Second method. 40 2.0 po. 4 3.05 ro. = 7.7625 ac. 776250 sq. lks. =77·625 sq. ch. 40. To reduce square links and square chains to acres, roods, and poles. RULE.-Reduce them to acres and decimal parts of an acre (Arts. 36, 38), and reduce the decimal parts to roods and poles by multiplying by 4 and 40 in succession, and cutting off five decimal places in each product. Ex. Reduce 15 sq. ch. 3125 sq. lks. to acres, roods and poles. Here, 15 sq. ch. 3125 sq. lks. =150000 sq. lks.+3125 sq. lks.= 153125 sq. lks. = 1.53125 ac. ao. 1.53125 4 ro. 2.12500. 40 po. 5.00000 = Hence, 15 sq. ch. 3,125 sq. lks. 1 ac. 2 ro. 5 po Ex. XII. (1) Reduce to square links, 10 ac. 3 ro. 5 po.; 9 ac. 2 ro. 6 po.; 2 ac. 3 ro.; 3 ac. 1 ro.; 4 ac, 20 po. (2) Reduce to square chains and square links, 2 ro. 15 po.; 3 ro. 1 po.; 3 ro. 5 po.; 1 ac. 2 ro. 8 po.; 3 ac. 3 ro. 25 po. (3) Reduce to square chains and dec. parts of a square chain, 3 ac. 2 ro. 1 po. ; 9 ac. 2 ro. 6 po.; 1 ac. 2 ro. (4) Reduce to acres, roods, and poles, 35625 sq. lks., 536250 sq. Iks., 687500 sq. lks., 151875 sq. lks., 56250 sq. lks., 53125 sq. lks., 24375 sq. lks. (5) Reduce to acres, roods, and poles, 25 sq. ch., 35·0625 sq. ch., 70-875 sq. ch., 60.0625 sq. ch. (6) Reduce to acres, roods, and poles, 78 sq. ch. 1250 sq. lks., 68 sq. ch. 7500 sq. lks., 100 sq. ch. 4375 sq. İks. 41. To find the acreage of a square field, when its side is given in chains and links. RULE.-1°. Reduce the side to links by Art. 29, 2o. Square the result, and reduce to acres, roods, and poles, by Art. 40. [The method of working is exemplified on the next page.] Ex. What is the acreage of a square field, whose side is 8 ch. 75 lks.? lks. (1) Pequired, in acres, roods, and poles, the area of a square field whose side measures 625 links. (2) The side of a square is 23 ch. 75 lks. Find its acreage. (3) Required the area of a square field whose perimeter measures 1000 links. (4) Find the area of a square field 22 ch. 25 lks. in length. (5) What is the rent of a square field 1375 links in length, at 50s. an acre? (6) Find the number of acres in the area of the base of the great pyramid of Egypt, whose side was 1250 links. 42. To find the number of square tiles, bricks, flags, &c., required to cover a square floor or other area, i.e., to find how many given smaller spaces will exactly fill up a larger one. RULE. Divide the greater space by the less, and the quotient will be the number required. [The areas must be both expressed in the same denomination.] Ex. How many square tiles 8 in. long will be required for paving a college dining-hall in the form of a square whose side measures 21 ft. 4 in. ? Here, 21 ft. 4 in. =256 in. Area of hall floor = 256 × 256=65536 sq. in. Ex. XIV. (1) How many sheets of paper, each 2 ft. square, will cover a floor 6 yds. square? (2) How many shrubs, each requiring 3 sq. ft. of space, can be planted in a garden 15 ft. square? (3) A square courtyard is 30 yds. long. How many flagstones, 2 ft. square, will be required for paving it? (4) A square field, 387.5 yds. long, is divided into allotments, each 15.5 yds. square. How many are there? (5) How many square tiles, 15 in. long, will be required for paving a square dining-hall whose side is 18 ft. 9 in. long? (6) How many persons can have square allotments, each measuring 500 links, out of a square common containing 15 acres? (7) A square court, whose side is 24 yds., is paved with 41472 sq. tiles. Find the area of each tile. 43. To find the side of a square when the area is given. RULE. Extract the square root of the number of square units in the area, and the result will give the number of linear units in the side. [See Fig. Art. 15.] 44. The square root of any number is that number which, when multiplied by itself, has the given number for product. Thus, 5 Thus, 5 is the square root of 25. The square root of any number up to 12 is known from the multiplication table. We now proceed to show how to find the square root of a number exceeding 12. 1o. Divide the given number into periods of two figures each, beginning with the units-figure. The number of periods will show the number of figures in the required root. 2o. Find the greatest number whose square is contained in the first period at the left; this is the first figure in the root, and must be placed in the form of a quotient at the right of the given number. 3°. Subtract its square from the first period, and to the remainder bring down the second period for a dividend. 4°. Double the part of the root already obtained for a divisor, and find how often it is contained in the dividend exclusive of its right-hand figure, and set the result both in the quotient and also to the right of the divisor. с 5°. Multiply the divisor, as it now stands, by this last quotient figure, and subtract the product from the dividend. 6°. To the remainder bring down the next period for a new dividend. Then find a new divisor by doubling the figures of the quotient, and proceed as before, till all the periods are brought down. Ex. 1. Find the side of a square containing 71289 sq. yds. 7,12,89 (267 4 46) 312 276 527) 3689 .. Side of square=267 yards. 45. The square root of decimal fractions is found precisely as that of whole numbers: only the periods of two figures are reckoned both ways from the decimal point. Ex. 2. A square field contains 3370.9636 sq. yds. Find the length of its side. 33,70 96,36 (58.06 25 108) 870 864 11606) 69636 .. Side of square=58·06 yards. 46. The square root of a vulgar fraction is found by taking the square root of the numerator and the denominator. Ex. 3. The area of a square is 738 sq. ft. Find its side. 7381361 The square root of 2809 is 53; and the square root of 361 is 19. .. Sq. root of 282 = 43215. 47. When the area is expressed in different de nominations, it must be reduced to its lowest name before the square root can be taken. [NOTE. Areas are expressed in Square Measure, lengths of sides in Linear Measure.] Ex. 4. Find the side of a square whose area is 85 sq. ft. 81 sq. in. Here, 85 sq. ft. 81 sq. in. =12321 sq. in. The square root of 12321 is 111, which is the number of linear inches in each side, and 111 linear in.=9 ft. 3 in. Ex. 5. Required the side of a square field containing 1 ac1 ro. 25 po. Here, 1 ac. 1 ro. 25 po.=1.40625 ac.=140625 sq. links. The square root of 140625 is 375, which is therefore the number of links in the side. Ex. XV. Find the sides of the squares whose areas are respectively:(1) 196 sq. yds.; 1024 sq. ft.; 289 sq. yds.; 5184 sq. ft.; 126736 sq. yds.; 502681 sq. in. (2) 5.29 sq. in.; 156 25 sq. ft.; 882-09 sq. yds.; 0121 sq. in.; 6.5536 sq. in.; 001156 sq. in. (3) 45 sq. in; 41 sq. yds.; 7598 sq. ft.; 61 sq. in.; 73 sq. in.; 732 sq. ft.; 39 sq. yds. (4) 7 sq. ft. 16 sq. in.; 1 sq. ft. 52 sq. in.; 3 sq. ft. 97 sq. in.; 8 sq. ft. 73 sq. in.; 30 sq. yds. 5 sq. ft. 1 sq. in. (5) Required in links the sides of the squares whose areas are respectively:-131 ac. 1 ro. 25 po.; 13 ac. 3 ro. 9 po.; 7 ac. 36 po.; 604 ac. 2 ro. 1 po.; 1 ro. 9 po.; 15 ac. 2 ro. 20 po.; 10 ac. 2 ro. 1 po.; 14 ac. 1 ro. 24 po. (6) The area of a square is 86436 sq. yds.; find the length of its side. (7) Required the side of a square whose area equals 14 sq. ft. 9 sq. in. (8) What is the length of the side of a square court which contains 43785.5625 sq. ft.? (9) A carpet contains 167 71545025 sq. ft. Find its breadth when it is as broad as it is long. (10) Find, in yards, the side of a square contain ten acres. [4840 sq. yds. = 1 acre.] field which shall (11) Required, in links, the side of a square field containing 10 acres of ground. (12) The area of a square is 133225 sq. yds. find its perimeter. (13) A square deer-park, including 1600 acres of land and a lake of 960 acres, is to be enclosed with a wall; what will be its length in yards? |