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Divide AE into equal parts, such that each shall be less than OI ; there will be at least one point of division, m, between 0 and I.
Let P be the parallelopipedon having for base ABCD, and for altitude Am ; since the altitudes, AE, Am, are to each other as two whole numbers, we shall have
sol. AG :P :: AE : Am. But, by hypothesis, sol. AG : sol. AL :: AE : A0; hence,
sol. AL : P :: AO : Am. But AO is greater than Am; hence, for the proportion to be correct, the solid AL must be greater than P. Now, on the contrary, it is less; hence, it is impossible that the fourth term of the proportion,
sol. AG : sol. AL :: AE : X, can be a greater line than AI. By like reasoning it may be shown that the fourth term cannot be less than AI; hence it is equal to AI; therefore rectangular parallelopipedons having the same base, are to each other as their altitudes.
Two rectangular parallelopipedons, AG, AK, having the same altitude, AE, are to each other as their bases, ABCD, AMNO.
Having placed the two solids by the side of each other, as the figa ure represents, produce the plane ONKL until it meets the plane DCGH in PQ; we will thus have a third parallelopipedon, AQ, which may be compared with each of the parallelopipedons, AG, AK. The two solids, AG, AQ, having the same base, AEHD, are to each other as their altitudes, AB, AO ; in like manner the two solids, AQ, AK, having the same a base, AOLE, are to each other as their altitudes, AD, AM. Thus we have the two proportions
sol. AG : sol. AQ : : AB : AO,
Multiplying together the corresponding terms of these two proportions, and omitting in the result the common multiplier, Sol. AQ, we have
sol. AG : sol. AK : : AB X AD : AO X AM, But AB x AD represents the base ABCD, and AO ~ AM represents the base AMNO; hence, two rectangular parallelopipedons having the same altitude are to each other as their bases.
Any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes, or as the products of their three dimensions.
For, having placed the two solids, AG, AZ, so that their surfaces have the common angle BAE, produce the planes necessary to complete the third parallelopipedon, AK, having the same altitude as the parallelopipedon AG. By the last proportion we shall have
sol. AK : sol. AZ : : AE : AX. Multiplying together the corresponding terms of these two proportions, and omitting in the result the common multiplier, Sol. AK, we obtain
sol. AG : sol. AZ : : ABCD X AE : AMNO X AX.
In the place of the bases, ABCD, AMNO, put AB X AD and AO X AM ; it will give
sol. AG : sol. AZ : : AB X AD X AE : AO X AM X AX.
Hence, any two rectangular parallelopipedons are to each other, etc.
The volume of a rectangular parallelopipedon, P, is equal to the product of its base, a xb, by its height, c, or equal to the product of its three dimensions, a X 6 X c.
For, comparing the parallelopipedon P with the
Q Now, if we suppose the cube. Q to be the unit 1 of measure of the parallelopipedon, and n the
Р linear unit on which it is constructed, then the ratio
ē the measure of the volume of the parallelopipedon P, and the
a b c ratios
are its three dimensions considered as abstract numn' n' n bers. Hence, volume P = a × b × c.
n n n
SCHOLIUM I. Let a = 3 inches, b= 2 inches, c= 5 inches, and let the cubic inch be the unit of solid measure of volumes.
Р Then the ratio
= 3 X 2 X 5
= 30 cubic inches. In order, then, to comprehend the nature of this measurement, it is necessary to reflect that the product of the three dimensions of a parallelopipedon is a number which signifies nothing of itself, and would be different if a different linear unit had been assumed. But if the three dimensions of another parallelopipedon are valued according to the same linear unit, and multiplied together, the two products will be to each other as the solids, and will serve to express their relative magnitude.
SCHOLIUM 2. The three dimensions of a cube being equal to each other, if the side is 1, the volume will be 1 X 1 X1=1; if the side is 2, the volume will be 2 X 2 X 2 = 28 = 8; if the side is c, the volume will be cxcxc=c. Hence it is that in arithmetic the cube of a number is the name given to the third power of the number.
SCHOLIUM 3. The side, x, of a cube equivalent to a given volume, V, is equal to the cube root of V. For x = V; and hence x = ov. The problem of the duplication of the cube, famous among the ancient Greek Geometers, consists in determining the side of a cube which shall be double a given cube. Now, if x and c be the sides of the two cubes, then 4 = 20°, or x = c.
Cņ 2, that is, the side of the required cube would have to be to the side of the given cube as the cube root of 2 is to unity. Now, the square root of 2 is easily found by a geometrical construction. But the cube root of 2 cannot be so found, that is, by the simple operations of elementary Geometry, which employ no other lines than straight lines and circumferences. The problem admits of solution, however, by other methods no less rigorous than those of elementary Geometry.
The volume of a parallelopipedon, and generally of any prism, is equal to the product of its base by its altitude.
First.—Any parallelopipedon is equivalent to a rectangular parallelopipedon having the same altitude, and an equivalent base (Prop. XIII.). Now, the volume of the latter is equal to its base multiplied by its altitude ; hence the volume of the former is likewise equal to the product of its base by its altitude.
Secondly.—Any triangular prism is half the parallelopipedon so constructed as to have the same altitude and a base twice as great (Prop. X.). But the volume of the latter is equal to its base multiplied by its altitude; hence, that of a triangular prism is equal to the product of its base (half that of the parallelopipedon) multiplied by its altitude.
Thirdly. —Any prism may be divided into triangular prisms having for their bases the different triangles which form the polygon which serves as its base, and for their common altitude the altitude of the prism. But the volume of each triangular prism is equal to its base multiplied by its altitude ; and, since the altitude is the same for all, it follows that the sum of all the partial prisms will be equal to the sum of all the triangles which constitute their bases, multiplied by the common altitude.
Hence, the volume of any polygonal prism is equal to the product of its base by its altitude.
Cor. 1. The volume of any prism is equal to the product of its right section by its lateral edge (Prop. IX.).
Cor. 2. First.Any two prisms are equivalent when they have equal bases and equal altitudes, or when the products of their bases by their altitudes respectively are equal.
Secondly.—Two prisms of the same altitude are to each other as their bases.
Thirdly.—Two prisms of equivalent bases are to each other as their altitudes,
THEOREM. convex surface of a regular pyramid S-ABCDE is equal to the perimeter of the base, ABCDE, multiplied by half the slant height, SH. For, the convex surface, S, is composed of five
SH isosceles triangles, each equal to SAE = AE X
SH Hence, convex surface S 5AE X
2 (AB + BC + CD ) perimeter, ABCDE.
SH Therefore, S = perimeter, ABCDE X
SCHOLIUM. The area of the base, ABCDE, being equal to the product of its perimeter by half the apothem, OH (Book IV., Prop. IX.), the whole surface of the pyramid is SH
OH perimeter ABCDE X
+ perimeter ABCDE X
SH + OH perimeter ABCDE X
The convex surface of the frustum, AD', of a regular pyramid, is equal to sum of the perimeters of its bases, ABCDE, abcde, multiplied by half the slant height, Hh, of the frustum.