Sidebilder
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PROBLEM II.

To cut off from a Triangle, any Part thereof, by a Line running parallel to one of it's Sides.

RULE.

As the area of the triangle,
Is to the square of one side the division line is to be on;
So is the area of the part to be taken off,
To the square of its proportion of that side :
The square-root of which will be the length required.

ExAMPLE.

Suppose the triangle ABC to contain 500 square perches, and it is required to cut off 120 towards A, by a line parallel to BC; the line AB being 40 perches long, and AC 36.

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A G B
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40 × 40 = 1600, square of AB.

2 As 500 . . 1600 : : 120 . . / 384 = 19.6 nearly, the distance from A to G.

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** i =0 ... 19.6 : ; 36 ... 17.64 Perches.

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PROBLEM III.

To cut off from a Triangle, any Part thereof, by a Line issuing from a given Point in one of it's Sides.

ExAMPLE.

In the annexed triangle, containing 800 square perches AB is 50 perches long, and AC 40, and it is required to cut off towards A, 500 square perches, by a line issuing from the point F36 perches from A.

RULE.

By Problem 1, cut off 500 square perches, by a line from B to E.

- AC AE Thus : As 800 . . 40 : : 500 . . 25.

AF AB AE AG Again. As 36 . . 50 : ; 25 . . $4.72 The answer.

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DIVISION OF LAND. “ 175

PROBLEM IV. To cut off, from a Square, or Oblong, any Part thereaf.

Let ABCD be a square, each side whereof is 20 ehains, and it be required to cut off, toward AB, 160 square chains, by a line parallel to AB.

JWote. By Problem 4, page 105, lay off the proposed quantity, thus:

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The method of dividing land, in other forms, will be shewn in examples as they occur.

ExAMPLE 1.
Taken in Perches.

A field, bounded as under specified, is to be divided into two equal parts, by a line parallel to AD or BC. Required the distance from AB, that the division line EF must run.

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JNote. The learner is requested to work all the operations in full, in this and the following examples.

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Continue AB and DC, until they intersect at G ; then in the triangle CBG, are given the side CB 60 perches,

and by

the bearings of the lines all the angles, viz. CBG

100°, CGB 39° 30, GCB 40° 30', to find the other requisites; and by Case 1, Oblique Angled Trigonometry, page

81, the

side BG is found to be 61.26 perches; then by .

Problem 10, page 111, find the area of the triangle CBG,

As radius 90° 40.00000 Is to CB A G0 1,77815 BG J. 61.26 1.78748 So is sine angle B 100° 9.993.35 13.55868 10.00000 - 1. **--> - . 3. To the double area 3619.8 3.55868 Area of CBG 1809.9 square perches. - Add 1087.31 = } the area of the map

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By the application of this method, a tract of land may

be divided accurately, in any Z

proportion, by a line run

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