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fecond man's base, which fet from E to F; and the third man's bafe must be 20.83, viz. from F to C. This F to C. This done draw an obfcure line from D to the oppofite angle B, and from E and F draw the lines EG and FH parallel to B D. Laftly from D draw DG and DH, which will divide the triangle into three fuch parts as were required.

PROBLEM XIV.

To divide a triangular piece of land according to any proportion given by a line drawn parallel to one of its fides.

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EXAMP. Let ABC be the triangular piece of land containing 60 acres, the bafe A Cis 50 chains; this piece of land is to be divided between two men by a line drawn parallel to BC, in fuch proportion, that one shall have 40 acres, the other 20.

1. Divide the bafe by this proportion,

A Ch.

As 60 50

A

Ch.

20: 16.67,

which lay from C to D; and then AD is 33-33.

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2. Find a mean proportional between A D and A C, viz. 40.82, which fet from A to E; then draw EF parallel to BC, and the triangle is divided as required.

PROBLEM XV.

To reduce a trapezium into a triangle, by lines drawn from any angle thereof.

Let A B C D be the trapezium to be reduced into a triangle and B the angle affigned.

A

F D

E

Draw the obfcure line B D, and draw CE parallel to it, produce the bafe A D to E, and draw BE, which will make the triangle ABE equal to the trapezium ABCD.

Now to divide this trapezium according to any affigned proportion, is no more than to divide the triangle ABE, as is before taught in problem 13th. which will also divide the trapezium.

EXAMPLE.

Suppofe the trapezium A B C D contains 124 acres, 3 roods, & perches, is to be divided between

between two men; the first to have 50 Ac. 2 R. 3 P. and the other 74 Ac. 1 R. 5 P. and the line of divifion to proceed from B.

1. Reduce the fhates into perches, and it will be 8083 for the firft man, and 11885 for the fecond man's fhare.

2. Meafure the bafe of the triangle, viz. AE 78 chains; then fay,

Per. Ch. Per.

::

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Ch. 1.

As 19968 78 8083 31.52, which fet off from A to F, draw the line BF and the trapezium is divided as required; the triangle ABF being the first man's fhare, and the trapezium BCDF the fecond's.

PROBLEM XVI.

To divide an irregular plot of any number of fides according to any proportion given, by a Straight line drawn thro' it.

Suppofe the field ABCDE contains 46 acres to be divided into two equal parts between two men, by a line proceeding from A.

B

E

F

J. Draw

1. Draw a line at pleasure thro' the figure, as AF; then caft up the content of either half, and fee what it wants, or what it is more than the true half should be.

2. Thus I caft up the content of AEF, and find it to be but 15 acres; whereas the true half is 23 acres; 8 acres being in the part ABCDF more than in AEF; therefore I make a triangle containing 8 acres, and add it to A E F, as the triangle AGF; then the line A G parts the figure into two equal parts.

Thus you may divide any piece of land of never fo many fides and angles, according to any proportion, by ftraight lines drawn thro' it, with as much certainty, and more expeditiously. than by any other way yet known..

Another EXAMPLE will make all plain.

Suppofe the following field, containing 27 acres, is to be divided between 3 men, each to have 9 acres, and the lines of divifion to run from a pond in the field fo that each may have the benefit of the water, without going over one another's land.

1. From the pond draw lines to every an

OE; and the figure is divided into 5 triangles, each of which meafure, and put down the contents feverally; which contents reduced into perches will stand thus,

gle, as CA, OB, OC, On every an

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The whole content being 4320 perches or 27 acres, each man's proportion being 144° perches.

From to any angle draw a line for the divifion line, as OA; then confider that the firft triangle AB is but 674 perches, and the fecond triangle BOC 390, both together make but 1064 perches, which is lefs than 1440 by 376. You must therefore cut off from the third angle COD 376 perches for the first man's dividing line, which you must find out thus: The bafe DC is 18 chains, the content of the triangle" 1238; then fay,

P. C. P.

C.
As 1238 18 :: 376: 5.45

which fet from C to F, and drawing OF, you have the firft man's part, viz. AOF.

Then fee what remains of the triangle OD, 376 being taken out you will find it to be 862. which is less than 1440 by 578: therefore from the triangle DOE cut off 578 perches, and the point of divifion will fall in G; draw the line OG, which with OA and OF divides the figure into three equal parts.

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