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Let ABCD be a field in form of a trapezium, the diagonal AC 64.4 perches, the perpendicular Bb 13.6, and Dd 27.2; required the contents.

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Note—The method of multiplying together the half-sums of the opposite sides of a trapezium for the contents is erroneous, and the more so the more oblique its angles are.

To draw the map, set off Ab 28 perches, and Ad 34.4, and there make the perpendiculars to their proper lengths, and join their extremities to those of the diagonal.

Note.—When one of the diagonals and the four sides of a trapezium are given, it is divided into two triangles whose sides are given; the area of each triangle may be found (by prob 8), and their sum will give the area of the trapezium.

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If there be drawn two diagonals cutting each other, the product of the diagonals multiplied by the natural sine of the angle of intersection of the diagonals will be double the area. And this rule is common to a square, rhombus, rhomboides, &c., as well as ACX BD X Nat. S. Z R

2

= the area. Pl. 14, fig. 9. Or, as radius: S. Z. R :: , ACX BD : the area.

Note.—Because the diagonals of a square and rhombus in

tersect at a right angle, whose sine is 1, therefore half the product of their diagonals is the area.

to all other quadrilateral figures; that is,

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- EXAMPLE.

Let the two diagonals be 40 and 30 chains, and at their intersection one of the less angles is 48°; the area is required. Then, since the natural sine of 48° is .743.1448, the area =

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Area 445.8869 2.649224 To find the area of a trapezium when three side sand the two included angles are given. EXAMPLE. In a quadrangular field the south side is 23.4, the east side 19.75, and the north side 20.5 chains; also the south-east and north-east angles are 73° and 87° 30'. What is the area?

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#AD 11.7 1.068 1859

BD * 1.4446378

S. Z ADB 25° 38' 9,6360969

140.9031 the area of the A ABD 2, 1489206

Their sum is 343.1479 sq. chains = 34A. 1R. and 10.3664P., the area required.

EXAMPLES FOR PRACTICE.

1. Required the area of a trapezium whose diagonal measures 120 perches, and the perpendiculars 24 perches and 40 perches. Ans. 24 acres.

2. Required the area of a trapezium whose diagonals are 85 and 24 four-pole chains, and at their intersection one of the less angles is 30°. Ans. 105A.

3. What is the area of a trapezium whose south side is 27.4 chains, east side 35.75 chains, north side 37.55 chains, and west side 41.05 chains; also the diagonal from south-west to north-east 48.35 chains : Ans. 123A, 11.867P.

PROBLEM XHI.
To find the area of a circle or an ellipsis.
RULE.

Multiply the square of the circle's diameter, or the product of the longest and shortest diameters of the ellipsis, by .7854 for the area. Or, subtract 0.104909 from the double logarithm of the circle's diameter, or from the sum of the logarithms of those elliptic diameters, and the remainder will be the logarithm of the area. Note.—In any circle the Diam. multiplied so S produces the circum. Circum. . } by 3.14159 {{..." the diam.

EXAMPLES.

1. How many acres are in a circle of a mile diameter 1

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2. A gentleman, knowing that the area of a circle is greater than that of any other figure of equal perimeter, walls in a circular deer-park of 100 perches diameter, in which he makes an elliptical fish-pond 10 perches long by 5 wide. Required the length of his wall, contents of his park, and area of his

pond. Answer. The wall 314.16 perches, enclosing 49A. 14P., of

which 394 perches, or # of an acre nearly, is appropriated to the pond.

PROBLEM XIII. The area of a circle given, to find its diameter. RULE. , To the logarithm of the area add 0.104909, and half the sum will be the logarithm of the diameter. Or, divide the area

by .7854, and the square root of the quotient will be the diameter.

ExAMPLE.

A horse in the midst of a meadow suppose
Made fast to a stake by a line from his nose:
How long must this line be that, feeding all round,
Permits him to graze just an acre of ground?

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It is customary to deduct 6 acres out of 106 for roads; the land before the deduction is made may be termed the gross, and that remaining after such deduction the neat.

RULE.

The gross divided quotes the neat.
The neat multiplied } by 1.06 { produces the gross.

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PROBLEM XV. To find the area of a piece of ground, be it ever so irregular, by dividing it into triangles and trapezia. PL. 7. fig. 4. .

We here admit the survey to be taken and protracted; by having, therefore, the map, and knowing the scale by which it was laid down, the contents may be thus obtained.

Dispose the given map into triangles by fine pencilled lines, such as are here represented in the scheme, and number the triangles with 1, 2, 3, 4, &c. Your map being thus prepared, rule a table with four columns, the first of which is for the number of the triangle, the second for the base of it, the third for the perpendicular, and the fourth for the contents in perches.

Then proceed to measure the base of number 1, from the scale of perches the map was laid down by, and place that in the second column of the table, under the word base; and from the angle opposite to the base open your compasses so as when one foot is in the angular point, the other, being moved backward and forward, may just touch the base line, and neither go the least above nor beneath it; that distance in the compasses, measured from the same scale, is the length of that perpendicular, which place in the third column under the word perpendicular.

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