tio which corresponds to the part, to be adjacent to CD, and the square of CD, by the other number of the ratio. Add the two products together, and divide the sum by the sum of the numbers expressing the ratio. The square root of the quotient will give FE. Then, As DC-AB: FE-AB:: AD: AF.* EXAMPLES. 1. Let the bearing of AB be N. 14° E. dist. 10 ch. BC, N. 55° E. dist. 18.67 ch.; CD, S. 14° W. dist. 20.98 ch. ; and DA, W. 12.70 ch. ; it is required to divide the trapezoid into two parts by a line FE, parallel to AB or DC, * DEMONSTRATION. Produce DA and CB, Fig. 104, to meet in P. Then (19.6) PDC: PAB:: CD2: AB2, or (17.5) ABCD:: PAB:: CD-AB2 : AB2, or (16.5) ABCD: CD2—AB2 : : PAB ; AB2. In like manner ABEF : FE-AB2 :: PAB: AB2. Hence (11.5) ABCD : CD2-AB2 :: ABEF: FE2 -AB2, or (16.5) ABCD: ABEF:: CD2-AB2: FE-AB2. Therefore m+n : m :: CD2—AB2: FE2-AB2, or multiplying extremes and means, m+n FE2 --m+n. AB2=m. CD2—m. AB2. But (m+n.) AB2=m. AB2+n. AB2. Therefore, adding equals to equals, we have m+n. FE2=m. CD2+n. AB2. Hence the truth of the rule is evident. Construction. Join CA, Fig. 103, and parallel to it, draw BH, meeting DA produced in H. Divide (prob. 17, page 34) HD in K so that HK may be to KD in the given ratio of ABEF to FECD, and draw KL parallel to BC. On CD, describe the semicircle CMD, and draw LM perpendicular to CD. With the radius CM and centre C, describe the arc MN, and from N draw NF parallel to KL. From F, draw the division line FE parallel to AB or CD. For join KC, Fig. 104, and draw KU parallel to AB. Then since BH is parallel to AC, the triangle AHC is equal to ABC; and adding ADC to each, we have CHD=ABCD. Now (4.6) PC: PE :: CD: FE (CM): : (cor. 8.6) CM (EF): CL (UK): : PF : PK. Therefore (11.5) PC: PE:: PF: PK and hence (15.6) the triangle PEF is equal to PCK. Consequently CKD= FECD. But it has been proved that CHD-ABCD; hence taking equals from equals we have HCK=ABEF. But (1.6) HCK: CKD :: HK : KD so that the part ABEF may be to the part FECD as 3 to 2. 2. AB2-200. 3. CD2=1320.4812 5)1520.4812 FE 304.0962=17.44 As 10.98 7.44: 12.70: AF-8.61 2. The boundaries of a trapezoidal field ABCD are given as follow; viz. AB, N. 80° W. 60 per.; BC, N. 39° W. 45.5 per.; CD, S. 80° E. 89.4 per. ; and DA, South, 30 per.; and it is required to divide it into two equal parts by a line FE parallel to AB or CD. What will be the length of the division line FE, and the distance AF? Ans. FE 76.13 per., and AF 16.46 per. PROBLEM XVII. The bearings and distances of the sides AB, BC, CD, DA, Fig. 105, of any quadrilateral tract of land being given, to divide it into two parts having a given ratio, by a line FE, running parallel to one of the sides as AB or CD. RULE. Call the side to which the division line is to be parallel, the parallel side; and the one opposite to this, the opposite side. From the bearings, find the angles. Take the difference between the sum of the angles adjacent to the As the product of the sines of the angles adjacent to the parallel side, Is to the product of the sines of the angles adjacent to the opposite side: So is the square of the opposite side, To a fourth term. Multiply this fourth term by that number of the ratio which corresponds to the part to be adjacent to the parallel side, and to the product add the product of the square of the parallel side by the other number of the ratio; and divide the sum by the sum of the numbers expressing the ratio. The square root of the quotient will be the length of the division line FE. Then, As the sine of P, Is to the sine of E; So is the difference between FE and the parallel side, To the distance of F from the adjacent end of the parallel side.* DEMONSTRATION. Produce DA and CB, Fig. 106, to meet in P. Draw DR and CG each parallel to AB; and let VW, also parallel to AB, make the triangle PVW equal to PCD. Then (15.6) PD : PV : : PW : PC. But (4.6) PD : PV :: DR : VW, and PW : PC :: VW: CG. Therefore (11.5) DR: VW:: VW: CG; and hence (17.6) DR× CG=VW2. But by trigonometry, As sin. CRD (sin. B): sin. C :: CD: DR. Hence (23.6.) sin. Axsin. B: sin. Cx sin. D:: CD2: DR× CG. But by the demonstration to the rule in the last problem, we have m+n. FE2=m. VW2+n. AB2. Hence the truth of the rule is evident. Construction. From C, Fig. 105, draw CG parallel to AB, and on it de EXAMPLES. 1 Let the bearing of AB be North, 12 ch.; BC, N. 56° E. 20.78 ch.; CD, S. 331° E. 22.21 ch.; and DA, S. 80 W. 30 ch.; it is required to divide the tract into two parts by a line FE, parallel to AB, so that the part ABEF may be to the part FECD as 3 to 5. Angle A=801, B=1234°, C=90°, D=66°%, and P=24°. Join CA, and draw BH parallel to it meeting DA produced in H. Divide HD in K, so that HK may be to KD in the given ratio of the part ABEF to FECD. Draw KL parallel to BC and LM perpendicular to CG. With the radius CM and centre C, describe the arc MN. Draw NF parallel to KL and FE parallel to AB. Then will FE be the division line. When the division line is to be parallel to CD, the semicircle must be described on CD; and the line CG need not be drawn. The demonstration of the construction is the same as for the last problem. 2. The boundaries of a field ABCD are given as follow: viz. AB, S. 101° W. 7.20 ch.; BC, S. 67° W. 12.47 ch.; CD, N. 23° W. 13.33 ch.; and DA, S. 89° E. 18 ch.; and it is to be divided into two parts by a line FE, parallel to the side AB, so that the part ABEF may be to the part FECD, as 3 to 4. Required the length of the division line FE, and the distance AF. Ans. FE 10.69 ch.; and AF 7.15 ch. 3. Given the boundaries of a field the same as in the preceding example, to divide it into two parts by a line FE, parallel to the side CD, so that the part ABEF may be to the part FECD, as 3 to 4. Required FE and AF. Ans. FE 10.14, and AF 10.16. PROBLEM XVIII. The bearings and distances of the sides AB, BC, CD, DA, Fig. 107, of any quadrilateral tract of land, being given, to divide it into two parts having a given ratio by a line FE, running a given course from some point in AD to another in BC. RULE. From A, draw AS parallel to BC, meeting FE in S. from the bearings find the angles A, B, C, D, E, and F.* Take the difference between the sum of the angles A and B, and 180° and call it P. Then, As the product of the sines of the angles E and F, : the product of the sines of the angles A and B, :: the square of AB, Also, As the product of the sines of the angles E and F, : the product of the sines of the angles C and D, It is iminaterial whether it is the angle BEF or CEF, that is found, als |
