Theodolite with Telescope fights, as every Practioner will soon find the Advantage ; as for the plain Table, it is a proper Instrument for Learners of this Art to use it for a small Inclosure or two, such as Gardens, or Ground-plots of Hufes, &c. But 'tis a a fhaine for any Artist to use this Inftrument to Survey a Gentlemans Eftate." I have Surveyed after those who have used this Instrument and I have increased upon him no less than two Acres in a Field of 20. This was occafioned by his going out in such a Morning that was Foggy, which damp: his Paper, and after he had works about two Hours the Sun fhone out and diminished his plot upon his plain Table to the abovementioned two Acres. The like Error has been committed by the Circumferencor. I hope Gentlemen and others will take this into their Confideration, and not suffer themselves to be imposed upon by having such Instruments used in Surveying their Estates.

The Dividing (or cuting off) both Right lined

and Irregular Figures, into as many parts as you shall require, equal or unequal.

To cut off from a Triangle any parts, as Esc. with a Line ifluing from any -Angle afligned.

III
2 3 42

RULE,

RULE.

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Triangles consistirg of equal Bases, and and in the same Parallel are equal; therefore take , , 8c. of the Line opposite

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À& to the Angle ; draw a Line, which thall in: chide a Triangle to contain the parts re- . quind.

Example Fig. 102. Admit ABC, to be a Triangle, whole part is required to

be cut oft, with a Line issuing cut of the Angle B, to cut the Line AC, and then will AB be one fide of the Triangle : Then let the part of the Triangles Base be taken, which endeth in D, and let the Line BD be drawn, which includeth the Triangle ABD, which is the

part of the Triangle ABC.

PROBLEM IT.

To cut off froin a Triangle any number of Measures, as. 4, 6, 8, 32, c. with a Line issuing out of the Argle assigned.

RU LE

First Measure the Area of the whole Tri. angle, then multiply the side opposite to

the Argle assigned by the parts to be cut off, and divide the product by the Area of the whole Triangle; the Quotient lhews how much you

Thall cut off, to make a Triangle to contain the required.

Example. Fig. 102. Let ABC be a Triangle given, and let the Proposition be to cut off 84 parts, with a Line issuing from the Angle B, and falling on the Line AC, and mak. ing BC one of the sides of the new Triangle ; first, the whole content of the Triangle ABC, is found to be 336" Having proceeded thus, let 84 be the Numerator, and 336 the Denominator, which being abbreviated thus, 176, 177, of the Content, then proceed in all respects as you did in the last Problem, and you fall find the Triangle BCD, to contain 84 parts of the Area of the Triangle ABC, which was required.

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42

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Arithmetically performed. Firft, The Content or Area of the whole Triangle ABC, being found to contain 336, and the Line AC 42; then fay by the Rule of Proportion, as the whole Area 336, is to 42 ; so is the lesser Area 84, to a fourth Number, which is found 10 in the same

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Parallel, Parallel, which fet from C towards A, which falleth in D: Then draw the Line BD, which Triangle BCD, contains 84 parts, the thing required.

PROBLEM III.

a

To cut off any number of parts, as 20, 40. 60, 8c. in a Triangle, Proportional to the Triangle given, with a Line paral. lel to any side given.

Firfi, Measure the whole Triangle, then Square any of the sides, in which you would have the Parallel to cut; that Square number enultiply by the parts given to be cut off, and divide the product by the Area of the whole Figure; out of which Quotient extract the Square Root, and it Thews how much you shall take of the fide of the Triangle, to make a new Triangle ; with which Measure found, fet from B to G.

Examples Fig. 103. Let ABC be a Triangle given, from which 112 is to be cut off with a Line Parallel to the Line CA; the Triangle being measured, and found to be 336,

then put 112 over for the Numerator, and 336 ander it tor the Denominator; and by abbreviating it, you shall find the same to be , then having described the Semi-Circle on the Line AB, divide the Line AB into -3 equal parts, and from one of them erect the Perpendicular DE; then take the distance from B to E, and set the fame from B, towards A, which endeth in. F; by which point draw a Line Parallel to AC: So the Triangle BGF doth contain 112, as was required.

Arithmetically performed. Fird, Square the side BC 20, which makes 400;

then say, as 335 is to 400, the Square of that fide, fo is 112 to 1331, whose Square Root is 11 near rational, which is the di

B G; Line drawn from

, have the Triangle BGF, which contains 112, as before.

G Parallel to CA

From a Triangle given, to lay the parts cut off in a Trapezium; if there be a Proportion given between the parts cut off, and the whole Figure.