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The Content of a Triangular Piece of Ground, and the Base given, to find the Perpendicular.

RULE.

Divide the content in perches, by half the base in perches; the quotient will be the perpendicular in perches.

ExAMPLE.

Let BC be a ditch 12.40 chains long; by which it is required to lay out a triangular piece of ground, that shall contain 4A. 1R. 10P. Required the perpendicular :

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PROBLEMI VIII.

To find the Content of a Trapezium.
RULE.

Multiply the sum of the perpendiculars into the base, and take half the product for the square measure, and divide as in Problem 1.

ExAMPLE.

Let ABCD be a field, in form of a Trapezium; the base AC 16.10 chains, the perpendicular Bb3.40 chains, and Dd 6.80 chains. Required the area o

JWote. Bb3.25 chains from A ; and Dd 5.57 chains from C.

Ch. Ch. Ch. Per.
Perpens. 6.80+3.40=10.20 = 40.8 their sum.
Base 16.10 = 64.4

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PROBLEM IX.

To find the Content of a Field, bounded by four Sides, two of which are parallel, but unequal.

RULE,

Multiply the sum of the parallel sides into their perpendicular distance; take half the product, and divide as in

Problem 1.

ExAMPLE.

Let ABCD be a field, and the parallel sides AD and BC respectively 7.20, and 12.25 chains; and their perpendicular distance 15.40 chains. Required the area:

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A B

PROBLEM X.

Two Sides of a Triangle, and their contained Jingle given, to find the Area.

RULE.

As radius,

Is to the log. of the two sides;

So is the sine of the contained angle (or its supplement to 180°, if obtuse)

To the log. of the double area.

ExAMPLE. In the triangle ABC, the lines AB and AC respective

ly, are 16, and 10.12 chains, and their contained angle 30 degrees. Required the area.

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As radius 90° 10.00000
AB 16 1.20412

Is to the si {

S to sides AC 10.12 1.00518

So is sine of the angle 30° 9.69897

11.90827 10.00000 To double area ;)80.96 1.90827 A. 4,048 4. ,192 40 P. 7,680

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JVote. The area, and two sides of an oblique angled triangle given, to find

the third side. Invert the preceding Rule, thus:
As the product of the given sides,
Is to radius; -

So is the double area,
To the sine of their contained angle,

Then, by Case 4, oblique Angled Trigonometry, find the third side.

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