Ex. 2. Required the contents of a tetrahedron, the side of which is 6. Ans. 25.452.

Centre of Gravity. Is in the common centre of the centres of gravity of the triangles made by a section through the centre of each side of the figures.

Hexahedron.*

To ascertain the Contents of a Hexahedron (Cube), see Fig. 72, and Rule, p. 155.

Octahedron.

To ascertain the Contents of an Octahedron, Fig. 84.

RULE.-Multiply

of the cube of the linear side by the square root of 2 (1.414213), and the product will be the contents required.

EXAMPLE.-What are the contents of the octahedron a b c d, Fig. 84, the linear side of which is 4?

Ex. 2. Required the contents of an octahedron, the side of which is 8?

Ans. 241.3226.

Centre of Gravity. Is in its geometrical centre.

* A hexahedron and a cube are identical figures, being solids having the same number of similar and equal plane faces.

Dodecahedron.

To ascertain the Contents of a Dodecahedron, Fig. 85. RULE. TO 21 times the square root of 5 add 47, and divide the sum by 40; then, the square root of the quotient being multiplied by 5 times the cube of the linear side, will give the contents required.

Or, √√√5x21+47×73×5=S.

40

EXAMPLE. The linear side of the dodecahedron a b c d e is 3; what are its contents?

Ex. 2. The linear side of a dodecahedron is 1; what is the capacity of it?

Ans. 7.6631.

Centre of Gravity. Is in its geometrical centre.

Icosahedron.

To ascertain the Contents of an Icosahedron, Fig. 86.

RULE. TO 3 times the square root of 5 add 7, and divide the sum by 2; then, the square root of this quotient being multiplied by of the cube of the linear side, will give the contents required.

Or,

√/5×3+7 13×5

S.

6

EXAMPLE.-The linear side of the icosahedron a b c d e f

Ex. 2. Required the contents of an icosahedron, the linear side of which is 1?

Ans. 2.1817.

Centre of Gravity. Is in its geometrical centre.

REGULAR BODIES.

To ascertain the Contents of any regular Solid Body.

When the Linear Edge is given.

RULE.-Multiply the cube of the linear edge by the multiplier in column A in the table on the following page, and the product will be the contents required.

EXAMPLE. What is the capacity of a hexahedron having sides of 3 inches?

33×1=27=tabular volume multiplied by cube of edge=contents required. When the radius of the Circumscribing Sphere is given.

RULE.-Cube the radius of the circumscribing sphere, and multiply it by the multiplier opposite to the figure in column B.

EXAMPLE. The radius of the circumscribing sphere of a hexahedron is 1.732 inches; what is the volume of it?

1.7323 × 1.5396=product of cube of radius and tabular multiplier=8= result required.

When the radius of the Inscribed Sphere is given.

RULE.-Cube the radius; and multiply it by the multiplier opposite to the figure in column C.

EXAMPLE.—The radius of the inscribed sphere of a hexahedron is 1 inch; what is its volume?

13×8=product of cube of radius and tabular multiplier=8=result re

5.05406 138 11 23

20 Icosahedron 2.18169 2.53615

NOTE. For further rules to ascertain the elements of polyhedrons, see Appendix, p. 262.

Centre of Gravity. Is in their geometrical centre.

Cylinder.

Definition. A figure formed by the revolution of a right-angled parallelogram around one of its sides.

To ascertain the Contents of a Cylinder, Fig. 87. RULE.-Multiply the area of the base by the height, and the product will give the contents required.

EXAMPLE.-The diameter of a cylinder, b c, is 3 feet, and its length, a b, 7 feet; what are its contents?

Area of 3 feet 7.068.

Then, 7.068×7=49.476=result required.

Ex. 2. What are the contents of a cylinder, the height of which is 5 feet, and the diameter 2 feet?

Ans. 15.708 feet.

Ex. 3. The circumference of the base of a cylindrical column is 20.42 feet, and the height of the column is 9.695 feet; what is its volume? Ans. 320.515 feet.

Centre of Gravity. Is in its geometrical centre.

Cone.

Definition. A figure described by the revolution of a rightangled triangle about one of its legs, which remains fixed.

To ascertain the Contents of a Cone, Fig. 88.

RULE.-Multiply the area of the base by the perpendicular height, and one third of the product will be the contents required.

EXAMPLE.-The diameter, a b, of the base of a cone is 15 inches, and the perpendicular height, c d, 32.5 inches; what are the contents of the cone?