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Or, 15.96 x 5.32 x 3.68 x 6.96=2174.71113216; the square root of which is 46.63 for the area as before.
2. What quantity of land is contained in a triangle, the 3 sides of which are, 80, 120, and 160 perches respectively ?
Answer 29A. 7P.
Two sides of a plane triangle and their included
angle given, to find the area.
To the log. sine of the given angle (or of its supplement to 180°, if obtuse) add the logarithms of the containing sides; the sum, less radius, will be the logarithm of the double area.
Plate V. Fig. 16.
Suppose two sides, AB, AC, of a triangular lot ABC, form an angle of 30 degrees, and measure one 64 perches, and the other 40.5 what must the content be ?
30. sine 9.69897
2)1296. log. 3.11260
160)648(4A, 8P. answer.
2. Required the area of a triangle, two sides of which are 49.2 and 40.8 perches, and their contained angle 144degrees ? Answer, 3A. 2R. 22P.
3. What quantity of ground is inclosed in an equilateral triangle, each side of which is 100 perches, either angle being 60 degrees ? Answer, 27A. 10P.
Demonstration of this problem.,
Plate XI. fig. 3.
Let AH be perpendicular to AB and equal to AC and HE, FCG, parallel to AB; then making AH (= AC) radius, AF (=CD) will be the sine of CAD, and the parallelograms ABEH (the product of the given sides) and ABGF the double area of the triangle) having the same base AB, are in proportion as their heights AH, AF; that is, as radius to the sign of the given angle; which proportion gives the operation as in the rule above.
To find the area of a trapezoid, viz. a figure
bounded by four right lines, two of which are parallel, but unequal.
RULE. Multiply the sum of the parallel sides by their perpendicular distance, and take half the product for the area.
EXAMPLES. 1. Required the area of a trapezoid, of which the parallel sides are, respectively, 30 and 49 perches, and their perpendicular distance 61.6 ?
Note. On this 10th problem are founded most of the calculations of differences by latitude and departure, and those by off-setts, following in this treatise.
Answer, 2433.2=15A. 33.2P. Plate VIII. fig. 10.
2. In the trapezoid ABCD the parallel sides are, AD, 20 perches, BC, 32, and their perpendicular distance, AB, 26; required the content ?
Answer, 4A. 36P.
To find the content of a Trapezium.
RULE. Multiply the diagonal, or line joining the remotest opposite angles, by the sum of the two perpendiculars falling from the other angles to that diagonal and half the product will be the area.
EXAMPLE Plate VII. fig. 3.
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 content ? Diagonal = 64.4 Multiply.
} 13.64+27.2 = 40.8
160)131376(8A. 33|P. Answer.
Note. The method of multiplying together the half sums of the opposite sides of a trapezium for the content 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.
To find the area of a circle, or an ellipses.
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 0.10491 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.
Nute. In any circle, the
3. Circum. div.
quotes the diam.
EXAMPLES 1. How many acres are in a circle of a mile diameter ? 1 Mile=320 per. log. 2.50515
410)8042|5. log. 4.90539
502A. 2R. 25P.
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, content of his park, and area of his pond?
Answer, the wall 314.16 perches inclosing 49A. 14P. of which 394 perches, or } of an acre nearly, is appropriated to the pond.
The area of a circle given, to find its diameter.
To the logarithm of the area add 0.10491, 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.
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? Area in perches 160 log. 2.20412
Answer, 7.13665 per. = 117 F. 9 In.