area of the trapezium A B C D in the usual way, which united areas are found to be 570000 square links, the line A D being 810 links. Hence 6 acres = 600000 square links Area ABCD and offsets = 570000 ditto 9)3000.0 81.0 9)333.3 37.037 2 74.074 links = perpend. Pr. Therefore, apply the perpendicular Pr= 74.074 links as in Prob. IV., and range the straight line AP; then shall the 6 acres be parted from the field or common as required. PROBLEM XII. To divide from a field or common, bounded by straight fences, any given quantity of land, by a line parallel to one of the straight fences. RULE.-Measure the length of the straight fence or line to which the division line is required to be parallel; divide the given quantity by that length, and set the quotient perpendicular to the given straight fence, at two points near the ends thereof, ranging a line through the extremities of the two perpendiculars, and measuring the length thereof; find the area of the trapezoid thus obtained, and take the difference between this area and the given area, which difference, being divided by the last measured line, will give the breadth to be set out perpendicularly from the last measured line, either inwardly or outwardly, accordingly as the difference is in excess or defect of the given quantity. EXAMPLE Let a A Bb be a portion of a field or common with straight fences, it is required to lay out 10a. 2r. 16p. by a line parallel to AB, the length of which is 2200 links. (1060000-1018225) 2025 20.6 links, the distance to be set perpendicularly, at both ends of mn, to the right thereof, because the area cut off by mn is less than the given quanity, and the straight line CD ranged through the extremities of these perpendiculars will be parallel to AB, and will cut off the required quantity. NOTE 1. The three last Problems are much used in the inclosure of commons, and other waste lands, and more especially Problem XII. as the laying out inclosures with parallel fences contributes very much to the economy of cultivation, by thus allowing the ridges to run from the ends of each inclosure, parallel to the two parallel fences. A NOTE 2. Those who are accustomed to operations in analytical trigonometry, will find the following formulæ much more expeditious for laying out land by means of parallel fences, as in Prob. XII., especially where a great number of successive inclosures are required to be laid out:-Let A B = a, σ = sine LA s=sine LB, S = sine (σ + s), and A = area ABCD, which is required to be cut off; then, Whence both B C and AD become known without any preliminary measure. ment, except that of the side A B and the angles A and B: and by adding the next area to be cut off to A, and calling the sum B, and then substituting B for A in the formulæ for B C, the next following distances Bb and Aa may be found; and so on for any number of inclosures.-When A and B are supple. mental angles, the lines Aa, Bb, will be parallel, which makes the Problem extremely simple; and when the sum of the angles at A and B are greater than two right angles, S becomes negative, and the lines Aa, Bb, will meet on the other side of A B. In this case the formulæ will become NOTE 3. Investigation of the preceding formula. As the formulæ given in Note 2, have never been previously given by any author, it will be proper to give their investigation here.-Let E be the point where Aa, Bb, would meet, if prolonged; (this point is not shewn in the figure), then A E = ∞, BE= ασ S' area of the triangle ABE➡ αξεσ and the area of the triangle CDE= A. Now because the triangles ABE, CDE, are similar, area A ABE: area A CDE:: BE2: CE2, that is, BC= BC. Also BE: BC:: AE: AD= ABC = The other modification of the formula for B C, when the point E falls on the other side of A B, is derived by making S negative. SECTION II. ON DIVIDING LANDS OR COMMONS, OF EITHER EQUAL OR VARIABLE VALUE, AMONG VARIOUS CLAIMANTS, IN PROPORTION TO THEIR SEVERAL CLAIMS. Where land is the property of joint purchasers, co-heirs, co-partners, &c.; or where a common is to be divided among the adjoining proprietors in proportion to the values of their several estates; it becomes requisite to adopt the methods of division given in the following Problems, in conjunction with the Problems in the preceding Section. The present Section will, therefore, present cases of considerable complexity; but in order that these may be clearly understood, the simplest cases shall be first presented to the student. PROBLEM I. To divide a rectangular piece of ground of equal value throughout, either equally or unequally, among any given number of claimants, by fences parallel to one of its sides. CASE I.-If the parts into which the rectangular space is to be divided be equal, it will be only required to divide two of the opposite sides of the rectangle into the given number of equal parts, and range the lines for the several fences to the consecutive points of division, and the land will then be divided as required. This appears to be too simple a case to require a diagram. ČASE II.-If the rectangular space is to be divided among several joint purchasers, who have paid unequal sums for the purchase thereof: then use the following RULE. As the sum of all the sums paid, is to the length of the side of the rectangle from which the division lines abut, so is any one person's sum to the breadth on the side of the rectangle due to that person; then divide both this side and the one opposite to it, by laying off the resulting breadths, and range lines to the corresponding points, and the rectangle will be divided by parallel lines, as required. NOTE. This is obviously a question of single fellowship. Divide the rectangle ABCD, the length of which is 1470 links and its breadth 684 links, among three joint-purchasers P, Q and R, who paid for the purchase thereof respectively EXAMPLE. D с d C B £120, £150, £220. Whole breadth 1470 which proves the work. NOTE. In this operation the breadth of the rectangle, which is common to all the three divisions, is not required to be used. PROBLEM II. To divide a triangle of equal value throughout, either equally or unequally, among several claimants, who shall all have the use of the same watering-place, situated at one of the angles of the triangular field. REESE LIBRARY OF THE UNIVERSITY A a b C being 1000, A C B EXAMPLE. It is required to divide the triangular field ABC among three persons, whose claims therein are as the numbers 2, 3 and 5, so that they may all have the use of a wateringplace situate at the angle C; AB 685, and C B 610 links. The rule in this Problem is the same as in the last. - : = A a, 200 which are the portions of the base A B, belonging to the respective claimants; therefore, if lines be drawn from a and b to C, the triangular field will be divided in the required proportion, each claimant having the use of the watering-place at C. NOTE. In solving this Problem it is not necessary to use the lengths of the sides AC, CB, because all the three triangles A Ca, a Cb, bCB have a com⚫ mon perpendicular; and, therefore, their areas are as their bases. PROBLEM III. To divide a triangular field of equal value throughout, either equally or unequally, among sundry claimants, by fences running from any given point in one of its sides. The method of solving this Problem will be best shewn by the following EXAMPLE. Divide the triangular field ABC, the sides of which measure 30 chains = AB, 23 = B C, and 19 = A C, equally among three persons, by fences running from an occupation road that meets the side AB at H, which is 14 chains from A, that all the three persons may have the use of the road at H. Divide AB into three equal parts in the points D, E; from H, (the point where the road meets A B) draw HC; parallel to which draw DF, EG meeting AC, BC respectively, in F and G; and join HF and HG, in which directions fences being made will divide the triangle as required. |
