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. 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 triangle or parallelogram, one side and an adjacent angle being given. RULE. For a triangle. As the rectangle of the given side and sine of the given angle, Is to twice the given area; So is radius, To the other side adjacent to the 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. As the rectangle of the given side and sine of the given angle, Is to the given area; So is radius, To the other side adjacent to the given angle*. EXAMPLES. 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° E.; it is required to lay off 10 acres by a straight line AD, running from the point A to the side BC. * DEMONSTRATION. It is demonstrated, prob. 3, sect. 1, Measuring Land, that rad. : sin. B:: AB × BD : 2. ABD (see Fig. 87.); therefore (1.6 cor.) rad. ×AB: sin B×AB :: AB×BD : 2.ABD, or (16.5) sin. BxAB: 2 ABD :: rad. ×AB: AB×BD :: 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. 2. Given the side AB, Fig. 15, of a parallelogram, equal 20 ch. and the angle A 63° 30'; required the side AC, that the content may be 21 acres. 3. Given one side of a triangle, equal 30 perches, an angle adjacent to this side 71° 15′, and the area 2 acres; required the other side adjacent to the given angle. Ans. 22.53 perches. 4. Given one side of a parallelogram, equal to 32.26 ch., an angle adjacent to this side 83° 30′, and the area 74 acres; required the other side adjacent to the given angle. Ans. 23.09 Ch. PROBLEM VI. The area and base of a triangle being given, to cut off a given part of the area by a line running from the angle opposite the base. RULE. As the given area of the triangle, To the base corresponding to that area*. EXAMPLES. 1. Given the area of the triangle ABC, Fig. 88, equal 650 square perches and the length of the base AB 40 perches; it is required to cut off 290 perches towards the angle A, by a line running from the angle C to the base. ABC. ADC. AB. AD. 1 As 650 290 :: 40: 17.85 per. 2. In a triangle ABC, there are given the area 27 A. 1 R. 16 P. and the base AB 35.20 ch., to cut off 10 acres towards the angle B, by a line CD running from the angle C to the base: the part BD of the base is required. Ans. 12.87 Ch. PROBLEM VII. The area and two sides of a triangle being given, to cut off a triangle containing a given area, by a line running from a given point in one of the given sides and falling on the other. RULE. As the given area of the triangle, Is to the area of the part to be cut off; To a fourth term. Divide this fourth term by the distance of the given point from the angular point of the two given sides; the quotient will be the distance of the required point from the same angle*. * DEMONSTRATION. From the demonstration to prob. 3, sect. 1, Measuring Land, we have, Fig. 89, rad. : sin. A :: ABXAC: 2ABC and rad. : sin. A :: AP×AG : 2APG; therefore (11 & 16.5) 2ABC: 2APG :: ABXAC: AP×AG, or (15.5) ABC: APG :: AB × AC : EXAMPLES. 1. Given the area of the triangle ABC, Fig. 89, 5 acres, the side AB 50 perches, the side AC 40 perches, and the distance of a point P from the angle A, 36 perches, it is required to find a point G to which, if a line be drawn from the point P, it shall cut off a triangle APG containing 3 A. 0 R. 20 P. As the triangle ABC 800 sq. p. : the triangle APG 500 :: ABXAC, SAB 50 Ar. Co. 7.09691 2.69897 1.69897 AC 40 1.60206 2. Given the area of a triangle ABC. 12 A. 1 R. 23 P. the side AB 20 ch., the side AC 16.25 ch., and the distance of a point P in the side AB, from the angle A 8.50 ch.; it is required to find the distance AG of a point G in the line AC, so that a line drawn from P to G may cut off a triangle APG containing 3 acres. Ans. 9.25 ch. PROBLEM VIII. The area and base of a triangle being given, to cut off a triangle containing a given area by a line running parallel to one of the sides. RULE. As the given area of the triangle, Is to the area of the triangle to be cut off; So is the square of the given base, |