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As HI happens to be a meridian, the area of HOMH divided by half OV (19.1) quotes HM (43.23) without finding the area of HOIH, as we did of ICDI in example 2d. and HM-HV = VM 8.03 = = dif. lat. of OM, which with its dep. VO=38.2. gives the bearing and distance as before..

EXAMPLE IV.

PL. 12. fig. 4.

A trapezoidal field ABCD, bounded as under specified, is to be divided into two equal parts by a right line EF parallel to AB or CD; required AF or BF?

Bou. Bearing. Per.
AB South. 30.
BC N. 80 W. 60.
CD N. 39 W. 45.5
DA IS. 80 E. 89.4
13A. 3R. 7P.

In the triangle CBG are given BC and all the angles (known by the bearings) to find BG, and thence the area by prob. 9. sect. 4. which+half the area of ABCD-area of EFG; then as the area of CBG to that of EFG, so is the square of BG to the square of FG, and FG-BG=BF.

Operation at large.

Angle G 39° 30', log. S. Co. Ar. 0.19649

Side BC 60 per. log.

Angle C 40° 30', sine

1.77815 add

9.81254

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By the application of this method a tract of land may be divided accurately, in any proportion, by a line running in any assigned direction.

Note. When the practitioner would wish to be very accurate, it will be much better to work by four-pole chains and links than by perches and tenths; one tenth of a perch square being equal to 61 square links.

EXAMPLE V.

The following Field-Notes (from A. Burns) are ↳ of a piece of land, which is proposed, as an example, to be divided into three equal parts by two rightlines running from the sixth and seventh stations; and proved, by calculating the content of the middle part.

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EXAMPLE VI.

PL. 8. fig. 5.

The plot ABCDEFGHA is proposed to be divided, geometrically, in the proportion of 2 to 3, by a right line from a given point in any boundary or angle thereof, suppose the point D.

Reduce the plot to the triangle cDe, as already taught; divide the base ce in the point N, so that eN be to Nc in the ratio of two or three, by prob. 14. page 53; draw DN, and it is done.

EXAMPLE VII.

PL. 12. fig. 3.

Example 2d may likewise be performed geometrically.

Produce CD both ways for a base, and reduce the whole to a triangle, making the vertical point; then bisect the base in N, and draw IN. But,

Notwithstanding this geometrical method is demonstrably true in theory, it is not as safe, on practical occasions requiring accuracy, as the calculation, even when performed with the greatest care; for which reason we will not enlarge on it here.

EXAMPLE VIII.

Suppose 864 acres to be laid out in form of a right-angled parallelogram, of which the sides shall be in proportion as 5 to 3; required their dimensions ?

For the greater side, multiply the area by the greater number of the given proportion, and divide

by the less, or, for the less side, multiply by the less number, and divide by the greater; the square root of the quotient will be the side required: thus,

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If it be required to lay out any quantity of ground, suppose 47A. 2R. 16P. in form of a parallelogram, of which the length is to exceed the breadth by a given difference, for instance 80 perches, then add the square of half this difference to the area, and take the square-root of the sum; to which add half the difference for the greater side, and subtract it therefrom for the less; thus,

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1600 half diff. add and subt.-40

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Any proposed quantity of ground may be laid

out or inclosed in the form

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