: find the volume.