Elements of Algebra

CHAP. X. OF ALGEBRA. 161 493. Farther, if it be required to compare two rooms with respect to the space, or contents, we observe, that that relation is compounded of three relations; namely, that of the lengths, breadths, and heights. Let there be, for example, a room A, whose length is 36 feet, breadth 16 feet, and height 14 feet, and a room B, whose length is 42 feet, breadth 24 feet, and height 10 feet; we shall have these three relations: For the length $36:42$ For the breadth $16:24$ For the height $14:10$ And cancelling the common measures, these become $4:5$. So that the contents of the room A, is to the contents of the room B, as 4 to 5. 494. When the relations which we compound in this manner are equal, there result multiplicate relations. Namely, two equal relations give a duplicate ratio, or ratio of the squares; three equal relations produce the triplicate ratio, or ratio of the cubes; and so on. For example, the relations $a:b$ and $a:b$ give the compound relation $a^2:b^2$; wherefore we say, that the squares are in the duplicate ratio of their roots. And the ratio $a:b$ multiplied twice, giving the ratio $a^3:b^3$, we say that the cubes are in the triplicate ratio of their roots. 495. Geometry teaches, that two circular spaces are in the duplicate relation of their diameters; this means, that they are to each other as the squares of their diameters. Let A be such a space, having its diameter 45 feet, and B another circular space, whose diameter is 30 feet; the first space will be to the second as $45 \times 45$ is to $30 \times 30$; or, compounding these two equal relations, as $9:4$. Therefore the two areas are to each other as 9 to 4. 496. It is also demonstrated, that the solid contents of spheres are in the ratio of the cubes of their diameters: so that the diameter of a globe, A, being 1 foot, and the diameter of a globe, B, being 2 feet, the solid content of A will be to that of B, as $1^3:2^3$; or as 1 to 8. If, therefore, the spheres are formed of the same substance, the latter will weigh 8 times as much as the former. 497. It is evident that we may in this manner find the weight of cannon balls, their diameters, and the weight of one, being given. For example, let there be the ball A, whose diameter is 2 inches, and weight 5 pounds; and if the weight of another ball be required, whose diameter is 8 inches, we have this proportion, $2^3:8^3::5:320$ pounds, M