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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.).

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If the length of the base is greater than that of the edge, tet 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.

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The solidity of the prism bhl, the solidity of the pyramid -bh(L-l); and their sum, bhl+1⁄2bh(L—l)=‡bh3l+÷bh2L ―bh2l=bh(2L+I).

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=÷bh(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.

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Let I. 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 I let a plane be passed, and it will divide the prismoid into two wedges,

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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(2l+L)=÷h(2BL+B1+2b1+bL).

But since M is equally distant from L and I, 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 what ever. 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 '). To this add the areas of the two bases, when the extin su. face 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 101⁄2 feet, and the circumference of its base 9 feet. Ans. 22.56.

PROBLEM XIV.

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 one. third 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 diamAns. 197111024 sq. miles.

eter being 7921 miles. 4. What is the surface of a sphere, the circumference of its great circle being 78.54? Ans. 1963.5.

PROBLEM XVI.

To find the surface of a spherical zone.

RULE.-Multiply the altitude of the zone by the circumference of a great circle of the sphere, and the product will be the surface (Book VIII. Prop. X. Sch. 1).

1. The diameter of a sphere being 42 inches, what is the convex surface of a zone whose altitude is 9 inches?

Ans. 1187.5248 sq. in. 2. If the diameter of a sphere is 12 feet, what will be the surface of a zone whose altitude is 2 feet?

PROBLEM XVII.

Ans. 78.54 sq. ft.

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RULE I.-Multiply the surface by one-third of the radius (Book VIII. Prop. XIV.).

RULE II.-Cube the diameter, and multiply the number thus found by : that is, by 0.5236 (Book VIII. Prop. XIV. Sch. 3).

1. What is the solidity of a sphere whose diameter is 12?

Ans. 904.7808.

2. What is the solidity of the earth, if the mean diameter be taken equal to 7918.7 miles? Ans. 259992792083.

PROBLEM XVIII.

To find the solidity of a spherical segment.

RULE.-Find the areas of the two bases, and multiply their sum by half the height of the segment; to this product add the solidity of a sphere whose diameter is equal to the height of the segment (Book VIII. Prop. XVII.).

REMARK.-When the segment has but one base, the other is to be considered equal to 0 (Book VIII. Def. 14).

1. What is the solidity of a spherical seginent, the diameter of the sphere being 40, and the distances from the centre to the bases, 16 and 10. Ans. 4297.7088.

2. What is the solidity of a spherical segment with one base the diameter of the sphere being 8, and the altitude of the segment 2 feet? Ans. 41.888.

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