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PROBLEM W.

To find the contents of a piece of ground in form of an oblique angular parallelogram, or of a rhombus or rhomboides.

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 FC20 chains. Required the contents.

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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.

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* 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, DAX ABX nat, sine of the angle A = the area.” Pl. 7, fig. 2. -

* Demonstration. For, having drawn the perpendicular DE the area by the first rule is ABxDE; but as radius 1 (S. 4- E): S. 4-4 4D :

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. 0R. 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 are an acre) gives 13.182 acres, or 13A. 0R. 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.

PROBLEM WI. To find the contents of a triangular piece of ground. Multiply the base by half the perpendicular, or the perpendicular by half the base; or take half the product of the base into the perpendicular. The reason of this is plain from cor. 2, theo. 12, geom.

EXAMPLE.
PL. 1. fig. 16.

Let ABC be a triangular piece of ground whose longest side or base BC is 24ch. 38l., and perpendicular AD, let fall from the opposite angle, is 13ch. 28l. Required the contents.

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Or the base and perpendicular may be reduced to perches, and the contents may thence be obtained, thus:

ch. l. perches.
Perp. 13.28=27.12
! By prob. 4, sect. 1, part 2,

Half the perp. 13.56

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perches. 2. Perp. 27. 12 Half-base 24.76

16272 18984 10848 5424

671,4912=4A. 0R. 31P. But square perches may be reduced to acres, &c. rather

more commodiously by dividing by 40 and 4, than by 160; thus,

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The map may be readily drawn, having the distance from either end of the base to the perpendicular given; as may be evident from the figure. -- -

PROBLEM VII.

The contents of a triangular piece of ground and the base given, to find the perpendicular. w

Divide the contents in perches by half the base in perches,

and the quotient will give you the perpendicular in perches, and so in chains.

ExAMPLE.
PL. 1. fig. 16.

Let BC be a ditch, whose length is 24ch. 40l., by which it is required to lay out a triangular piece of ground, whose contents shall be 4A; IR. 10P. Required the perpendicular. * -- * * * * -- ch. l. Perches. Base 24.40=49.6 Half the base=24.8

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This perpendicular being laid on any part of the base, and lines run from its extremity to the ends of the base, will lay out the triangle (by cor. to theo. 13, geom.) so that the perpendicular may be set on that part of the base which is most convenient and agreeable to the parties concerned.

PRACTICAL QUESTIONS.

Ex. 1. What is the area of a parallelogram whose length is 12.25 and its height 8.5 four-pole chains : Ans. 10A. 1R. 26P. Ex. 2. What is the area of a square whose side is 70.25 two-pole chains 2 - Ans. 124A. 1R. 1 P. Ex. 3. What is the area of a rhombus whose side is 60 perches, and its height 45 perches : Ans. 16A, 3R. 20P.

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