REMARK. In this example, the links are treated as decimal parts of the chain; the result, therefore, is in square chains and decimal parts of a square chain. 2. What is the area of a triangle whose sides are 30 and 40 chains, and their included angle 28° 57'? Ans. 29A OR 7P. THIRD METHOD. Measure the three sides of the triangle. Then, add them together and take half their sum. From this half sum subtract each side separately. Then, multiply the half sum and the three remainders together, and extract the square root of the product: the result will be the area (Geom. Mens. Prob. II). Or, after having obtained the three remainders, add together the logarithm of the half sum and the logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area. 1. Find the area of a triangular piece of ground whose sides are 20, 30, and 40 chains. 45 × 25 × 15 × 5=84375: and √84375=290.4737=the area. Ans. 29A OR 8P. 2. What is the area of a triangle whose sides are 2569, 110. To find the area of a piece of land in the form of a trapezoid. Measure the two parallel sides, and also the perpendicular distance between them. Add the two parallel sides together, and take half the sum; then multiply the half sum by the perpendicular, and the product will be the area (Geom. Bk. IV. Prop. VII). 1. What is the area of a trapezoid, of which the parallel sides are 30 and 49 chains, and the perpendicular distance between them 16 ch 60 l, or 16.60 chains? 30+49=79; dividing by 2, gives gives for the area in square chains, 30 49 39.5 16.60 655.700 Ans. 65A 2R 11P. 2. Required the content, when the parallel sides are 20 and 32 ch, and the perpendicular distance between them 26 ch. Ans. 67A 2R 16P. PROBLEM IV. 111. To find the area of a piece of land in the form of a quadrilateral. Measure the four sides of the quadrilateral, and also one of the diagonals: the quadrilateral will thus be divided into two triangles, in both of which all the sides will be known. Then, find the areas of the triangles separately, and their sum will be the area of the quadrilateral. 1. Suppose that we have measured the sides and diagonal AC, of the quadrilateral ABCD, and found AB 40.05 ch, CD=29.87 ch, = BC 26.27 ch, AD=37.07 ch, and required the area of the quadrilateral. B Ans. 101A 1R 15P. REMARK. Instead of measuring the four sides of the quadrilateral, we may let fall the perpendiculars Bb, Dg, on the diagonal AC. The area of the triangle may then be determined by measuring these perpendiculars and, the diagonal AC. The perpendiculars are Dg=18.95 ch, and Bb 17.92 ch. PROBLEM V. 112. To find the content of a field having any number of sides. Measure the sides of the field and also the diagonals: the three sides of each of the triangles into which the field will be thus divided will then be known, and the areas of the triangles may then be calculated by the preceding rules. Or, measure the diagonals, and from the angular points of the field draw perpendiculars to the diagonals and measure their lengths: the base and perpendicular of each of the triangles will then be known. 1. Let it be required to determine the content of the field ABCDE, having five sides. Let us suppose that we have measured the diagonals and perpendiculars, and found AC=36.21 ch, EC-39.11 ch, Bb 4.08 ch, Dd=7.26 ch, Aa= E α B 4.19 ch; also Ea=4.00 ch, Ed=13.60 ch, Ab-20.30 ch: 113. To find the content of a long and irregular figure, bounded on one side by a straight line. Suppose the ground, of which the content is required, to be of the form ABEeda, bounded on one side by the right line AE, and on the other by the curve edca. At A and E, the extremities of the right line AE, erect the two perpendiculars Aa, Ee, and on each of them measure the breadth of the land. Then d a b divide the base into any convenient number of equal parts and measure the breadth of the land at each point of division. Add together the intermediate breadths and half the sum of the two extreme ones: then multiply this sum by one of the equal parts of the base line, and the product will be the required area very nearly (Mens. Prob. VI). 1. The breadths of an irregular figure, at five equidistant places, being 8.20 ch, 7.40 ch, 9.20 ch, 10.20 ch, and 8.60 chains, and the whole length 40 chains, required the 2. The length of an irregular piece of land being 21 ch, 5.15 ch, 3.55 ch, 4.12 ch, 5.02 ch, and 6.10 chains: required the area. Ans. 9A 2R 30P. REMARK. If it is not convenient to erect the perpendiculars at equal distances from each other, the areas of the trapezoids, into which the whole figure is divided, must be computed separately their sum will be the required area. PROBLEM VII. 114. To find the area of a piece of ground in the form of a circle. Measure the radius AC: then multiply the square of the radius by 3.1416 (Mens. Prob. A X). 1. To find the area of a circular piece of land, of which the diameter is 25 ch. PROBLEM VIII. Ans. 49A OR 14P. 115. To find the content of a piece of ground in the form of an ellipsis. Measure the semi-axes AE, CE. Then multiply them together, and their product A F EG B by 3.1416. 1. To find the area of an elliptical piece of ground, of which the transverse axis is 16.08 ch, and the conjugate axis 9.72 ch. Ans. 12A 1R 4P. REMARK I. The following is the manner of tracing an ellipse on the ground, when the two axes are known. From C, one of the extremities of the conjugate axis as a centre, and AE half the transverse axis as a radius, describe the arc of a circle cutting AB in the two points |