EXAMPLE.

It is required to lay out a square piece of ground which shall contain 12A. 3R. 16P Required the number of chains in each side of the square; and to lay down a map of it, by a scale of 40 perches to an inch.

A. R. P.
12. 3, 16.

4

51

40

Ch. L.

2056(45.34+perches-22. 331 by prob. 6.

From a scale where 4 of the large, or 40 of the small divisions are an inch, take 45.34, the perches of the side, of which make a square.

PROB. III.

To find the content of an oblong piece of ground.

Multiply the length by the breadth, for the

content.

1

EXAMPLE.

PL. 1. fig. 3.

Let ABCD be an oblong piece of ground, whose length AB is 14C. 25L. and breadth 8C. 37L. required the content in acres, and also to lay down a map of it, by a scale of 20 perches to an inch.

160)506.9200(3. 0. 27. content.

To draw the map.

Make an oblong (by schol. to prob. 9. geom.) whose length, from a scale of 20 to an inch, may be 29 perches, and breadth, 17.48 perches.

PROB. IV.

The content of an oblong piece of ground, and one side given, to find the other.

Divide the content in perches, by the given side in perches, the quotient is the side required in perches; and thence it may be easily reduced to chains.

EXAMPLE.

There is a ditch 14 Ch. 25 L. long, by the side of which it is required to lay out an oblong piece of ground, which shall contain 3A. OR. 37P: what breadth must be laid off at each end of the ditch to enclose the 3A. OR. 37P?

The map is constructed like the last.

PROB. V.

To find the content of a picce of ground, in form of an oblique angular parallelogram; or of a rhombus, or rhamboides.

RULE I.

Multiply the base into the perpendicular height. The reason is plain from theo. 13. geom.

EXAMPLE.

PL. 7. fig. 2.

Let ABCD be a piece of ground in form of a rhombus, whose base AB is 22 chains, and perpendicular DE or FC, 20 chains. Required the content.

The converse of this is done by prob. 4. and the map is drawn, by laying off the perpendicular on that part of the base from whence it was taken; joining the extremity thereof to that of the base by a right line, and thence completing the parallelogram.

RULE II.

As rad. (Viz. S. of 90°, or Tang. of 45°)
Is to the sine of any angle of a parallelogram ::
So is the product of the sides including the angle:
To the area of the parallelogram.

*That is DA× AB × nat. sine of the angle A the area. Pl. 7. fig. 2.

==

EXAMPLE.

How many acres are in a rhomboides, whose less angle is 30°, and the including sides 25.35 and 10.4 four-pole chains?

=

Ans. 13A. OR. 29.12P.

(Rad.) 1:.500000 (Nat. S. of 30°):: 263.640 25.35 x 10,4): 131.82 = the area in four-pole chains; which, divided by 10, (because 10 square chains is an acre,) gives 13.182 acres, or 13A, OR. 29.12P.

NOTE. Because the angle of a square and rectangle are each 90°, whose sine is 1, this rule for them, is the same as the first.

* Demonstration. For, having drawn the perpendicular DE, the area by the first rule is ABX DE; but as radius 1 (S. LE): S. ZA :: AD: DE=S.LAX DA; therefore, DE XAB⇒ÀB × S.LAX DA the area; or, 1: S. LA:: DAXAB : S. L A× DAXAB-the area of the parallelogram. Q. E. D.