PROBLEM II.

To find the solidity of a parallelopipedon.

RULE. Multiply continually the length, breadth, and depth together for the solidity.*

1. What is the solidity of the parallelopipedon ABCDEFG, the length A B being 10 feet, the breadth A G 4 feet, and thickness AD 5 feet?

ABXAG XAD 10 x 4 x 5 = 200 feet.

2. A piece of timber is 26 feet long, 10 inches broad, and

8 inches deep; required its solid content. Ans. 14 feet.

3. A piece of timber is 10 inches square at the ends and 40 feet long; required its content. Ans. 277 feet.

4. A piece of timber 15 inches square at each end, and 18 feet long, is to be measured, required its content, and how far from the end must it be cut across, so that the piece cnt off may contain 1 solid foot.

Ans. The solidity is 28-125 feet; and 7.68 in length will make one foot.

5. What length of a piece of square timber will make one solid foot, being 2 feet 9 inches deep, and 1 foot 7 inches. broad?

Ans. 2-756 inches in length will make one solid foot.

PROBLEM IIL

To find the solidity of a prism.

RULE. Multiply the area of the base by the perpendicular height, and the product will be the solidity.*

1. What is the solidity of a prism, A B C FIE, whose

base C A is a pentagon, each side of which being 375, and height 15 feet?

When the side of a pentagon is 1, its area is 1.720477, (Table II.); therefore 1-720477 x 3.752 = 24.1942 = the area of the base in square feet; hence 24.1942 × 15 = 362.913 solid feet, the content.

2. What is the solidity of a square prism, whose length is 5 feet, and each side of its base 1 foot?

Ans. 97 solid feet. 3. What is the solidity of a prism, whose base is an equilateral triangle, each side being 4 feet, and height 10 feet? Ans. 69-282 feet. 4. What quantity of water will a prismatic vessel contain, its base being a square, each side of which is 3 feet, and height 7 feet? Ans. 63 feet.

PROBLEM IV.

To find the solidity of a cylinder

RULE. Multiply the area of the base by its height, and the product will be the solid content.*

1. What is the capacity of a right cylinder ABGC, whose height, and the circumference of its base, are each 20 feet?

First

20

3-1416

the diameter, half of which multiplied by

half the circumference will give the area of the base, (Prob.

2. What is the content of the oblique cylinder ABFE, the circumference of whose base is 20 feet, and altitude A C 20 feet?

25

25

; then

.7854

•7854

As before, the area of the base is X 20636-61828, the solid content, as before. 3. The length of a cylindrical piece of timber is 18 feet, and its circumference 96 inches; how many solid feet in it? Ans. 91-676 feet.

4. Three cubic feet are to be cut off a rolling stone 44 inches in circumference; what distance from the end must the section be made? Ans. 33.64 inches.

See Appendix, Demonstration 64.

PROBLEM V.

To find the content of a solid formed by a plane passing parallel to the axis of a cylinder.

RULE. Find by Prob. XXVIII. Sec. II. the area of the base, which, multiplied by the height, will give the solidity.*

1. In the cylinder A B G C, whose diameter is 3, and height 20 feet; let a plane L N pass parallel to the axis, and 1 foot from it; what is the solidity of each of the two prisms into which the cylinder is divided. (See the last figure.)

SC

3

&C = &−1) ÷ 3 =

GC

===·1663 the tabular versed

sine, to which, in the Table of Circular Segments, corres

Then 329 which X 08604117 = 7·7437053 = seg. DCN.

Also 9 × 69935699 = 6·29421291 = seg, D G N.

Hence 20 X 7.7437053 = 15.4874 the slice LKACND; and 20 × 6·29421699 = 125·88434 = the slice L KB GND.

2. Suppose the right cylinder, whose length is 20 feet, and diameter 50 feet, is cut by a plane parallel to, and at the distance of, 21-75 feet from its axis; required the solidity of the smaller slice. Ans. 1082.95 feet.

PROBLEM VI.

To find the solidity of a pyramid.

RULE. Multiply the area of the base by the one-third of the height, and the product will be the solidity.†

* See Appendix, Demonstration 64.
+ See Appendix, Demonstration 65.

1. What is the solidity of a square pyramid, each side of its base being 4 feet, and height 12 feet?

4 x 4 16 the area of the base:

Then 16 x 64 feet, the solidity.

2. Each side of the base of a triangular pyramid is 3, and height 30; required its solidity. Ans. 38.97117.

3. The spire of a church is an octagonal pyramid, each side at the base being 5 feet 10 inches, and its perpendicular height 45 feet; also each side of the cavity, or hollow part, at the base is 4 feet 11 inches, and its perpendicular height 41 feet; it is required to know how many solid yards of stone the spire contains.

Ans. 32-19738 yards.

4. The height of a hexagonal pyramid is 45 feet, each side of the hexagon at the base being 10; required its solidity. Ans. 3897 1143.

PROBLEM VII.

To find the solidity of a cone.

RULE. Multiply the area of the base by one-third of the height, and the product will be the solidity.*

1. The diameter of the base of a cone is 10 feet, and its perpendicular height 42 feet; what is its solidity?

102 = 100 × 7854 = 78.54; then 78.54 × 42 = 1099-56 feet.

2. The diameter of the base of a cone is 12 feet, and its perpendicular height 100; required its solidity. Ans. 3769.92 feet. 3. The spire of a church of a conical form measures 37.6992 feet round its base; what is its solidity, its perpendicular height being 100 feet? Ans. 3769.92.

4. How many cubic yards in an upright cone, the circum ference of the base being 70 feet, and the slant height 30? Ans. 134.09.

See Appendix, Demonstration 66.