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. 120 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, 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, 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 FE and AH 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 Sq. P. As 3 : 5 :: 138240 : 230400 Sq. P. PROBLEM IV. having the length to exceed the breadth by a given dif- 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) DF=DC X DE + FC2=DC X 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 perpendi. cular 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.)FGo=EG +EF2=the given area+the square of half the given difference of the sides, the truth of the construction is plain from the preceding demonstration. : EXAMPLES. 1. It is required to lay out 47 A. 2 R. 16 P. in a rectangle, of which the length is to exceed the breadth by 80 perches. 2)80 P. 47 A. 2 R. 16 P.=7616 Per. 1600 9216=96 length 136 breadth 56 2. It is required to lay out 114 A. 2 R. 33.4 P. in a rectangular form, having the length to exceed the breadth by 15.10 chains. Ans. Length 42.25 Ch. Breadth 27.15 Ch. PROBLEM V. To lay out a given quantity of land in the form of a tri angle or parallelogram, one side and an adjacent angle being given. RULE. For a triangle. given angle, To the other side adjacent to tire given angle. Then having two sides and the included angle given, the other angles and side, if required, may be found by For a parallelogram. given angle, EXAMPLES. a 1. Let AB, BC, Fig. 87, be two sides of a tract of land the bearing of AB is S. 87°1 W. dist. 16.25 ch. and the bearing of BC, N. 27°1 E. ; it is required to lay off 10 acres by a straight line AD, running from the point A to the side BC. Bearing of BA, N. 87°E. BC, N. 27 E. Angle B, 60° As ABxsin. B{AB 16.25 ch. sin : twice the given area 200 sq. ch. . :: rad. Ar. Co. 8.78915 0.06247 2.30103 10.00000 : : : BD 14.21 ch. 1.15265 * DEMONSTRATION. It is demonstrated, prob. 3, sect. 1, Measuring Land, that rad. : sin. B :: AB X BD : 2. ABD (see Fig. 87.); therefore (1.6 cor.) rad. XAB : sin BXAB :: ABXBD : 2.ABD, or (16.5) sin. BXAB: 2 ABD :: rad. XAB: ABXBD :: rad : BD. Since ABDF is equal to 2. ABD, the truth of the rule for the parallelogram is evident. This problem may be constructed as follows; take AB equal the given side and draw BC making the angle B equal the given angle; make BE perpendicular to AB and equal twice the given area of the triangle divided by the given side, or equal the given area of the parallelogram divided by the given side ; and parallel to AB, draw EF cutting BC in D, join DA, then will ABD be the triangle required, or complete the parallelogram ABDF, for the one required. The reason of the construction is plain. |