Solution.-Let it be required to divide the polygon ABCDE into two parts in the ratio of m to n by a line from C. From B draw a parallel to CA, and from D, a parallel to CE, meeting AE produced at the points G and F. Divide GF, as at H, into two parts to each other as m to n. Draw CH as the line sought. Similarly, a polygon may be divided into any number of parts having a given ratio. Cor.-To part off a given area a, we may find GH = a × GF area Pol To part off a given area, independently of the area of the polygon, we proceed as follows: Let ABCDE represent a many.sided field. From the corner C, run a random as near in the direction of the required line of division as can be determined by the eye, conveniently to an opposite corner, as A. Find the area of the part, as CAB, thus parted off, and find the difference between it and the area required to be parted off. Let this difference be denoted in the figure by the triangle ACH d. We shall then 2d CASE 2.-By lines from a given point in a side. viding line required, so that ABPH : Pol. :: m : m + n. Run a random, as PA or PF, parting off an area which is to be computed. Call it a. Find the difference between a and m X Pol. m + n Call the difference d, and apply the formula of Case 1, Cor., to find AH or FH. The consideration whether a is greater or less than required will indicate the direction from A or F in which the distance found by the formula should be measured. Examples.-1 Given the following Field Notes, it is Sta. Bearing. Dist. N. 201⁄2 E. 5.83 1 2 S. 794° E. 10.15 3 S. 272 W. 9.45 4 N. 614° W. 5 N. 152° W. 8.28 required to locate a line from Station 1, which shall divide the field into two equivalent parts. 2. Required a line which shall part off from the same field S A. adjoining the first course. 3. Required a line running from the middle of the first course, which shall divide the same field into parts having the ratio of 2 to 3. 4. Two men own land situated between a road XX and a line YY', and di vided by a line AA'. It is required to run a line BB', at right angles with the road which shall part off areas of equal value from the two portions. Solution.*- Let T be the triangle AOB, and T', the triangle A'OB'. Let of T'. value per acre of T, and v' *By George Key, a pupil of the author. = value per acre Let angle OAB = A, and angle OA'B' = and let AB = x, AA' c, and A0 = 2. We shall then have 22 sin A cos A A' be known; T''v'. By conditions of the problem, Tv = T'r. Whence, from (1) and (2), Then T = FIG. 88. B 100. The essential parts of the Transit, as shown in the cut, are the telescope with its axis and two supports, the circular plates with their attachments, the sockets upon which the plates revolve, the leveling head, and the tripod on which the whole instrument stands. The telescope is from ten to eleven inches long, firmly secured to an axis having its bearings nicely fitted in the standards, and thus enabling the telescope to be moved in either direction, or turned completely around if desired. The different parts of the telescope are shown in Fig. 88. The object-glass, composed of two lenses, so as to show objects without color or distortion, is placed at the end of a slide having two bearings, one at the end of the outer tube, the other in the ring CC, suspended within the tube by four screws, only two of which are shown in the cut. The object-glass is carried out or in by a pinion working in a rack attached to the slide, and thus adjusted to objects either near or remote as desired. The eye-piece is made up of four plano convex lenses, which, beginning at the eyeend, are called respectively the |