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.

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

Then as

AB.. AG :: AC.. AD.

10 19.6 36.. 17.64 Perches.

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 F 36 perches from A.

By Problem 1, cut off 500 square perches, by a line

PROBLEM IV.

To cut off, from a Square, or Oblong, any Part thereof.

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

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

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.

Stations. Bearings. D. Pr. N. L.JS. L. JE. DJW. D.

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

D

A

F

B

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..

Then to find GF, by Problem 2, page 173.

CGB sqr. BG- EGF

As 1802.9.. 3753:: 2897.21

GF

= 77.65 perches, nearly,

Then {

77.65 61.26 = 16.39 perches, the anwer. Draw FE parallel to BC, and it is done as required.

By the application of this method, a tract of land may be divided accurately, in any proportion, by a line run