This method might have been employed in Cases 2 and 3. Those given in the rules are, however, more concise, and are therefore to be preferred. 355. Though the methods illustrated above will serve to supply omissions in all cases where not more than two of the dimensions are unknown, yet it will not be amiss again to impress on the young practitioner the necessity, in all cases in which it is practicable, of determining each side independently of every other. The rules for supplying omissions should only be used in cases where one or more of the data have been accidentally omitted, or have become defaced on the notes. However accurate the field-work may be, there is always a liability to error, and if one side is determined by the rest no means are left of detecting any error. When a side cannot be measured directly, the best way is to determine it by some of the trigonometrical methods, taking the angles and base-lines with great care. In this way a degree of accuracy may be obtained equal to that of the sides measured directly. The latitudes and departures may then be balanced as usual. SECTION IX. CONTENT OF LAND. 356. FROM the bearings and distances of the sides of a tract of land, or from the angles and the lengths of the sides, the area may be found, however numerous the sides may be. This may be done by Problem 4, which is entirely general, it being applicable whatever the number of sides may be, provided they are straight lines. As, however, there are other more concise methods applicable to triangles and quadrilaterals, those are first given. If one or more of the boundaries is irreguiar, instead of multiplying the number of sides by taking the bearings of all the sinuosities of the boundary, it is better to run one or more base lines and take offsets, as directed in chain surveying. The content within the base lines is then to be calculated, and the area cut off by the base lines, being found by the method Art. 256, is to be added to or subtracted from the former area, according as the boundary is without or within the base. As has been already remarked, (Art. 257,) when the tract bounds on a brook or rivulet, the middle of the stream is the boundary, unless otherwise declared in the deed. Lands bordering on tide water go to low-water mark. When the stream, though not tide water, is large, the area is generally limited by the low-water mark, or by the regular banks of the stream. If the farm bounds on a public road, the boundary is, except in special cases, the middle of the road, and the measures are to be taken accordingly. 357. Problem 1.-Given two sides and the included angle of a triangle or parallelogram, to determine the area. Say, As radius is to the sine of the included angle, so is the rectangle of the given sides to double the area of the triangle, or to the area of the parallelogram. DEMONSTRATION.-We have, (Fig. 155,) by Art. 137, As rad. sin. A :: AC: CD :: AB.AC: AB.C.D CD, (Cor. 1.6); but AB. CD = 2 ABC. Fig. 155. с Ex. 2. Given AB N. 37° 14' W. 17.25 chains, and BC N. 74° 29′ W. 10.87 chains, to determine the area of the triangle ABC. Ans. 5 A., 2 R., 28 P. = Ex. 3. Given AB = 23.56 chains, AC 16.42 chains, and the angle A 126° 47'. Required the area of the triangle. Ans. 15 A., 1 R., 38.7 P. 358. Problem 2.-The angles and one side of a triangle being given, to determine the area. Say, As the rectangle of radius and sine of the angle opposite the given side is to the rectangle of the sines of the other angles, so is the square of the given side to double the area. and DEMONSTRATION.-We have (Fig. 155) (23.6). r. rsin. A :: AC: CD (Art. 137), sin. B sin. C:: AC: AB (Art. 139). sin. B: sin. A. sin. C :: AC2: AB. CD, or 2 ABC. EXAMPLES. Ex. 1. Given AB = 21.62 chains, and the angle A= 47° 56' and B = 76° 15', to find the area. Ex. 2. Given AB 17.63 chains, and the angle A= 63° 52′ and B 73° 47', to find the area. Ans. 19 A., 3 R., 22 P. Ex. 3. Given one side 15.65 chains, and the adjacent angles 63° 17′ and 59° 12', to determine the area of the triangle. Ans. 11 A., 0 R., 22 P. 359. Problem 3.-To determine the area of a trapezium, three sides and the two included angles being given. RULE. 1. Consider two adjacent sides and their contained angle as the sides and included angle of a triangle, and find its double area by Prob. 1. 2. In like manner, find the double area of a triangle of which the two other adjacent sides and their contained angle are two sides and the included angle. 3. Take the difference between the sum of the given angles and 180°, and consider the two opposite given sides and this difference as two sides and the included angle of a triangle, and find its double area. 4. If the sum of the given angles is greater than 180°, add this third area to the sum of the others; but if the sum of the given angles is less than 180°, subtract the third area from the sum of the others: the result will be double the area of the trapezium. DEMONSTRATION.-Let ABCD (Figs. 156, 157) be the trapezium, of which AB, BC, and CD, and the angles B and C, are given. Join BD, and draw DE and CG perpendicular to Then will 2 ABD - AB.DE =AB (EFDF) =AB. EFAB.DF = 2 ABC 2 CDH. B E A Fig. 156. H Whence 2 ABCD 2 BDC+2 ADB = 2 BCD+ 2 ABC ± 2 CDH: the plus sign being used (Fig. 157) when the sum of the angles is greater than 180°. EXAMPLES. = Ex. 1. Given AB 6.95 chains, BC 8.37 chains, CD = = 5.43 chains, ABC = 85° 17', and BCD = 54° 12', to find the area of the trapezium. Ex. 2. Given AB S. 27° E. 12.47 chains, BC N. 66° E. 11.43, and CD N. 8° W. 9.16 chains, to find the area of the trapezium. Ans. 14 A., 0 R., 1.56 P. Ex. 3. Given AB S. 45° W. 8.63 chains, BC S. 86° 30' E. 9.27 chains, and CD N. 34° E. 11.23 chains, to find the area of the trapezium. Ans. 6 A., 2 R., 9 P. |