The content being given, and the sum of the length and breadth. Let e represent this content, and 8 this sum. Then the longer side 8+ √(82 — 4 c). = Example. Let the content be 6.4 acres, and the sum 20 chains. The above formula gives the sides of the rectangle 16 chains and 4 chains as before. (494) To lay out triangles. The content and the base being given, divide the former by half the latter to get the height. At any point of the base erect a perpendicular of the length thus obtained, and it will be the vertex of the required triangle. The content being given and the base having to be m times the height, the height will equal the square root of the quotient obtained by dividing twice the given area by m. The content being given and the triangle to be equilateral, take the square root of the content and multiply it by 1.520. The product will be the length of the side required. This rule makes the sides of an equilateral triangle containing one acre to be 4801 links. A quarter of an acre laid out in the same form would have each side 240 links long. An equilateral triangle is very easily set out on the ground, as directed in Art. (90), under "Platting," using a rope or chain for compasses. (495) The content and base being given, and one side having to make a given angle, as B, with the base Fig. 342. the bearing of the side BC being N. 70° E. Here the angle B is found from the Bearings (by Art. (243), reversing one of them) to be 30°. Hence BC=53.33. The figure is on a scale of 50 chains to 1 inch =1:39600. Any right-line figure may be laid out by analogous methods. (496) To lay out circles. Multiply the given content by 7, divide the product by 22, and take the square root of the quotient. This will give the radius, with which the circle can be described. on the ground with a rope or chain. A circle containing one acre has a radius of 1781 links. A circle containing a quarter of an acre will have a radius of 89 links. (497) Town lots. House lots in cities are usually laid off as rectangles of 25 feet front and 100 feet depth, variously combined in blocks. Part of New-York is laid out in blocks 200 feet by 800, each containing 64 lots, and separated by streets, 60 feet wide, running along their long sides, and avenues, 100 feet wide, on their short sides. The eight lots on each short side of the block, front on the avenues, and the remaining forty-eight lots front on the streets. Such a block covers almost precisely 33 acres, and 17 such lots about make an acre. But, allowing for the streets, land laid out into lots, 25 by 100, arranged as above, would contain only 11.9, or not quite 12 lots per acre. Lots in small towns and villages are laid out of greater size and less uniformity. 50 feet by 100 is a frequent size for new villages, the blocks being 200 feet by 500, each therefore containing 20 lots. (498) Land sold for taxes. A case occurring in the State of New-York will serve as an application of the modes of laying out squares and rectangles. Land on which taxes are unpaid is B sold at auction to the lowest bidder; i. e. to him who will accept the smallest portion of it in return for paying the taxes on the whole. The lot in question was originally the east half of the square lot ABCD, containing 500 acres. At a sale for taxes in 1830, 70 acres were bid off, and this area was Fig. 343. N C set off to the purchaser in a square lot, from the north-east corner. Required the side of the square in links. Again, in 1834, 29 acres more were thus sold, to be set off in a strip of equal width around the square previously sold. Required the width of this strip. Once more, in 1839, 42 acres more were sold, to be set off around the preceding piece. Required the dimensions of this third portion. The answer can be proved by calculating if the dimensions of the remaining rectangle will give the content which it should have, viz. 250-(70 +29 +42)=109 Acres. The figure is on a scale of 40 chains to 1 inch =1:31680. 66 (499) New countries. The operations of laying out land for the purposes of settlers, are required on a large scale in new countries, in combination with their survey. There is great difficulty in uniting the necessary precision, rapidity and cheapness. "Triangular Surveying" will ensure the first of these qualities, but is deficient in the last two, and leaves the laying out of lots to be subsequently executed. Compass Surveying" possesses the last two qualities, but not the first. The United States system for surveying and laying out the Public Lands admirably combines an accurate determination of standard lines (Meridians and Parallels) with a cheap and rapid subdivision by compass. The subject is so important and extensive that it will be explained by itself in Part XII. CHAPTER II. PARTING OFF LAND. (500) It is often required to part off from a field, or from an indefinite space, a certain number of acres by a fence or other boundary line, which is also required to run in a particular direction, to start from a certain point, or to fulfil some other condition. The various cases most likely to occur will be here arranged according to these conditions. Both graphical and numerical methods will generally be given.* The given lines will be represented by fine full lines; the lines of construction by broken lines, and the lines of the result by heavy full lines. The given content is always supposed to be reduced to square chains and decimal parts, and the lines to be in chains and decimals. A. BY A LINE PARALLEL TO A SIDE. (501) To part off a rectangle. If the sides of the field adjacent to the given side make right angles with it, the figure parted off by a parallel to the given side will be a rectangle, and its breadth will equal the required content divided by that side, as in Art. (493). If the field be bounded by a curved or zigzag line outside of the given side, find the content between these irregular lines and the given straight side, by the method of offsets, subtract it from the content required to be parted off, and proceed with the remainder as above. The same directions apply to the subsequent problems. Fig. 344. D (502) To part off a parallelogram. If the sides adjacent to the given side be parallel, the figure parted off will be a parallelogram, and its perpendicular width, CE, will be obtained as above. The length of one of the parallel B (503) To part off a trapezoid. When the sides of the field adjacent to the given side are not parallel, the figure parted off will be a trapezoid. When the field or figure is given on the ground, or on a plat, begin as if the sides were parallel, dividing the given content by the base AB. The quotient will be an approximate breadth, CE, or DF; too small if the sides converge, as in the figure, and vice versa. Measure CD. Calculate the content of ABDC. E Fig. 345. F H D Divide the difference of it and the required content by CD. Set off the quotient perpendicular to CD, (in this figure, outside of it,) and it will give a new line, GH, a still nearer approximation to that desired. The operation may be repeated, if found necessary. (504) When the field is given by Bearings, deduce from them, as in Art. (243), the angles at A and B. The required sides will then be given by these formulas: B Fig. 346. When the sides AD and BC diverge, instead of converging, as in the figure, the negative term, in the expression for CD, becomes positive; and in the expressions for both AD and BC, the first factor becomes (CD-AB). The perpendicular breadth of the trapezoid or = = BC. sin. B. = AD. sin. A; Example. Let AB run North, six chains; AD, N. 80° E.; BC, S. 60° E. Let it be required to part off one acre by a fence parallel to AB. Here AB=6.00, ABCD-10 square chains, A = 80°, B=60°. Ans. CD=4.57, AD=1.92, BC= 2.18, and the breadth = 1.89. The figure is on a scale of 4 chains to 1 inch=1:3168. B. BY A LINE PERPENDICULAR TO A SIDE. (505) To part off a triangle. Let FG be the required line. When the field is given on the ground, or on a plat, at any point, as Fig. 347. E DEB. Then the required distance BF, from the angular point |