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EXAMPLES. 1. The area of a rhom
C boid is required to be 6 acres, and its longer and shorter sides to be respectively 1200 and 625 links, required the place and length of its perpen
500 = perpr. = DE
N 140625 = 375 links = A E. 2. Lay out the figure when its sides are each 4 chains, that is, when it is a rhombus, its given area being la. lr. 24p. Ans. 350 links = perpendicular, and its distance 193-6 links.
PROBLEM VII. To lay out a proposed quantity of land in a regular polygon.
RULE. Divide the area of the proposed polygon, by its corresponding area in the following table, and the square root of the result will be the length of its sides. Multiply the side just found by the corresponding radius, in the column marked radii, and the result will be the radius of the circle that circumscribes the required polygon. A table of regular polygons, with their names, areas, and radio
of their circumscribing circles, the sides of the polygons being unity.
0•433 0:577 4 Square...
0•707 5 Pentagon
3.634 1.152 8 Octagon
4.828 1.306 9 Nonagon
6.182 1.462 10 Decagon
7.694 1.618 11 Undecagon
9.365 1•775 12 Duodecagon...
11.196 1.932 NOTE. If the square of the side of a polygon be multiplied by its corresponding area in the preceding table, the product will be the area of tho polygon.
EXAMPLES. 1. Lay out an acre of land in a regular heptagon.
100000 Here ♡
= 165.88 links = AB.
3.634 Also 165.88 x 1.152 = 191•09 links = A O. Construction of the heptagon on the ground.—Provide a strong cord, equal in length to the radius of the circumscribing circle of the heptagon, which in this case is 191•09 links, or very nearly 191 1 links; fix one of its ends at o, as a centre,
and tie the other end to a pointed arrow; stretch the cord to A, and with the radius O A describe the dotted circle, keeping the arrow vertical, and making a visible circular mark with it. Then take the distance AB = 165.88 links on the cord, which being applied seven times within the circle, will just reach round the circumference, thus forming the heptagon
required. 2. Lay out 4 acres of land in a hexagon.
Ans. 392.38 links radius of circumscribing circle and side of the hexagon, for both are equal in this figure.
RULE.--Divide the given area by •7854, and half the square root of the quotient will be the radius of the circle.
EXAMPLES. 1. Lay out 4 acres of land so that its boundary may be a
1 400000 Heren = 356.8 links = A O.
2 7854 With the radius A O, the circle may be described on the ground, as in the last
Problem. 2. Required the radius of a circle containing half an acre of land.
Ans. 126 links. NOTE. The rules given in the two preceding Problems are the inverse of the rules for finding the areas of the same figures in works on mensuration; and their invostigations are given in various works on analytical trigonometry.
PROBLEM IX. To lay out a given quantity of land in the form of an ellipse.
RULE I.-When one of the diameters is given. Divide the given area successively by :7854 and the given diameter, and the quotient will be the other diameter.
RULE II.—When the two diameters are required to be in a given proportion or ratio. Divide the given area successively by •7854 and the product of the terms of the ratio, and the square root of the quotient, multiplied separately by the terms of the ratio will give the two diameters.
EXAMPLES. 1. Lay out 4 acres in an elliptical form, with a transverse diameter TR of 9 chains. By Rule I. 400000
links = CO, the 7854 x 900
conjugate diameter. Construction of an ellipse.-Lay off from the centre C of the ellipse, on the transverse TR, the distances CF, Cf, each equal to the square root of the difference of the squares of the two semi-diameters ; take a strong flexible cord, equal to TR, and fix its ends at F, f; extend the cord to I, so that it may take
F the position FIf, and keeping the cord continually stretched, with an arrow trace the elliptical curve TICRO, which will be the required boundary.
In this case CF Cf = N4502 2832 = 350 links.
2. Lay out an ellipse to contain 2a. Or. 32p., the proportion of the diameters of which shall be as 7 to 4. By Rule II.
= 100 links; and 100 x 7 = 700
4x7x •7854 links, the transverse diameter, and 100 X 4 = 400 links, the conjugate.
Note 1. The investigation of the rules used in this Problem, are given in various works on Conic Sections,
NOTE 2. The three last Problems are chiefly of use in laying out ornamental grounds, as the inner area of squares in cities and towns, shrubberies, plantations in parks, &c.
PROBLEM X. To divide any proposed quantity of land from a triangle, by a line parallel to one of its sides.
RULE.—Since the areas of triangles are as the squares of their like or homologous sides, we shall have Area A ABC : area a abc :: A C2 : a C?. C
See Theor. VIII., Chap. I.
EXAMPLES. 1. Let the base AB = 2600 links, AC= 2001), and BC 1600, cut off
Ea. Ir. 24p. by the line ab 16 parallel to A B.
From the three sides А
the area of the triangle ABC is found = 1599218 square links, and 6a. 1r. 24p. = 640000 square links,
the difference = 959218 = area of A a b C. Then
per Rule, 1599218 : 959218 :: 20002 : 2399217 ; and V2399217 = 1548.94 links = aC; whence 2000 -- 1548.94 = 451.06 links = Aa. Again by similar triangles 2000 : 1600 : : 451.06 : 360•84 links = Bb. Therefore, measure on the ground the distances A a, Bb, just found, respectively, from the points A and B; and range the line a b, which will divide the given quantity A B ba from the triangular field A B C, as required.
NOTE. In this manner a triangle may be divided into any number of equal or unequal parts, by lines parallel to any one of its sides, by successively subtracting the sums of 2, 3, 4, &c., areas from the area of the triangle, and making the remainders successively the second terms of the proportion.
2. The sides of a triangular field are 900, 750, and 600 links; it is required to cut oa. Br. 28p. therefrom, by a straight fence parallel to its least side.
Ans. The distance from the ends of the least side, on the largest and intermediate sides, are respectively 2114 and 176 links.
PROBLEM XI. To lay off any given quantity of land from an irregular field or common.
RULE.—First find the area of the crooked or irregular part by means of offsets, which area must be deducted from the given area, after which the remaining area must be laid out by means of Problems II., IV., and V.
CASE I. - When one side of the
A с field is crooked and two straight and parallel.
EXAMPLE. Let a Ac B 6 be a portion of a field or common, the side AcB being crooked and A a, B b parallel, the length of A B is 2400 links, and the area of the offset piece A c B is found to be 227500 square links; it is required to cut off 12 acres from the field by a line parallel to A B. 12 acres = 1200000
links offset piece
105.2=-AC=BD Therefore, measure the distances AC=BD = 405} links on the ground, and range the straight line C D, then shall the space
D 2 AcBDC contain 12 acres.
Case II.-When all the sides of the field or common are crooked.
EXAMPLE. Let a ABCDd be a portion of a field or common, the
q boundary of which B is crooked, it is required to part therefrom 6 acres, by a straight fence running from D towards A.
Measure the four lines AB, BC, CD, DA, taking the off sets on the three first, and finding the