And for the cafting up of the Content thereof the directions formerly given, in meafaring by the Chain only: And those which follow in the next Propofition will fufficiently inform you.

But, by way of Advertisement, In this Example where you went on the outfide of the Wood, you must first find the Content of the whole Quadrilateral Figure PQRS, and then of the several small Triangles and Trapezias made without the Wood, as the Trapezia, Pa Br, and the reft: Which being deducted from the Content of the whole Quadrilateral PQRS, will leave the Content of the whole Wood included therein.

PROP. VII.

The Plat of a Field being laid upon Paper, and the Scale by which it was laid down known, to find how much the faid Field containeth in Acres, Roods and Perches.

T

Here are feveral ways to effect this; but I fhall here deliver only one, which shall be general, and that which by all (or most) Surveyors is practised.

Every irregular Plat or Figure, before the quantity or content of it can be found, must first be reduced into fuch Regular Figures, for the mensuration whereof there are certain Rules: The most meet and convenient Figures into which Irregular Plats may be reduced, are Triangles and Quadrilaterals, called Trapezias, which is done by drawing of lines from Angle to Angle crofs the Plat. And here note, That of how many fides foever your Plat confifteth, into fo many Triangles, wanting two,, will the Plat be reduced, and no lefs, as the Figure VII. denotes, where the Plat confifteth of 7 Sides, and it is reduced into 5 Triangles.

L

The manner of reducing the Plat into Triangles.

FIG. VII.

ET the Figure ABCDEFG be a Plat which contains feven fides; to reduce which into Quadrilaterals and Triangles, do thus:

Firft, Draw the line BF, fo is part of the Plat reduced into the Quadrilateral ABFG.

Secondly, Draw the line FD, fo is another part of the Plat reduced into the Trapezia BCDF; and the other part of the Plat is comprehended in the Triangle F D E.

Thus the whole Plat being reduced, contains two Trapezias, viz. K and L, and the Triangle M, in number five, (for Trapezia contains two Triangles) which are less by two than the number of fides. every

Το

is, 13

To caft up the Quantity or Content of the Plat.

Firft begin with the Triangle M, whofe bafe, F B, is 13. 75. that Chains 75 Links, and his Perpendicular E Q, 7 Chains 12 Links. Now the Quantity, Area, or Content of every Triangle, is found, By multiplying the length of the bafe by half the length of the perpendicular.

So here the bafe 13. 75. and the perpendicular 7. 12. the half whereof is 3. 56. if you multiply 13.75. by 3.56. (as if they were whole Numbers, tho in rea lity the 75 and 56 are Fractions) the Product will be 4.89500, and that is the Quantity of the Triangle FDE.

I 3.75 3.56 8250 6.875

4125

4.89500

Secondly, For the Trapezia or Quadrilateral L, in which Trapezia you may fee that the line CF is a common base to the two Triangles BFC, and C D F, for the perpendiculars B P, and DO, of both Triangles, fall upon it.

Now the Quantity of any Trapezia is found, By multiplying the common bafe (here FC) by half the Sum of the two Perpendiculars which fall upon it (here B P, and D O.)

So in this Trapezia, the common base C F, is 14 C. 90 L. and the fum of the two Perpendiculars B P, and DO, is 13 Chains 80 Links, the half whereof is 6.90. If you multiply 14.90. by 6. 90. the Product will be 10. 28100.

Thirdly, For the Trapezia K, multiply 12. 66. (the length of the base BG) by 4. 55. half the fum of the two perpendiculars AH and FR, the Product of that Multiplication will be 5.76030, which is the Quantity, Area, or Content of the Trapezia K.

14.90

6.90

134100 8940

Thus have you the Content of the two Trapezias and the Triangle. Now,

To find the Quantity of the whole Field, in Acres, Roods, and Perches.

Add the Products of the feveral Multiplications together, the Sum of them is the whole Quantity.

So

The

Which is 20 compleat or entire Acres, and parts
of an Acre; and to know how much that is, multiply
93630 by 4 (because 4 Roods make an Acre the
Product is 3 compleat, and parts of a Rood; and
to know how much that is, multiply 74520 by 40 (be-
caufe 40 Perches are contained in one Rood) the Pro-
duct will be 29.80800, which is 29 compleat Perches,
and 80800
parts of a Perch, which is inconfiderable, for
160 whole Perches make but one Acre.

93630

4

3.74520

74520

40

29.80800

Thus have you the Area, Quantity, or Content of this Field caft up, and you find it to contain 20 Acres, 3 Roods, and 29 Perches. And in this manner may you caft up the quantity of any Irregular Plat whatsoever, remembring that in all your Multiplications you cut off 5 figures towards your right hand with a prick of your Pen, as in this Work I have done all along.

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