Examples. 1. It is required to multiply 12ft. 9in. by 6ft. 4in. NOTE. In this example the highest denomination of the multiplier, vix. 6 feet, was multiplied into the lowest denomination of the multiplicand, viz. 9 inches; and the product was 54', in which was 4 of the next denomination, vis: 4 twelves, and 6 left which was set down, and the 4 was carried to the next denomination; I then multiplied the 6 feet into the 12 feet, and added the 4 to the product, and set the whole down: I then multiplied the 4 inches into the 9 inches, (removing the product one place to the right) and found the product to be 36; three twelves, none over; I then multiplied the 4 in. into the 12 ft. and added in the 3 I carried which made 51, which was 4 twelves and 3 over; I then added the products together, the sum is the product required. 2. It is required to multiply 3ft. 2' 3" by 3ft. 2' 3". Ans. 10ft. 1' 11" 0" 9"", METHOD SECOND. RULE.-Multiply the highest denomination in the multiplier into all the denominations of the multiplicand, and set down the whole products, &c. then multiply the next highest denomination in the multiplier into all the denominations of the multiplicand, and remove the products one place to the right, and set them all down; proceed thus through all the denominations, observing to set down the whole products each time of multiplying; then add up the several products in the order in which they stand, and carry the same as in Compound Addition, the sum will be the product required. 4t 1. What is the product of 12 ft. 9 in. multiplied by 6 ft. 4 in. pro. 80 9 0 Ans. 2. What is the product of 3 ft. 2′ 3′′ multiplied by 3 ft. 2' 3" ? 6.9 646"" 339 16 509. pro. 10 1 11 0 9 Ans.. 3. What is the product of 12 ft. 11 in. 11 sec. multiplied by 12 ft. 11 in. 11 sec. Ans. 168 ft. 9 10 0 1!!! NOTE. This method need not be confined to 12, as some imagine, but may be extended to any denomination, as rods, yards, feet, and inches, &c. Examples. 1. Required to multiply 22 yds. 2 ft. 11 in. by 7 yds, 2 ft. 2 in. yds. ft. in. 22 2 11 2. It is required to multiply 10 rods, 20 links, by 12 rods and 20 links, allowing 25 links to a rd. ? Ans. 138 rods, 6 links. NOTE. In the last example I multiplied 12 rds. by 20 links and set down the product, viz. 240; I then multiplied the 12 rds. by the 10 rds. and set down the product, viz. 120; I then multiplied the 20 lin. by the 20 lin. and removed the product one place to the right and set it down, viz, 400; I then multiplied 20 lin. by 10 rds. and set down the product, viz. 200; 1 then added the products together and carried by 25. 3. It is required to multiply 7 ft. 3 in. by 3 ft. Ans. 23 ft. 6′ 9′′-23% ft. 3 in. NOTE.-The 6 inches are 9 9 and the 9′′ of a foot, and 34 +12=18 as expressed above; any question that is solved by duodecimals may be solved by vulgar, or decimal fractions. in. 4. It is required to multiply 8 ft. 3 in. by 4 ft. 3 Ans. 35 ft. 0' 9"=35 ft. The same question by vulgar fractions. 8 ft. 3 in.8 and 84 is equal to 33. 4 ft. 3 in.4, 8. ft. 3 in.-8.25 ft. and 4 ft. 3 in.=4.25 ft. and 9 in. The same question solved by vulgar fractions. 3,91 and 6×14=25-45 ft. Ans. 9 25 rt SUPERFICIAL MEASURE. DEFINITION. Superficial measure is that which respects length and breadth, without regard to thickness; the dimensions are various according to the nature of the thing measured; land is measured by superficial measure, and its dimensions are generally taken in acres, rods and links: boards are also measured by superficial measure, and the dimensions are taken in feet and inches, &c. Artificers' work is calculated by different dimensions. Glazing by the square foot. Masons' Hat work, such as plastering by the square foot or yard. Painting, paving, &c. by the yard. Partitioning, flooring, roofing, tiling, &c. are calculated by the square of 100 feet. Brick work is generally calculated by the solid foot. GENERAL RULE. Multiply the length by the breadth, the product is the superficial content, or area, in the same measure, or dimension, as that which the dimensions were taken in; if yards, then the area is yards, if feet then the area is feet, &c. CASE I. To measure, or find the area of a board, or any other plane surface whose width is equal; such a figure is called a parallelogram. RULE. Multiply the length by the breadth, and such dimension as the length and breadth are taken in, such will be the dimension of the area; if feet, the area will be square feet, &c. NOTE.-It is obvious that if the length of any plane surface is multiplied into the breadth, that the product will be the area in square measure, as may be more fully understood by the following figure. A 27 B 27 5 It is evident from the preceding figure, that if the length B C 27, be multiplied by the width A Bor DC 5, the product will be 135, and will be the number of squares contained in the figure; and it need not be argued that if these dimensions were taken in feet, that the area would be square feet, if taken in rods, or yards, that the area would be rods, or yards, &c. Examples. 1. What is the area of a floor that is 22 feet 61 in. by 14 ft. 9 in ? feet. 2. How many feet of boards will cover a floor of a hall, that is 41 ft. 9 in.; by 30 ft. 6 in. allowing the floor to be 2 boards thick? Ans. 2546 ft. 9 in. feet are in a board that is 3. How many square 21 feet long, and 11 in. wide? Ans. 19 3'193 ft. 4. How many square feet of boards will it take to lay a single floor, that is 28 ft. 6 in. long, and 14 ft. 6 in. wide? Ans. 413 ft. 3'. CASE II. To measure a board or any other plane, when it is wider at one end than the other, and of a true taper. RULE. Add together the width of the two ends, and half the sum is the mean width; or take the width in the middle (which is the same as the half of the sum of the width of the two ends,) then multiply the length by the mean width, the product is the answer, or area required. |