Theodolite with Telescope fights, as every Practioner will foon find the Advantage; as for the plain Table, it is a proper Inftrument for Learners of this Art to use it for a fmall Inclosure or two, fuch as Gardens, or Ground-plots of Houfes, &c. But 'tis a fhane for any Artift to use this Inftrument to Survey a Gentlemans Eftate. I have Surveyed after those who have used this Inftrument,and I have increased upon him no lefs than two Acres in a Field of 20. This was occafioned by his going out in fuch a Morning that was Foggy, which dampt his Paper, and after he had workt 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 fuffer themfelves to be impofed upon by having fuch Inftruments used in Surveying their Eftates. The Dividing (or cutting 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, &c. with a Line ifluing from any Angle affigned. 2349 RULE. RULE. Triangles confiftir g of equal Bafes, and and in the fame Parallel are equal; therefore take,,, &c. of the Line oppofite to the Angle; draw a Line, which fhall incde a Triangle to contain the parts required... Example. Fig. 102. Admit ABC, to be a Triangle, whole part is required to be cut off, with a Line iffuing 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 Bafe 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 II. To cut off from a Triangle any number of Measures, as 4, 6, 8, 32, &c. with a Line iffuing out of the Argle affigned. RULE. Firft Measure the Area of the whole Triangle, then multiply the fide oppofite to the Angle affigned by the parts to be cut off, and divide the product by the Area of the whole Triangle; the Quotient fhews how much you shall cut off, to make a Triangle to contain the required. Example. Fig. 102. Let ABC be a Triangle given, and let the Propofition be to cut off 84 parts, with a Line iffuing from the Angle B, and falling on the Line AC, and making BC one of the fides of the new Triangle; firft, 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, 46, 47, of the Content; then proceed in all refpects as you did in the laft Problem, and you fhall find the Triangle BCD, to contain 84 parts of the Area of the Triangle ABC, which was required. 2 21 Arithmetically performed. 4 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; fo is the leffer Area 84, to a fourth Number, which is found 10 in the fame Parallel, I 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. To cut off any number of parts, as 20, 40. 60, &c. in a Triangle, Proportional to the Triangle given, with a Line parallel to any fide given. RULE. Firft, Measure the whole Triangle, then Square any of the fides, in which you would have the Parallel to cut; that Square number multiply 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 fhews how much you fhall take of the fide of the Triangle, to make a new Triangle, with which Measure found, fet from B to G. Example. 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 under it for the Denominator; and by abbreviating it, you fhall find the fame to be, then having defcribed 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 diftance from B to E, and fet the fame from B, to wards 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. Firf, Square the fide BC 20, which makes 400; then fay, as 336 is to 400, the Square of that fide, fo is 112 to 133, whofe Square Root is 11 near rational; which is the diftance from B to G; fo a Line drawn from G Parallel to CA, you have the Triangle, BGF, which contains 112, as before. PROLEM IV. 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. |