all the angles of the field are visible. The bearing and distance of the stations, and the bearings of the angles, from each station, are as follow. What is the area of the field? The station G bears from the station F, N 43° W. LAYING OUT AND DIVIDING LAND. PROBLEM I. To lay out a given quantity of land in a square form. RULE. Reduce the given quantity to chains or perches, and extract the square root, which will be the length of a side, of the same denomination to which the given quantity is reduced. EXAMPLES. 1. Required the side of a square that shall contain 9 A. 3 R. 28 P. 40)28 Per. 4)3.7 R. 9.925 A. 99.25 Ch. Ch. 99.25(9.96 Ch. the length of a side. 81 189)1825 1986)12400 484 2. Required the side of a square tract of land that shall contain 325 acres. Ans. 57 Chains. PROBLEM II. To lay out a given quantity of land in a rectangular form, having one side given. RULE. Divide the given content by the length of the given side, the quotient will be the length of the required side. EXAMPLES. 1. It is required to lay out 120 acres in a rectangular form, the length of one side being given, equal 100 perches. Acres. 4 480 40 1,00)192,00 192 perches, the length of the other side. 2. The length of a rectangular piece of land is 8 chains; what must be its breadth, that the content may be 5 acres. Ans. 6.25 chains. U PROBLEM III. To lay out a given quantity of land in a rectangular form, having the length to the breadth in a given ratio. RULE. As the less number of the given ratio, Is to the greater ; So is the given area, To a fourth term*. The square root of this fourth term will be the length required. Having the length, the breadth may be found by the preceding problem. Or it may be found in the same manner as the length. Thus As the greater number of the given ratio, Is to the less; So is the given area, To a fourth term. The square root of this fourth term will be the breadth required. EXAMPLES. 1. It is required to lay out 864 acres in a rectangular form, having the length to the breadth in the ratio of 5 to 3. * DEMONSTRATION. Let ABCD, Fig. 85, be a rectangle, and ABFE and AHGD be squares on the greater and less sides respectively then (1.6) AD: AE(AB) :: the rectangle AC : square AF. Also AB : AH (AD) :: the rectangle AC: square AG. Hence the ર. As 3: 5 :: 138240 : 230400 230400-480 Perches, the length required. Sq. P. Sq. P. As 5 3: 138240: 82944 82944-288 Perches, the breadth required. 2. It is required to lay out 27 A. 3 R. 20 P. in a rectangular form, having the length to the breadth in the ratio of 9 to 7. Ans. Length 75.725 P. Breadth 58.897. P. PROBLEM IV. To lay out a given quantity of land in a rectangular form, having the length to exceed the breadth by a given difference. RULE. To the given area, add the square of half the given difference of the sides, and extract the square root of the sum; to this root add half the given difference for the greater side, and subtract it therefrom for the less*. DEMONSTRATION. Let ABCD, Fig. 86, be a rectangle; in DC let DE be taken equal DA or BC, and let EC be bisected in F; then (6.2) DF2=DC × DE + FC2=DC × AD+FC2=the rectangle AC+ the square of half the difference of the sides DC, DA; also DF+FC=DC, the greater side, and DF-FC=DE or DA, the less side. This problem may be neatly constructed thus: take EC equal the given difference of the sides and bisect it in F ; make EG perpendicular to EC and equal to the square root of the given area, and with the centre F and radius FG describe the arc DG meeting CE produced in D; make DA perpendicular to DC and equal to DE, and complete the rectangle ABCD, which will be the one required. Since (47.1.)FG2=EG2+EF2=the given area+the square of half the given difference of the sides, the truth of the construction is |