First, divide either dimension of the floor by the given breadth of carpet reduced to feet, for the approximate number of breadths; then taking the nearest whole number, divide the same dimension by the number of breadths, which will give the required approximate breadth of carpet.

Example.

What breadth of carpet, nearly a yard wide, will, without waste, cover a floor 19 ft. by 13 ft. ? and how many yards will it take?

4

First, 1341; then, taking 4 for the number of breadths,1331 ft., or 3 ft. 3 in. for the required breadth of carpet. Hence, there will be 4 breadths, each 19 ft., or 61 yds. long, and 4 × 63 = 25 yds. the required length. Or, as 13 ft. 13 yds., thus:

=

yds. yds. yds.

yds. yds. yds.

76

19:13:19; Or, (189) 1:4:: 19:46-251, as before. Again, 1961, and, taking 6 for the number of breadths, we have 1931 ft., or 3 ft. 2 in. for the required breadth of carpet; that is, there will be 6 breadths, each 13 ft. or 4 yds. long, and 6 4 26 yds. the required length. Or, as 19 ft. = 19 yds., thus:

yds. yds. yds.

=

yds. yd.

12: 19:: 13, Or (189) I: 6:: 13, or *1 : 2 :: 13:26 yds.,

as before.

From which we infer, that if we would carpet the floor lengthwise, with carpet nearly yard wide, we must choose a carpet 3 ft. 3 in., or rather more than 1 yd. 0 qr. 1 n. wide, of which it will require 4 breadths, each 6 yds. 1 ft. in length, making 25 yds. Also, if we would carpet it crosswise, we must choose a carpet 3 ft. 2 in., or rather less than 1 yd. 0 qr. 1 n. wide, of which we require 6 breadths, each 4 yds. 1 ft. in length, making 26 yards.

423. Having seen (402) that when the measures or sides of two equal parallelograms are placed in reciprocal proportion, those of the one become the means, and those of the other the extremes; it follows that when the area of any given parallelo

*See Article 210.

gram is divided by one side of another parallelogram, of equal area*, the quotient will be the other side of that other parallelogram. We must, however, observe that the given area and given side must be of the same name; that is, if the given area is expressed in square yards, the length of the given side must be expressed in yards; if the given area is expressed in square feet, the length of the given side must be expressed in feet, &c. Also, that the given area must be a multiple of the given side, if we wish the result to be a whole number.

Suppose that from one side of a large field we would cut off a piece of ground of a given breadth, say 120 ft., which shall contain just an acre.

The area of an acre in s. f. is 43560, and

feet, the required length.

435610

1210

363

Or, to ascertain whether the length will be a whole number, we may resolve 43560 into its prime factors 2 × 2 × 2 × 3 × 3 × 5 × 11 × 11, and then see, if by the involution of any of these factors, we can form 120. We find that 2 × 2 ×2 × 3 X5=120; consequently, the product of the other factors 3 × 11 × 11=363 ft. is the required length.

424. When the first of three given numbers is to the second as the second is to the third, we say that the numbers are in continual proportion. The third is called a third proportional to the two first, and the second a mean proportional between the first and third. When we say, as the first is to the second, so is the second to the third, the second, being both antecedent of the last ratio, and consequent of the first, forms of itself the two means, the product of which is the square of the second; consequently, (405,) to find a third proportional to two given numbers, we square the last and divide by the first. Or, which is the same thing, we multiply the last by its ratio to the first. We must, however, observe, that the second must be a multiple or a measure of the first, or both must be composed of powers of the same prime numbers, otherwise (179) the third proportional cannot be a whole number.

6 x 6

Thus, to find a third proportional to 2 and 6, we say 2

18, the number sought. For a third proportional to 8 and 9, which are prime to each other, we have 281 = 10,

Answer.

*We are here speaking of right-angled (rectangular) parallelograms.

425. As in the preceding article, the product of the extremes is equal to the product of the means; the product of the first and third is equal to the square of the second; wherefore, to find a mean proportional between two numbers, we take the square root of their product.

Thus, to find a mean proportional between 3 and 27, we say, 3 x 27 81, and 1/819, the required mean.

=

426. Let A, B and C, be any three numbers, the ratio of A

is, the ratio of A to C is compounded of the ratios A to B and B to C.

This is called duplicate ratio. Therefore, when three numbers are in continual proportion, the ratio of the first to the third is duplicate of the ratio of the first to the second, or of the second to the third; that is to say, it is the square of either of those ratios.

428. When there are four numbers A, B, C, D, the ratio of A to D is compounded of the ratios A to B, B to C, and C to D. Thus

B

X

B

A

X

=
D

·; and, in general, whatever may

be the number of quantities, the ratio of the first to the last is compounded of the ratios, the first quantity to the second, the second to the third, the third to the fourth, and so on to the last.

429. When four numbers are in continual proportion, as the ratio of the first to the fourth involves three equal ratios, it is called the triplicate ratio or cube of any one of those ratios; when there are five numbers, the ratio of the first to the fifth is the quadruplicate or fourth power of any one of those ratios; when there are six numbers, the ratio of the first to the sixth is quintuplicate of the ratio of the first to the second, &c.; and in general, when a series of numbers is in continual proportion, the ratio of the first to the last is equal to the ratio of the first to the second raised to a power signified by the number of terms, less one.

430. The astronomer Kepler found that the squares of the times in which any two planets revolve round their primary

are to one another as the cubes of their distances from that primary; wherefore, if we know the times of the revolution of any two planets in the solar system, and the distance of one of them from the Sun, the distance of the other may be found by a simple proportion, and by extracting the cube root; or, if we know the distance of each from the Sun, and the time of the revolution of one of them, the time of the revolution of the other may be found by a proportion, and by extracting the square

root.

Examples.

1. Let the time of the revolution of the planet Mercury about the Sun be 87 d. 23 h. 14 m. 32,7 sec.; that of the revolution of Venus 224 d. 16 h. 41 m. 27,5 sec.; also the distance of Venus 68514044 miles; required the distance of Mercury from the Sun.

Square of revol'n. of Venus in seconds. Square of revol❜n. of Mercury in seconds. As 376891262347656,25 : 57767185263445,29

Cube of distance of Venus.

Cube of distance of Mercury.

::321616859411343801397184:49295121849560890257289,

and extracting the cube root of this last term, we have 36666375,479+ miles, the required distance of Mercury from the Sun.

2. Let the time of the revolution and distance of the planet Venus be as in the above example; also the time of the revolution of the Earth 365 d. 5 h. 48 m. 51 sec. ; required the distance of the Earth from the Sun.

Ans. 94719068,71 + miles.

3. Let the time of the revolution of the Earth, also the time of the revolution and the distance of the planet Mercury, be as in the above examples; required the distance of the Earth from the Sun. Ans. 94719068,71 + miles.

4. Let the time of revolution and the distance of the Earth, also the distance of Mercury, be as above; required the time of the revolution of Mercury.

Ans. 87 d. 23 h. 14 m. 32 + sec.

Compound Proportion.

431. The name Compound Proportion may properly be applied to every proportion in which several ratios are compounded, that is, multiplied together.

This rule is frequently, and we think improperly, called the Double Rule of Three, or Rule of Five; because, though in

many questions five terms are given, in others we find seven, nine, eleven, &c.

In common with Simple Proportion, one term is given, which is of the same kind as the answer; and of the rest, two and two are of the same kind; consequently, each couple forms a ratio.

432. To solve a question: First, write the term which is like the answer by itself; then arrange the two terms of each ratio as in Simple Proportion; that is, exactly as if the result depended on them alone.

Having arranged all the terms, cancel the antecedent and consequent of any ratio, or the antecedent of any one and consequent of any other, not forgetting the term first written, which may be cancelled with any of the antecedents, as in Simple Proportion. Lastly, multiply the remaining consequents and term first written together, and divide the result by the product of the remaining antecedents.

Examples.

1. If 40 masons can build a wall 1350 feet long, 3 ft. thick and 12 ft. high, in 9 days, when the days are 10 hours long how many masons will build a wall 5400 ft. long, 21 ft. thick and 15 ft. high, in 15 days, when the days are only 8 hours long?

ny

As the question requires masons, we write 40, the given number of masons, by itself. Then, for the ratios, we say 5400 ft. will require more than 1350 ft.; wherefore, we write 1350 for the antecedent and 5400 for the consequent; we also accompaeach of these with its respective thickness and height. We then say, it will require fewer men to build the wall in 15 d. than in 9 d.; consequently, we write 15 for the antecedent and 9 for the consequent. Lastly, as it will require more men when the days are only 8 hours long, we write 8 for the antecedent and 10 for the consequent.

We then find that 1350 cancels 5400 and gives 4 for the quotient, which we write opposite 5400. Then 24 cancels 121, giving 5 for the quotient; and 8 times 5 cancels 40.