PROBLEM VI.

To find the solidity of the frustum of a pyramid.

RULE.-Add together the areas of the two bases of the frustum and a mean proportional between them, and then multiply the sum by one-third of the altitude (Book VII. Prop. XVIII.).

1. To find the number of solid feet in a piece of timber, whose bases are squares, each side of the lower base being 15 inches, and each side of the upper base 6 inches, the altitude being 24 feet. Ans. 19.5.

2. Required the solidity of a pentagonal frustum, whose altitude is 5 feet, each side of the lower base 18 inches, and each side of the upper base 6 inches. Ans. 9.31925.

Definitions.

1. A wedge is a solid bounded by five planes: viz. a rectangle ABCD, called the base of the wedge; two trapezoids ABHG, DCHG, which are called the sides of the wedge, and which intersect D each other in the edge GH; and the two triangles GDA, HCB, which are called the ends of the wedge.

When AB, the length of the base, is equal to GH, the trapezoids ABHG, DCHG, become parallelograms, and the wedge is then one-half the parallelopipedon described on the base ABCD, and having the same altitude with the wedge.

The altitude of the wedge is the perpendicular let fall from any point of the line GH, on the base ABCD.

2. A rectangular prismoid is a solid resembling the frustum of a quadrangular pyramid. The upper and lower bases are rectangles, having their corresponding sides parallel, and the convex surface is made up of four trapezoids. The altitude of the prismoid is the perpendicular distance between its bases.

PROBLEM VII.

To find the solidity of a wedge.

RULE. To twice the length of the base add the length of the edge. Multiply this sum by the breadth of the base, and then by the altitude of the wedge, and take one-sixth of the product for the solidity.

Suppose AB, the length of the base, to be equal to GH, the length of the edge, the solidity will then be equal to half the parallelopipedon having the same base and the same altitude (Book VII. Prop. VII.). Hence, the solidity will be equal blh (Book VII. Prop. XIV.).

to

If the length of the base is greater than that of the edge, let a section MNG be made parallel to the end BCH. The wedge will then be divided into the triangular prism BCH-M, and the quadrangular pyramid G-AMND.

The solidity of the prism =bhl, the solidity of the pyramid -bh(L); and their sum, bhl+}bh(L—1)=}bh3l+}bh2L jbh2l=bh (2L+7).

If the length of the base is less than the length of the edge, the solidity of the wedge will be equal to the difference between the prism and pyramid, and we shall have for the solidity of the wedge,

bhl—bh

(l—L)=}bh3l—}bh2l+}bh2L=}bhḥ(2L+1).

1. If the base of a wedge is 40 by 20 feet, the edge 35 feet, and the altitude 10 feet, what is the solidity?

Ans. 3833.33. 2. The base of a wedge being 18 feet by 9, the edge 20 feet, and the altitude 6 feet, what is the solidity?

Ans. 504.

PROBLEM VIII.

To find the solidity of a rectangular prismoid.

RULE. Add together the areas of the two bases and four times the area of a parallel section at equal distances from the bases: then multiply the sum by one-sixth of the altitude.

B b

Let L and B be the length and breadth of the lower base, 7 and b the length and breadth of the upper base, M and m the length and breadth of the section equidistant from the bases, and h the altitude of the prismoid.

Through the diagonal edges L and ' let a plane be passed, and it will divide the prismoid into two wedges,

1

M

B

I

having for bases, the bases of the prismoid, and for edges the lines L and l'=l.

The solidity of these wedges, and consequently of the prismoid, is

Bh(2L+1)+bh(27+L)=h(2BL+Bl+2bl+bL).

But since M is equally distant from L and 7, we have 2M=L+1, and 2m-B+b; hence, 4Mm=(L+) x (B+b)=BL+Bl+bL+bl.

Substituting 4Mm for its value in the preceding equation, and we have for the solidity

th(BL+bl+4Mm).

REMARK.-This rule may be applied to any prismoid whatever. For, whatever be the form of the bases, there may be inscribed in each the same number of rectangles, and the number of these rectangles may be made so great that their sum in each base will differ from that base, by less than any assignable quantity. Now, if on these rectangles, rectangular prismoids be constructed, their sum will differ from the given prismoid by less than any assignable quantity. Hence the rule is general.

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 solidity. Ans. 3700.

2. What is the solidity of a stick of hewn timber, whose ends are 30 inches by 27, and 24 inches by 18, its length being 24 feet? Ans. 102 feet.

OF THE MEASURES OF THE THREE ROUND BODIES.

PROBLEM IX.

To find the surface of a cylinder.

RULE.-Multiply the circumference of the base by the altitude, and the product will be the convex surface (Book VIII. Prop. 1.). To this add the areas of the two bases, when the entire surface is required.

1. What is the convex surface of a cylinder, the diameter of whose base is 20, and whose altitude is 50?

Ans. 3141.6.

2. Required the entire surface of a cylinder, whose altitude is 20 feet, and the diameter of its base 2 feet.

Ans. 131.9472.

PROBLEM X.

To find the convex surface of a cone.

RULE.-Multiply the circumference of the base by half the side (Book VIII. Prop. III.): to which add the area of the base, when the entire surface is required.

1. Required the convex surface of a cone, whose side is 50 feet, and the diameter of its base 8 feet. Ans. 667.59. 2. Required the entire surface of a cone, whose side is 36 and the diameter of its base 18 feet. Ans. 1272.348.

PROBLEM XI.

To find the surface of the frustum of a cone.

RULE.-Multiply the side of the frustum by half the sum of the circumferences of the two bases, for the convex surface (Book VIII. Prop. IV.): to which add the areas of the two bases, when the entire surface is required.

1. To find the convex surface of the frustum of a cone, the side of the frustum being 12 feet, and the circumferences of the bases 8.4 feet and 6 feet. Ans. 90.

2. To find the entire surface of the frustum of a cone, the side being 16 feet, and the radii of the bases 3 feet and 2 feet. Ans. 292.1688.

PROBLEM XII.

To find the solidity of a cylinder.

RULE.-Multiply the area of the base by the altitude (Book VIII. Prop. II.).

1. Required the solidity of a cylinder whose altitude is 12 feet, and the diameter of its base 15 feet. Ans. 2120.58. 2. Required the solidity of a cylinder whose altitude is 20 feet, and the circumference of whose base is 5 feet 6 inches. Ans. 48.144.

PROBLEM XIII.

To find the solidity of a cone.

RULE.-Multiply the area of the base by the altitude, and take one-third of the product (Book VIII. Prop. V.).

1. Required the solidity of a cone whose altitude is 27 feet, and the diameter of the base 10 feet. Ans. 706.86.

2. Required the solidity of a cone whose altitude is 10 feet, and the circumference of its base 9 feet.

PROBLEM XIV.

Ans. 22.56.

To find the solidity of the frustum of a cone.

RULE.-Add together the areas of the two bases and a mean proportional between them, and then multiply the sum by onethird of the altitude (Book VIII. Prop. VI.).

1. To find the solidity of the frustum of a cone, the altitude being 18, the diameter of the lower base 8, and that of the upper base 4. Ans. 527.7888.

2. What is the solidity of the frustum of a cone, the altitude being 25, the circumference of the lower base 20, and that of the upper base 10? Ans. 464.216.

3. If a cask, which is composed of two equal conic frustums joined together at their larger bases, have its bung diameter 28 inches, the head diameter 20 inches, and the length 40 inches how many gallons of wine will it contain, there being 231 cubic inches in a gallon? Ans. 79.0613.

PROBLEM. XV.

To find the surface of a sphere.

RULE I.-Multiply the circumference of a great circle by the diameter (Book VIII. Prop. X.).

RULE II.-Multiply the square of the diameter, or four times the square of the radius, by 3.1416 (Book VIII. Prop. X. Cor.).

1. Required the surface of a sphere whose diameter is 7. Ans. 153.9384.

2. Required the surface of a sphere whose diameter is 24 inches. Ans. 1809.5616 in. 3. Required the area of the surface of the earth, its diameter being 7921 miles. Ans. 197111024 sq. miles.

4. What is the surface of a sphere, the circumference of its great circle being 78.54? Ans. 1963.5.