Example. Fig. 104. Let the Triargle ABC be given, from whence is to be cut off in a Trapizium: Fift then on the Line CB describe à Semi-Circle ; then divide CB into 5 equal parts, and of 3 of thein from B, erect the i Perpendicular DE, then setting one Foot of your Compasses in B, extend the other to E, which distance fer off from B towards C, which endeth in F; by which point draw a Line Parallel to AC: So I include the Tri. angle GFB, to contain of the whole : And consequently the Trapezium ACFG doth contains of the same as was required. Arithmetically performed. First take from the whole, the remainder will be }; then by the last Problem inake the Triangle GFB to contain ș, and then it will follow, that the Trapezium ACFG muft contain of the whole Triangle,which was required. PROBLEM Y. To divide a Triangle, when the Line of Partition goes not parallel with any lide, take this Eximple. Fig 105. Let ABC be a Triangle to be divided into two parts which shall bear Proportion to one another, às 3 and 2, by a Line drawn from the point D in the Bife or Line AC. Fro:n the limited point draw a Line to the Angle B.; then divide the Base AC into five equal parts, and from the third point of Division draw the Line to E, Parallel to BD. Laftly, from E draw the Line ED, so shall the Trapeziumn ABED be in Content as three to two, to the new Triangle DEC. I have now done with the Division of Tri. angles, when I have added these three Advertisements. First, You must be sure to take very exactly the distance of every Point, where a dividing Line cutteth any fide, to one of the ends of the fame fide, as in the laft Figure, the distance BE and AD, which diftances beirg applied to the Scale by which the Figure was protracted with, will thew at how many Chains and Links end, you are to make your dividing Line on the Field itselt. Second, The proportions by which you are to divide, are not always so formally given as in the former . Example, but are sometime to be found out by Arithmetical working, as in this case. I 3 Suppose Suppose a Triangle of 6 Acres, 2 Roods, 31 Perches, must be divided, so as the one of the 2 parts shall be 4. Acres 3 Roods, and s Perches, and the other consequently 1 Acre, 3 Roods, and 26 Perches; reduce both mealures into Perches, and the one will be 765, and the other 306. There Sum is 1671; which by their common measure being reduced into their lowest terms of Pro portion in whole numbers, will be 5, 2, and 7, which shews that the Triangle being di. vided in 7 Equal parts, the one must have 5 of thofe 7 parts, and the other 2. And obferve that it will be sufficient to find the common measure betweep the Sum of the Terms, and either of the Terins; the Method whereof is shewed in every Arithmatical Book for reducing Fractions into their lowest Terins. Third, In these and all other Divisions of Land, where a strict Proportion of Quantity is to be observed, you must have refpect to the Rules hereafter delivered. But if there be any useful Pond or Well to draw the Line of Division through ; but if there be an unuseful Pond, Lake or Puddle; or if there be any Boggy or barren Ground, that muft be cast out in the Divifion ; meafure that first, and substract it from the Content of the whole Close, and then lay the just Quantity of the remainder on that fide that is free from it, that the other inay have his just part also, besides that 'which is useless. PROBLEM VI. To cut off from a Square any part propounded in a Parallelogram. Divide the parts to be cut off, by the side of the Square; the Quotient shews how much of the side of the Square you shall take for the Breadth of the Parallelogram; at which distance draw a Parallel Line, which thall include the Parallelogram required. Example. Fig. 106. Admit ABCD to be a Square given, whose fide is 20; from whence 160 parts is to be cut off with a Line Parallel to CD, so then CD makes one of the sides of the Parallelogram; then work as is before taught, and draw the Line EF; which includeth the Parallelogram (DEF, and contains 160, the parts required. Note, And if you would have cut off , 15,8 C. then you must divide the Square Side to be cut off into these proportional parts, and so by those parts draw a Parallel Line, which would have included a Parallelo1 4 gram gram to have contained the parts Proportionable, PROBLEM VII. From an Irregular Figure to cut off any parts required. R U L E. Measure so many Triargles lying next to the part assigned, till you have something too much, then by the first and second Problem of this part, cut of the overplufs from the last Triangle, so fhill you have a Figure to contain the parts required. . Example Fig. 107. Admit ABCDEFG to be a Plot, from whence 480 parts are to be cut off with a Line issuing trom C, and to be tcwards the fide AB : First, let the Trape. zium ABCG be measured, whose Area is 370, which added to 136, the Area of the Tria argle CFG maketh 506, which is too much by 26, which 26 let be cut off from the Triangle CFG, by the Line Cb; fo doth the Figore ABC 6 G, contain 483, the parts required to be cut off. Note, This Problem is very material in the Practice of Surveying, in dividing and |