get Again, multiplying the multiplicand by feet, we ft. x 7 ft. = 21 sq. ft. = sq. ft.+ sq. ft., and ft. X 12 ft. = 3 sq. ft. Adding these last together, we get (2) 3 sq. ft.+ sq. ft.+ sq. ft., or 3 1′ 9′′ square feet. Adding (1) and (2) together, which must give the whole area, we get 53 sq. ft.+ sq. ft.+ which is 53 5' 9" sq. ft. 5 here are of a square foot, i.e., sq. ft., X 144, or 5 × 12 square inches. 9" is of a square foot, or 9 square inches. Hence, to reduce the square primes and seconds in the product to square inches, multiply the primes by 12, and add the seconds to the product. Hence the result is 53 square feet 69 square inches. 10. It will be seen from the above that feet X feet give (square) feet, feet X primes give (square) primes, feet x seconds (square) seconds, and so on. Primes x primes give (square) seconds, primes X seconds (square) thirds, seconds x seconds give (square) fourths, and so on. Hence we see that, in multiplying any denominations together, the denomination of the product is got by adding together the accents placed above each number which we multiply, observing that the numbers expressing feet have no accent or index above them. The above remarks will sufficiently explain the following Rule for Cross Multiplication. Place the several terms of the multiplier under the corresponding terms of the multiplicand. Multiply each term of the multiplicand by each term of the multiplier separately, beginning with the lowest denomination in the multiplicand and the highest in the multiplier, and placing the first figure of each line one place to the right of that of the preceding line, under its corresponding denomination. Finally, add the several lines together, carrying 1 for each 12 both in multiplying and adding. The sum will be the answer required. EXAMPLE 2. Find the area of a board 9 ft. 7′ 2′′ long, and 3 ft. 4′ 7′′ wide. 9 ft. 7' 2" 3 ft. 4' 7" 28 sq.ft.9' 6" 3 sq.ft.2' 4" 8" 5' 7" 2" 2"" 32 sq.ft. 5' 5" 10" 2"" Answer. We now subjoin some other Examples connected with square and cubic measure, such as are of frequent occurrence, not confining ourselves to the method of duodecimals. EXAMPLE 3. Find the cost of carpeting a room 24 ft. 6 in. long by 18 ft. 4 in. broad, with carpet which is of a yard broad, and costs 3s. 9d. a yard. = 24 X 18 number of square feet in the floor, And each foot of carpet costs Is. 3d., or Is. Therefore, required cost = EXAMPLE 4.-What is the height of a room which contains 223 cub. yds., 7 cub. ft., 624 cub. in. of air, the area of the floor being 41 sq. yds. 12 sq. in. ? Since the cubical contents are obtained by multiplying the three dimensions of length, breadth, and height together, and the area of the floor is obtained by multiplying the length and breadth together, we shall evidently get the height of the room by dividing the cubical contents by the area of the floor. 223 cub. yds. 7 cub. ft. 624 cub. in. = 6028,624 cub. ft. 369 sq. ft. (1.) Transform 12345678 from the decimal to the duodecimal scale, and also to the scale of 7. (2.) Transform 58367 from the decimal to the duodecimal scale. (3.) Transform 57t39 from the duodecimal to the decimal scale. (4.) Transform 67535 from the scale of 8 to the scale of 6. (5.) Transform 79e658 from the duodecimal scale to the scale of 8. (6.) Transform tetet from the duodecimal to the decimal scale. (7.) How many square feet are there in a board 15 ft. 7 in. long, and I ft. 10 in. wide? (8.) How many cubic feet in a block 18 ft. 5 in. long, 4 ft. 2 in. wide, and 3 ft. 6 in. thick? ་་ (9.) Find the area of a space 16 ft. 3′ 4′′ by 6 ft. 5′ 8′′ 10"'. (10.) Find the area of a space 18 ft. 0′ 5′′ 10"' by 4 ft. 8′ 7′′ 9′′. (11.) What will it cost to plaster a room 20 ft. 6 in. long, 18 ft. wide, and 10 ft. high, at 6d. a square yard? (12.) How many bricks 8 in. long, 4 in. wide, and 2 in. thick, will make a wall 50 ft. long, 10 ft. high, and 2 ft. 6 in. thick? (13.) Find the cost of carpeting the following rooms: (a) 18 ft. 4 in. long, 13 ft. 6 in. broad, with carpet & of a yard wide, at 2s. 9d. a yard. (b) 14 ft. 6 in. long, 10 ft. 4 in. broad, at 10s. 6d. a square yard. (c) 16 ft. 11 in. long, 13 ft. 3 in. wide, with carpet 2 ft. 3 in. wide, at 4s. 7d. a yard. (14.) A Turkey carpet 11 ft. 6 in. long by 9 ft. 8 in. wide is laid down in a room 14 ft. long by 12 ft. 6 in. wide; find the amount of oil-cloth necessary to complete the covering of the floor, and its cost at 5s. per square yard. (15.) The length of a room is twice its breadth, and its area is 1152 square feet; what is its length? (16.) Find the cost of carpeting a room 17 ft. 6 in. long by 13 ft. 9 in. wide with carpet yard wide, at 4s. 8d. a yard. (17.) If a cubic foot of water weighs 1000 ounces, what must be the depth of a rectangular tank, which is 35 ft. long and 10 ft. broad, that it may just contain 1000 tons of water? (18.) A roller being 6 ft. 6 in. in circumference and 2 ft. 3 in. wide, makes 12 revolutions as it moves from one end of a grass-plot to the other, and passes 10 times from one end to the other; find the area of the grass-plot. (19.) What length must be cut off a plank 1} ft. broad and 9 in. deep, in order that it may contain 11 cubic feet? (20.) A block of wood in the form of a rectangular parallelepiped measures along its edges 18 ft., 5 ft., and 3 ft. respectively; find its value if a cubical block of the same wood whose edge is II in, is worth 3s. 6d. 21.) On laying down a bowling-green with sods 2 ft. 6 in. long by 9 in. wide, it is found that it requires 75 sods to form one strip the whole length of the green, and that a man can lay down one strip and a quarter each day; find the area laid down in 8 days. (22.) If a cubic foot of water weighs 1000 oz., find to what depth a ton of water will cover an acre. (23.) A square foot of paper weighed 104'68 grains, and when 320 figures had been written on it, weighed 105 155 grains. If a strip of this paper 5 inches wide be taken to have written on it the circulating period of (which contains 100102 figures) in two lines at the rate of 5 figures in an inch, find the weight of the whole, and the length of the paper, and express this act in terms of the height of Salisbury spire, which is 400 feet. CHAPTER XIX. THE METRIC SYSTEM OF WEIGHTS AND MEASURES- 1. It will be seen, from the previous chapters, that there is no uniform system of subdivision of the various magnitudes and quantities with which we are daily concerned. Divisions into 2, into 4, 6, 8, 10, 12, 14, 16, 20, 24, 28, &c. parts, occur. The actual condition of our system of weights and measures, however, is, in fact, much more complicated and irregular than would appear from our statement. For throughout the kingdom there are, besides those we have mentioned, numerous local weights and measures; and, what is worse, very frequently the same name indicates a different amount. Thus, there are several different "stones" and different bushels. The inconveniences of our present system have long been felt, and plans for their remedy have engaged the attention of almost all the more eminent scientific men of the country. So far the agitation has only resulted in an Act of Parliament, passed July 29th, 1864, permitting the employment of the system of Weights and Measures called the Metric System, which we are about to explain, to be legally used in commerce, at the option of the parties concerned. The advantages of the system K |