PRACTICAL EXAMPLES. The following examples should be examined by resolving the bearings and distances as explained at page 147, and when corrected, plotted carefully, for the purpose of finding the areas. Ex. 1. Required the area of a tract of land, whose bearings and distances are as follows, viz.:-1st, N. 15° 42′ E., 6·20 chains; 2nd, N. 52° 18' E., 6.75 chains; 3rd, S. 78° 48' E., 5.96 chains; 4th, S. 5° 51′ E., 4.84 chains; 5th, S. 49° 15' W., 4.75 chains; 6th, S. 4° 57′ E., 3.98 chains; 7th, S. 71° 24' W., chains; and 8th, N. 46° 18′ W., the place of beginning. Ans. To the 7th Bearing, the distance is 5.55 chains. to Ex. 2. Given the following bearings and distances to find the area, viz.:-1st, N. 10° 21' W., 4.50 chains; 2nd, N. 9°48' E., 5.20 chains; 3rd, N. 75° W., 3'00 chains; 4th, N. 20° 3′ E., 4.86 chains; 5th, S. 45° E., 5-20 chains; 6th, N. 3° 18′ W., 2.50 chains; 7th, E., 7.00 chains; 8th, S. 12° W., 3-94 chains; 9th, S. 43° E., 4.15 chains; 10th, S. 46° 57′ W., 8-20 chains; 11th, S. 29° E., 9.15 chains; 12th, S. 48° W., 4.56 chains; 13th, N. 19° W., 3·42 chains; 14th, S. 28° W., 8.54 chains; 15th, N. 53° W., 4.60 chains; and 16th,,,, to the place of beginning. Ans. 16th Bearing is N. 12° 18' E. And the Distance, 11.08 chains. CHAP. IV. DIVISION OF LAND. PROBLEM I. To lay out a given area in a rectangular form, having the length to the breadth in a given ratio. RULE. Divide the area by the product of the ratios; take the root, and multiply each side by its ratio. EXAMPLE 1. There are 2000 acres of land to be laid out in a rectangular form, whose sides are to each other, as 4 is to 5; what will their lengths be? m 4, n=5 and mn=20. A= 20,000 sq. chains. EXAMPLE 2. One side of a rectangular field is double A. R. P. the other, what are the sides, when the area is 20. 0. 19.' perches. Ans. 10.03 chs. and 20·06 chs. EXAMPLE 3. A man has a farm of 150 acres, of a rectangular form, the depth of the farm is 50 chains. He is desirous of adding 50 acres to it, having the same depth, to make up 200. What will be the length of frontage of his farm, after the addition? Ans. Half a mile. PROBLEM II. When one of the sides is a certain length longer than the other. To obtain the smaller side, square half the difference of the two sides, add it to the given area, and taking the root of their sum, from this root substract half the difference. EXAMPLE. Given 464 acres, it is required to lay it out in a rectangular form, the one side being 6 chains longer than the other. Ans. One side 65.18 chs. ; the other 71.18 chs. PROBLEM III. From a given block of land, with parallel sides, the angle of inclination of the front and sides being given, to cut off a given area, by a line parallel to the sides. RULE. Find first the perpendicular depth of the given block, by constructing the figure and measuring off the perpendicular, then, as all parallelograms upon equal bases, and between the same parallels, are equal, divide the given area by this depth for the frontage required. EXAMPLE 1. The concession road, of a certain township, bears N. 74° E., while the side lines bear N. 9° W. The length of the side lines is 66-66 chains, and the fronts of the lots are 30 chains. Required the frontage that must be taken, to cut off 100 acres, by a line parallel to the side lines. The side lines, bearing N. 9° W., and the fronts, N. 74° E., the included angle, or angle of inclination, of the front to the sides, is 9° 74°, or 83°. The perpend. depth is found to be 66.16 chains whence and 100 acres = 1000 square chains 1000 66.16 15.12 the frontage required. Ans. 15.12 chains. EXAMPLB 2. When the fronts of the lots bear S. 80° W., and the side lines N. 15° E., the side lines being 101.50 chains long, and the concession road frontage 20.00 chains, what will be the frontage required, to divide the whole lot between A and B, giving A 50 acres more than B, and and what quantity will each have? A, Ans. {1, 117 acres, and 12-72 chains frontage. PROBLEM. IV. From a given triangle, to cut off any given area by a line drawn from the vertex to the base. Triangles, of equal altitudes, are proportional to their bases (see Theorem 17, page 15); therefore making A, the given area of the triangle; a, the part to be cut off; and X, the required portion of the base, we have EXAMPLE 1. There is a Gore of land between two townships, whose area is 425 acres, and base is 85 chains. It is required to cut off 400 acres by a line drawn from the vertex. As 425: 400 :: 85: (x) chains. Where x = length of new base, = 80 chains. EXAMPLE 2. From a Gore of land, having a base of 60 chains, containing 125 acres, required the base, to cut off 50 acres. Answer 16 chains. EXAMPLE. 3. From a triangle, with a base of 74.54 chains, containing 35 acres, required the base to cut off 860 square yards. Answer 8 yards nearly. |