The square root of which will be the base of the required triangle*. EXAMPLES. 1. Given the area of the triangle ABC, Fig. 90, 500 square perches, and the base AB 40 perches; it is required to cut off 120 sq. per. towards the angle A, by a line DG running parallel to the side BC. 2. Given the area of a triangle ABC, 10 acres, and the base AB 25 ch., to find BD a part of the base, so that a line DG running from the point D, parallel to the side AC, may cut off a triangle BDG containing 4 acres. Ans. BD=16.77 ch†. The truth of this rule is manifest from 19.6. This problem may be neatly constructed as follows: Let ABC, Fig. 90, be the given triangle and AB the given base; on AB describe the semicircle AEB, and take AF to AB in the ratio of the part to be cut off, to the whole triangle; draw FE perpendicular to AB, meeting the semicircle in E, join AE, and make AD equal to AE; from D draw DG parallel to BC and the thing is done. For, join EB and we have, by similar triangles, AB : AE ;: AE: AF; therefore (20.6. cor. 2.) AB : AF :: AB2:AE2 (AD3) :: [19.6.] ABC: ADG. If it be required to produce two sides of a given triangle so far that the triangle formed by these sides produced, and a line drawn between them pa, rallel to the third side, may contain a given area, it may be done by the above rule. Thus, Fig. 90, ADG: ABC :: AD3: AB3. X PROBLEM IX. The bearings of two adjacent sides, AD, AE, Fig. 91, of a tract of land being given, to cut off a triangle ABC containing a given area by a line BC running a given course. RULE. From the given bearings of the lines, find the angles A, B, and C; then, As the rectangle of the sines of the angles A and B, To the square of the side AB*. In like manner the other sides may be found; or having found one side, the others may be found by trig. case 1. EXAMPLES. 1. Let the bearing of AD, Fig. 91, be N. 87° 30' E., and of AE, N. 27° 30' E.; it is required to cut off 10 acres by a line BC running N 38° W. * The truth of this rule is evident from the demonstration to prob. 4, sect. 1, Measuring Land. Construction. Draw AD, AE, Fig. 92, according to the given bearings, and in AD take AF equal the square root of the given area, and on it describe the square AFGH; make IE=AI, and draw ED of the same bearing as the division line BC, meeting AD in D; on AD describe a semicircle, and produce GF to meet it in K, join AK and make AB equal to it; draw BC parallel to DE, and ABC will be the triangle required. For join IF, EF and KD, then (31.3. and cor. 8.6) AD: AK (AB) :: AK (AB): AF; or (cor. 19.6.) AD: AF :: ADE : ABC, but (1.6.) AD : AF :: ADE : AFE; therefore (11.5.) ADE: ABC :: ADE : AFE, and consequently (9.5.) ABC=AFE: but because AI=IE, AFE=2.AFI=(41.1.) AFGH; therefore ABC=AF AD, N 87° 50' E DA, S 87° 30′ W EA, S 27 °30′ W AE, N 27 30' E BC, N 38 00 W CB, S 38 00 E 2. Given the bearing of one side of a tract of land, S. 53° 15' E., and the bearing of an adjacent side taken at the same angle, N. 55° 00′ E., to cut off 4 acres by a line running N. 4° 00' W.; required the distance on the first side. Ans. 9.76 ch. PROBLEM X. The bearings of three adjacent sides, EA, AB, BF, Figs. 93, 94, of a tract of land, and the length of the middle side AB, being given, to cut off a trapezoid ABCD containing a given area, by a line DC, parallel to AB. RULE. Suppose the sides EA, FB, produced to meet in G, and from the given bearings find the angles G, A and B; As the rectangle of the sines of the angles A and G, Is to the rectangle of radius and sine of the angle B; So is twice the area to be cut off, To a fourth term. To or from the square of GA, according as the sum of the given included angles EAB and ABF is greater or less than 180°, add or subtract this fourth term; the square root of the sum or remainder will be the distance GD, and the difference between GD and GA, will be the distance AD*. : * DEMONSTRATION. For, Fig. 93, 94, GAB: GDC :: GAa: GD2: (19.6.) and by division GAB: ABCD :: GA2: GAaGD3, or (15.16.5.) 2.GAB: GA3: 2. ABCD GA GD; but by the demonstration to prob. 4, sect. 1, Measuring Land, sin. A × sin. G : rad. × sin. B. :: 2.GAB : GA3; consequently (11.5.) sin. A sin. G: rad. X sin. B:: 2.ABCD :: GA GD2; hence the truth of the rule is manifest; for if the difference of the squares of GA and GD), be added to the square of GA in Fig. 93, or subtracted from it in Fig. 94, the sum or remainder will be the square of GD. Construction. Let EA, AB, BF, Fig. 95, represent the given sides. On AB make the parallelogram ABHL equal to twice the area to be cut off, produce EA, FB, to meet in G, and on GL describe the semicircle GML; from A, draw AM perpendicular to GL, meeting the semicircle in M, join GM and make GD equal to it; from D, draw DC parallel to AB, then will ABCD equal half the parallelogram ABHL. For join BL, then (1.6.) GAB: ALB :: GA: AL, or 2.GAB: 2.ALB (ABHL) :: GA: AL; but (13.6.) GA: AM :: AM: AL, or (cor. 2.20.6.) GA: AL :: GA3: AM3 (GD3—GA3) :: (by demonstration to the rule) 2.GAB : 2. ABCD : therefore (11.5.) 2.GAB: AB ́HL :: 2.GAB: 2.ABCD; consequently 2.ABCD=ABHL. Nearly in the same manner the area may be laid off on the other side of AB. Thus on AB make the parallelogram ABhl, equal twice the area to be cut off; on GA, describe the semicircle AmG, and from 7 draw Im perpendicular to GA, meeting the semicircle in m; join Gm, make Gd equal to it, and draw de parallel to AB, then will ABcd contain the given area. The demonstration is nearly the same as above. Note. When the side Al of the parallelogram ABhl exceeds GA, it is evident the given area cannot be laid off on this side AB, because it will exceed Note. This problem admits of several other methods of solution, but that contained in the above rule is, perhaps, as simple as any. EXAMPLES. 1. Let the bearing of EA, Fig. 93, be West, AB, N. 10° E., dist. 15 ch., and BF, N. 58° 30′ E.; it is required to cut off 10 acres by a line DC running parallel to AB. EG, N 90°00'W GE, N 90°00'E FG, S 58°30'W |