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Through a given point in a rectangular parallelogram, to draw, if possible, a right line which shall divide the same into two assigned parts.
Through the given point P draw ePf parallel to AB, or DC; find the area of the rectangle eB: then say, as the difference of the squares of Pe and Pf is to the difference between the area to be cut off, and the rectangle Be, so is Pe to en, or Pf to fm.
Let AB=60; AD=40; Pe=10; and Ae=15, to draw nPm, dividing the rectangle into two equal parts.
60x15 -900 the area of Be
300 their diff.
As 2400=diff. sqrs. : 300 :: 50 : 6.25=fm
: 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 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.
Given AD N 27° E. dist. 81.13 chains.
DC S37 E. 42.00
A 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 1000 - 59.70=33.50=per. Pn; and As S. 4 A 63° : Pn :: Rad. : AP=370 60 links. hence BP=51.21, 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. LA: En=26.60 hence the
C. triangle AED=1079.029—1351.733 half the content 275.704 * 2-DE (7.89)=72.55=Fm.
As sine < mDF,43° 21', : Fm:: Rad. : DF=11.00
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,
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 2 AGB : is to AB : : so is S. ZA : to BG=137.55; and as Rad. : BG :: S, ZB (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:Gh2, hence Gb=119.58, which taken from
GB leaves Bb - 17.97. In like manner Aa=20.11.