EXAMPLE. Let AB=60; AD-40; Pe=10; and Ae=15, to draw nPm, dividing the rectangle into two equal parts. 60x15900 the area of Be 20x60-1200 the 300 their diff. 2500 to be cut off 100 As 2400 diff. sqrs.: 300: 50: 6.25=fm PROB. VII. :: 10: 1.25=en From any given point in the boundary of a farm, to run a line that shall cut off a given number of acres from that farm. Estimate as nearly as convenient where the line will run then by Prob. 1. pt. 2 ; find the bearing and 2; length of that line, and the area it cuts off, the difference between which, and the area required, will be had by some one of the problems in this chapter. 38 EXAMPLE I. D C P B Given AD N27° E. dist. 81.13 chains. To draw a line from B to cut off 100 towards A. I is evident the line will fall on AD, and consequently the part cut off will be a triangle. Therefore 2 x 100059.70-33.50-per. Pn; and As S. 4 A 63° : Pn : : Rad. : AP=37C 60 links. hence BP-54.21, bearing North 51° 50′ West. From the middle of AB, in the last example, to draw a line dividing the whole survey into two equal parts. As radius AE:: S. ZA: En=26,60 hence the triangle AED=1079.029—1354.733 half the content 275.704 × 2÷DE (7.89)=72.55=Fm. As sine mDF, 43° 21'. : Fm :: Rad. : DF=11.00 CD. In the same manner, it might be divided into three, four, or any number of parts, by lines drawn from given points in any one of the bounding lines, PROB. VIII. To cut of any number of acres from a farm by a line parallel to one of the sides. Suppose it were required to cut off 100 acres from the farm, in the foregoing examples by a right line parallel to the side AB. Produce the sides AD, BC adjacent to that side which is to be parallel to the division line, till they meet in G; from G on AB let fall the perpendicu.ar GE. The angle AGB is 22°45'; therefore as Sine AGB is to AB : : so is S. ≤ A : to BG=137.55; : and as Rad.: BG :: S, ≤ B [85°45′] to GE=137.17; this multiplied by AB and divided by 2 gives 4094.5.245 for the area of the triangle aGb, from which take 1000 the area to be cut off, and there remains 3094.52, &c. for the area of the triangle aGb; then as triangle ABG GB:: abG: Gb2, hence Gb=119.58, which taken from GB leaves Bb-17.97. In like manner Aa-20.11. and 17 chains 97 links from B to b, the line ab will cut off 100 acres and be parallel to AB. Put AB-a; Co. tang. of A=s; Co. tang. of B=r, and the perpendicular afx; then as 1: :: s: sx=Aƒ; and as 1: x::r: rx=Bd; therefore, fd, or its equal ab is—a—sx+rx and 1⁄2 AB + 1⁄2 ab—a—1⁄2 sx+1/ rx put n-s-r; and R for the area to be cut off. 2a 2R x= and 2 Then ax-nx=R; hence we have x2 n n by completing the square x-a-a-2nR which gives n x=17.92; hence Aa=20. 11, and Bb=17. 97, the same as before. Note. If the angles at A and B were both obtuse, then would n=8+r, and x-a2+2nR-a but if one be obtuse, and the other acute, and the Co. tang. of the ob |