141. The exactness of the surveys will be seen from a comparison of the lines of verification as actually measured, with the lengths of the same lines as determined by calculation. These would be affected by the amount of all the errors in measuring the base lines, in taking the angles, in computing the sides of the triangles, and in making the necessary reductions for the irregularities of surface. A base of verification measured on Romney Marsh in England, was found to differ but about two feet from the length of the same line as deduced from a series of triangles extending more than 60 miles. A base of verification connected with the meridian passing through France, was found not to differ one foot from the result of a calculation which depended on the measurement of a base 400 miles distant. A line of verification of more than 7 miles, on Salisbury Plain, differed scarcely an inch from the length as computed from a system of triangles extending to a base on Hounslow Heath.* *See Note K. SECTION III. LAYING OUT AND DIVIDING LANDS. ART. 142. To those who are familiar with the principles of geometry, it will be unnecessary to give particular rules, for all the various methods of dividing and laying out lands. The following problems may serve as a specimen of the manner in which the business may be conducted in practice. PROBLEM 1. To lay out a given number of acres in the form of a SQUARE. 143. REDUCE THE NUMBER OF ACRES TO SQUARE RODS OR CHAINS, AND EXTRACT THE SQUARE ROOT. This will give one side of the required field. (Mens. 7.) Ex. 1. What is the side of a square piece of land containing 124 acres? Ans. 141 rods. 2. What is the side of a square field which contains 58 acres? PROBLEM II. To lay out a field in the form of a PARALLELOGRAM, when one side and the contents are given. 144. DIVIDE THE NUMBER OF SQUARE RODS OR CHAINS The quotient will (Mens. 7.) BY THE LENGTH OF THE GIVEN SIDE. be a side perpendicular to the given side. Ex. What is the width of a piece of land which is 280 rods long, and which contains 77 acres? Ans. 44 rods. Cor. As a triangle is half a parallelogram of the same base and height, a field may be laid out in the form of a triangle whose area and base are given, by dividing twice the area by the base. The quotient will be the perpendicular from the opposite angle. (Mens. 8. PROBLEM III. To lay out a piece of land in the form of a parallelogram, the length of which shall be to the breadth in a given RATIO. 145. As the length of the parallelogram, to its breadth; So is the area, to the area of a square of the same breadth. The side of the square may then be found by problem I, and the length of the parallelogram by problem II. If BCNM (Fig. 42.) is a square in the right parallelogram ABCD, or in the oblique parallelogram ABC'D', it is evident that AB is to MB or its equal BC, as the area of the parallelogram to that of the square. Ex. If the length of a parallelogram is to its breadth as 7 to 3, and the contents are 52 acres, what is the length and breadth ? PROBLEM IV. The area of a parallelogram being given, to lay it out in such a form, that the length shall exceed the breadth by a given DIFFERence. 146. Let x=BC the breadth of the parallelogram ABCD (Fig. 42.) and the side of the square BCNM. d=AM the difference between the length and breadth. a=the area of the parallelogram. Then a=(x+d)xx=x2+dx. (Mens. 4.) Reducing this equation, we have That is, to the area of the parallelogram, add one fourth of the square of the difference between the length and the breadth, and from the square root of the sum, subtract half the difference of the sides; the remainder will be the breadth of the parallelogram. Ex. If four acres of land be laid out in the form of a parallelogram, the difference of whose sides is 12 rods, what is the breadth? PROBLEM V. To lay out a TRIANGLE whose area and angles are given. 147. CALCULATE THE AREA OF ANY SUPPOSED TRIANGLE WHICH HAS THE SAME ANGLES. THEN AS THE AREA OF THE ASSUMED TRIANGLE, TO THE AREA OF THAT WHICH IS REQUIRED; SO IS THE SQUARE OF ANY SIDE OF THE Former, TO THE SQUARE OF THE CORRESPONDING SIDE OF THE LATTER. If the triangles B'CC' and BCA (Fig. 43.) have equal angles, they are similar figures, and therefore their areas are as the squares of their like sides, for instance, as AC2: CC/2. (Euc. 19. 6.) The square of CC' being found, extracting the square root will give the line itself. To lay out a triangle of which one side and the area are given, divide twice the area by the given side; the quotient will be the length of a perpendicular on this side from the opposite angle. (Mens. 8.) Thus twice the area of ABC (Fig. 45.) divided by the side AB, gives the length of the perpendicular CP. 148. This problem furnishes the means of cutting off, or laying out, a given quantity of land in various forms. Thus, from the triangle ABC, (Fig. 43.) a smaller triangle of a given area may be cut off, by a line parallel to AB. The line CC' being found by the problem, the point C' will be given, from which the parallel line is to be drawn. 149. If the directions of the lines AE and BD, (Fig. 44.) and the length and direction of AB be given; and if it be required to lay off a given area, by a line parallel to AB; let the lines AE and BD be continued to C. The angles of the triangle ABC with the side AB being given, the area may be found. From this subtracting the area of the given trapezoid, the remainder will be the area of the triangle DCE; from which may be found, as before, the point E through which the parallel is to be drawn. If the trapezoid is to be laid off on the other side of AB, its area must be added to ABC, to give the triangle D'CE'. 150. If a piece of land is to be laid off from AB, (Fig. 45.) by a line in a given direction as DE, not parallel to AB; let AC parallel to DE be drawn through one end of AB. The required trapezium consists of two parts, the triangle ABC, and the trapezoid ACED. As the angles and one side of the former are given, its area may be found. Subtracting this from the given area, we have the area of the trapezoid, from which the distance AD may be found by the preceding article. 151. If a given area is to be laid off from AB, (Fig. 46.) by a line proceeding from a given point D; first lay off the trapezoid ABCD. If this be too small, add the triangle DCE; but if the trapezoid be too large, subtract the triangle DCE'. PROBLEM VI. To divide the area of a triangle into parts having given ratios to each other, by lines drawn from one of the angles to the opposite base. 152. DIVIDE THE BASE IN THE SAME PROPORTION AS THE PARTS REQUIRED. If the triangle ABC (Fig. 47.) be divided by the lines CH and CD; the small triangles, having the same height, are to. each other as the bases BH, DH, and AD. (Euc. 1. 6.) PROBLEM VII. To divide an irregular piece of land into any two given parts. 153. Run a line at a venture, near to the true division line required, and find the area of one of the parts. If this be too large or too small, add or subtract, by the preceding articles, a triangle, a trapezoid, or a trapezium, as the case may require. A field may sometimes be conveniently divided by reducing it to a triangle, as in Art. 125, (Fig. 35.) and then dividing the triangle by problem VI. |