side of the triangle being 1280 links, and the perpendicular on it, from the opposite corner, 750. Ans. Length of base 320 links. 38. PROBLEM XI.-To cut off any portion from a triangular field, by a line drawn from a point in one of its sides. Let ABC be the given triangle, and D the given point. Cut off a part ACE, by last problem, of the required content. Join DE, and through C draw CF parallel to DE; draw DF, and it is A the line required. E F For triangle DFE = DEC (Pl. Geom. I. 37), and hence triangle ADF = ACE: the required area. A 39. PROBLEM XII.-To cut off a part from a triangle, by a line parallel to one of its sides. Let ABC be the given triangle, and BC the side to which the required line is to be parallel. Let A area of the given triangle ABC, a = area of the required triangle ADE, S= the side AB, s the side AD. & S2 Then A: a S2: s2, and s2 = A = or s = B E ... D B Hence finds, and make AD equal to it, and through D draw DE parallel to BC, and it will be the required triangle. EXAMPLE. The area of a triangle ABC is 5 acres 2 roods 15 poles, the side AB is 1525 links; required the length of AD, so that the triangle ADE shall contain 2 acres I rood 10 poles. A5 ac. 2 ro. 15 pls. =559375 links a=2 ac. 1 ro. 10 pls. = 231250 α Hence, s=√ = 1525√ 231250 980-5 lks. = AD. EXERCISE. The area of a triangle is 12.96 acres, its side AB is 1200 links; required the length of AD, so that the triangle ADE shall contain 3.24 acres. Ans. 600 links. 40. PROBLEM V.-To cut off from a quadrilateral any portion of surface, by a line drawn from one of its angles, or from a point in one of its sides. Let ABCD be the quadrilateral. 1. Let A be the angle from which the line is to be drawn. B A Draw the diagonal DB, and cut off a part, DE, from it that has the same proportion to DB that the required part has to the quadrilateral; draw AE, EC; then AECD is equal to the required area. Rectify the crooked boundary AEF by drawing AF (Prac. Geom. art. 130), and it is the required line which cuts off the part AFD = AECD = the given area. 2. When it is required to draw the dividing line from a point, G, in one side. Let ABCDEF be the given polygon. 1. Let A be the point from which the line is to be drawn. E P. F Draw AF, by the preceding case, then join GF, and through A draw a line parallel to GF, cutting CD in H, and a line joining G and H will be the required line. H 41. PROBLEM VI.-To cut off any part of the area of a given polygon, by a line drawn from any of its angles, or from a point in one of its sides. F Draw the diagonals FB, FC, FD. Cut BF in G, so that A: α = B BF: BG, A and a being the areas of the polygon and of the part required. Join AG and GC. Cut DF in H, so that A: a=DF: DH, and join CH and HE. Then the crooked boundary AGCHE evidently cuts off an area equal I D E to that required, for the triangle AGB is the same part of ABF that a is of A, and BGC the same part of FBC, and so on. Hence, rectify the crooked boundary AGCHE, by drawing from A the straight line AI, and AICB is the required part. 2. When the line is to be drawn from a point P in one of the sides. Draw AI, as in the first case, then from P draw another line to be determined, as GH in the preceding problem. 42. PROBLEM VII.-To cut off any proposed portion from a field with curvilineal boundaries, by a line from a point in its boundary, or by a line parallel to a given line. E Let ABCE be the given field. 1. When the line is to be drawn from a point in the boundary A. A Draw a trial line AC, and measure the area of the part cut off, AEDC. If it is too great, divide the excess in square links by the length of AC in links, and make the perpendicular GF to AC equal to twice the quotient; draw GD parallel to AC; join AD, and AD is the required line, For the area of the triangle ADC, considering CD as a straight line, is = AC GF, and therefore equal to the M N R V W B excess. 2. When the dividing line is to be parallel to a given line MN. Draw a trial line PQ parallel to MN, to cut off a portion PBQ equal to the required part, and measure it. If it is too small, find the defect in square links, and divide it by the length of PQ in links, and make the perpendicular VW to PQ equal to the quotient; and through W draw RS parallel to PQ, and it will be the required line when PR and QS are either parallel or equally inclined to PQ. When they are not so, a small correction may be required to be made, by drawing RS a little nearer or a little farther from PQ. 43. PROBLEM VIII.-To divide a field into portions having any required ratio, by lines drawn from a point within the field. B A Let ABCDE be the given field, and P the given point. Divide the area of the whole field into the required number of parts having the required ratio. Draw any line PÅ, and draw PC by last problem, so that APCB shall be equal to one of the required parts; then draw PE, so that APE shall be equal to another, and so on for the remaining parts. If the field, for example, is to be divided into three parts proportional to the numbers n, n', n", and if their areas be denoted by a, a', a", and A = area of the whole field, then A = a + a + a" and if N = n + n'+n", N:n A:a, and a An N An' N An" N:n" A:a", and a" = N From which expressions, a, a', a", can be found. Thus, if A 25 acres, and n, n', n", be 2, 3, and 5, 25 x 2 a = 10 25x3 respectively, then a = =5 acres, =7.5 10 = acres, and a" = N: n'A: a', and a' = = 25 × 5 10 Hence, in this example, and EPCD 12.5 acres. = F 12.5 acres. make APCB 5 acres, APE 7·5, INCLINED LANDS. 44. When the surface of a field is inclined, it is not that surface, but the surface of its projection on a horizontal plane, that is taken as its area. This projection is just the quantity of surface on a horizontal plane, determined by drawing perpendiculars upon it from every point in the boundary of the field; or, in other words, by projecting its boundaries on a horizontal plane; and a plan of this projection only is made; it is impossible to construct a plan of a curved surface on one plane. The surface of this plan is manifestly less than that of the inclined or curved surface, but the advantage in point of quantity of the latter surface is counterbalanced by the circumstance that scarcely any more grain will grow on an inclined or curved surface than upon the horizontal plane or projection that may be considered as its base; for grain grows in a vertical direction, and the stalks therefore have no more space on an inclined than on a plane surface, although the ears may have freer air; but any trifling advantage of this kind is compensated by the additional labour of cultivating it. An exception may perhaps properly be made in regard to common pasture, which probably grows in the same abundance on curved surfaces as on level ground of equal extent. The area of the horizontal projection can easily be computed by measuring the angle of acclivity of the field at different places. Thus, if ABCD is a vertical section of the field, then if AB is measured, and the angle of elevation A, the horizontal projection AE of AB is found thus: A E D B G H = Rad: cos AAB: AE and AE AB⚫ cos A when rad = 1. Thus, if AB = 1200 links, and angle A 15° 40', AE = 1200 × 96284901155-4188, or 1155 links, is the length of AE on the plan, which must also be taken for its length in computing the area. So if BC and angle CBF be measured, BF, or its projection EG, can be found; then AG = AE+ EG, is the projection of AB and BC. In the same way the other dimensions of the projection can be found; and if a theodolite is used for measuring any of the angles contained by lines measured on the field, these being horizontal angles on the instrument, are just the angles of the projection, and are to be used unaltered for constructing the plan. SURVEY OF A ROAD AND ADJOINING FIELDS. 45. Construct a plan of a road and adjoining fields the subjoined field-book. om |