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, such as Gardens, or Ground-plots of Houses, &c. But 'tis a shame for any Artist to use this Instrument to Survey a Gentlemans Estate. 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 fuch a Morning that was Foggy, which dampt his Paper, and after he had workt about two Hours the Sun shone 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 themselves to be impofed upon by having fuch Instruments used in Surveying their Estates. The Dividing (or cutting off) both Right lined and Irregular Figures, into as many parts as you hall require, equal or unequal. To cut off from a Triangle any parts, as,&e. with a Line issuing from any Angle assigned. 2342 RULE. RULE. Triangles confifting of equal Bafes, and and in the fame Parallel are equal; therefore take 4,&c. of the Line oppofite to the Angle; draw a Line, which shall inchiede a Triangle to contain the parts re quired... C Example. Fig. 102. Admit ABC, to be a Triangle, whote part is required to be cut off, 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 II. To cut off from a Triangle any number of Measures, as 4, 6, 8, 32, &c. with a Line issuing out of the Argle affigned. RULE. First Measure the Area of the whole Ti angle, then multiply the side opposite to the Argle affigned by the parts to be cut off, and divide the product by the Area of the whole Triangle, the Quotient shews 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 Proposition be to cut off 84, parts, with a Line issuing from the Angle B, and falling on the Line AC, and making BC one of the fides 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, of the Content; then proceed in all refpects as you did in the laft Problem, and you shall find the Triangle BCD, to contain 84 parts of the Area of the Triangle ABC, which was required. 4& 42 2 1 Arithmetically performed. Firft, The Content or Area of the whole Triangle ABC, being found to contain 336, and the Line AC 42; then say by the Rule of Proportion, as the whole Area 336, is to 42; so is the leffer Area 84, to a fourth Number, which is found 10 in the same 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. First, 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 shews how much you shall take of the fide of the Triangle, to make a new Triangle; with which Measure found, set 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 shall find the fame 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 EGF doth contain 112, as was required. : Arithmetically performed. L First, Square the fide BC 20, which makes 400; then say, as 336 is to 400, the Square of that fide, so is 112 to 133, whose Square Root is 11 near rational; which is the distance from B to G; so 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. |