Or still more concisely, thus; reducing each separate portion of the wall to the thickness of half a brick, adding, and dividing by 3. once for all. (PRINCIPLE XXVI.) *Where do these Multipliers come from? In the West of England, and in other districts, freestone walls are computed by the superficial perch or rod, of 2 square yards (18 ft. long, by 1 ft. high); the price per perch being regulated by the thickness agreed on. EXERCISE CXII. E. Find cost of undermentioned rectangular pieces of brickwork, at stated prices per rod of 272 su. ft., 11⁄2 brick thick. 6. 45 6 long, 14 7. 85 9. long, 24 courses high, 3. br. at £15 5's. * These prices are for labour only. at £2. 2's.* ESTIMATING WEIGHT OF LIVE CATTLE. EXAMPLE WORKED OUT. A grazier, wishing to esti mate roughly the weight of an ox, considers the superficial extent of the hide as equivalent to that of a rectangle having the girth and length of the animal for its two dimensions. He then fixes on 22 lb. as the probable. average weight under a superficial foot of the hide. What is the weight of the beast on these suppositions? Example completed. Area of rectangle 6 ft. 5 in. by 4 ft. 2 in. = 22 lb. X 263 5 cwt. 1. qr. 03 lb. .. Estimated weight But, in actual business, the would be taken as , which would occasion an error of only, for 3 = 94. .. Estimated weight = (nearly) 22 lb. × 263 — 5. cwt. 1. qr. 04 lb. EXERCISE CXII. F. = Find weight, when dimensions and estimated weight per superficial foot are as follow. This method of estimating the weight of cattle is constantly employed by graziers and butchers. Its accuracy, in any instance, depends mainly on the precision with which the weight per foot is fixed; a matter in which experience must guide. The girth is measured immediately behind the shoulder. The length is taken from the point of the shoulder to the root of the tail, or nearly, according to the form of the animal. AREA AND ONE DIMENSION GIVEN TO FIND THE OTHER DIMENSION. Calculations in Superficial Measure deal with Area, Length, and Breadth. When any two of these quantities are given, the third may be found. For, in the same sense in which we say that Area Length x Breadth, (Page 270) = we say also that Area Length = We must here repeat that these formulæ are only convenient, and not accurate. For, Area, consisting of superficial Units, cannot be divided by (that is, distributed into lots of) Lineal Units. Hence, Area cannot be divided by Length or Breadth. (See Pages 105' and 218) The subjoined examples worked out will exhibit the true meaning of the foregoing formulæ. FIRST EXAMPLE WORKED OUT. How wide is a rectangle whose Area is 64-75 super. chains, and Length 12.95 lin. chains? The Breadth, which we are to discover, will be equal to as many lineal chains as there are, in the rectangle, longitudinal strips each 1 chain wide. And each such strip contains 12.95 super. chains. (Why?) The question, therefore, resolves itself into this: How many rectangular strips, each containing 12.95 super. ch. can be formed out of a rectangle whose Area is 64.75 super. chains? If Area and Breadth be given to find Length, then, Length = as many lineal chains as there are transverse strips, one chain long. And each such strip contains 5 sup. chains, because the rectangle is 5 chains wide. SECOND EXAMPLE WORKED out. The Area of a rectangular court is 68 sup. ft. 3' 3", and its length 9. ft. 5 in. Required its Breadth. If Area and Breadth be given to find Length, then Area of transv. su. in. su. in. stp. 1. in. long.=87)9831 Area of rectangle. .. Length of Rectangle 113 lin. in. 9. ft. 5. in. = For other Parallelograms the process is precisely the same as for rectangles. THIRD EXAMPLE WORKED OUT. How long is the Base of a Parallelogram whose Area is 47· ac. 1· r. 30′ po., and Perpendicular Height 16.5 chains? Example completed. Area of Parallelogram = 47.4375 ac. = 474.375 sup. ch. Area of transv. strip, 1 ch. long (= perp. ht. of Par.) = 16·5 su. ch. .. No. of such strips 474-37516.5 = 28.75 .. Length of Parallelogram = 28.75 lineal chains. FOURTH EXAMPLE WORKED OUT. What is the Perpendicular height of a Triangle whose Area is 3' ac. 3. r. 3 po., and Base 173 chains? But, Perpend. Height of Triangle is the same as that of Parallelogram, for they are supposed to be on the same base, and the Parallelogram double of the Triangle. (Page 272.) .. Perpendicular Height of Triangle 41 lineal chains. = *17-75)37-71875 Ar. of Tri., or of Par, same height, but half as long. 2.125 strips*, Perp. Height of that Parallelogram. 2. 4:25 strips in Par. which is double of Triangle Perp. Ht. .. Perpendicular Height of Triangle 41 lineal chains. EXERCISE CXIII. Describe, accurately, in writing, each Divisor, Dividend, and Quotient. Prove each Example by working out the Area, with the Dimension found. Given Area A. 1 Of a Parallelogram, 22 ac. 12.8 po., and Perp. Ht. 640 links. Find Base. 2. Of a Rectangle, 1643 ac. 2 r. 18.016 po., and Length 19348 links. Find Breadth. 3. Of a triangular meadow, 5 r. 10 po., and Base 28 po. Find Perp. Height. 4. Of an oblong room, 269 su. ft. 6' 9", and length 18. ft. 11 in. Find Breadth. The strip here spoken of is a rectangle on same base as the Triangle, and one chain in Perpendicular Height. 5. Of a triangle, 28 su. po. 24 yd. 1. ft., and Breadth 64 feet. Find Length. 6. How wide is an oblong passage of which 17 ft. are exactly covered by 20 ft. of canvas 4 ft. 3' wide? 7. In exchange for a rectangular strip of land, (required for roadmaking,) 6241 links long, and 72 links wide, a triangular plat is offered which may be 474 links in perpendicular breadth. What length of Base should it have? B. 1. How many yards of carpet 20 inches wide will cover the floor of a rectangular room 16 ft. 8 in. long, by 13 ft. 4' wide, making no allowance for matching the pattern, in the several lengths? 2. Supposing the room referred to in last example to be 9 ft. 3 in. high, how many pieces of paper 21 inches wide will cover the walls, each piece containing 11 lineal yards? 3. What length of rectangular board 8 superficial foot? 4. How much planking 10 in. wide will long, by 12. ft. 4 in. wide? 5. How many yards of velvet trimming in. wide will be equal to a floor a room 14 ft. 11 in. yd. wide can be cut from 6. How much silk velvet § yd. wide will furnish 21 yd. of trimming in. wide? C. 1. The length of a rectangle being 64 ft. 10 in., and its Area 815. su. ft. 9" 10", what is its Breadth ? 2. From a rectangular piece of building-ground, which is 364 ft. wide, what Length must be cut off to make an Area of 2081 su. ft. 4" 3""? 3. Given Area of Rect. 888 su. ft. 10" 10"", and L. 95 ft. find B. 4 The space occupied by a certain oblong piece of common is known to be 613 ac. 20 po., and its uniform width 45 chains. Required its length. 5. If a triangle 15 po. in perp. height, have an Area of 5′ r. 10' po., what is its Base? 6. What is the average width of a road, of which 2 miles have an Area of 5 ac. 4′ po. 28′ yd. 3. ft.? 7. The walls of a rectangular chamber cover an area of 51 sup. yd. The apartment is 15 feet long, and 9 ft. high. How wide is it? 8. Find Breadth of a rectangle of which Area is 1824 su. ft., and Length 85 ft. 9. From a strip of land 81. ft. in perpendicular width, what length must be cut off to be worth 18's. 6d. rent, at 9'd. per customary superficial pole? 10. A rectangular field, 1125 links by 500 links, is to be divided into two equal longitudinal strips, and these again subdivided into rectangular allotments of 30 superficial statute poles each. How many allotments will there be, and what will be the length and breadth of each? |