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Perch. S. Base 49.52

Perp. 27.12

9904 4952 34664 9904

1342.9824

A. R. P. 671.4912=4.0.31.

The map may be readily drawn, having the distance from either end of the base, to the perpendicular given; as may be evident from the figure.

PROBLEM VII.

The content of a triangular piece of ground, and

the base given, to find the perpendicular.

Divide the content in perches, by half the base in perches; and the quotient will give you the perpendicular, in perches and so in chains.

EXAMPLES.

Plate I. fig. 16.

Let BC be a ditch, whose length is 24C. 40L. by which it is required to lay out a triangular piece of ground, whose content shall be 4A. IR. 10P. Required the perpendicular.

Ch. L. Perches.
Base 24.40 = 49.6
Half the base = 24.8

A. R. P.
4. 1. 10.
4.

17.
40

24.8)690(27.28 perches.

1940

2040

560

64

82

Perches. Ch. L. Answer perp. 27.28 = 13.45. . This perpendicular being laid on any part of the base, and lines run from its extremity to the ends of the base, will lay out the triangle (by cor. to theo. 13. sect. 1.) so that the perpendicular may be set on that part of the base which is most convenient and agreeable to the parties concerned.

LEMMA

If from half the sum of the sides of any plane tri

angle ABC, each particular side be taken ; and if the half sum, and the three remainders be multiplied continually into each other, the square root of this product will be the area of the triangle.

Plate VIII. fig. 9.

DG;

Bisect any two of the angles, as A and B, with the lines AB, BD meeting in D; draw the perpendiculars DE, DF, DG.

The triangle AFD is equiangular to AED ; for the angle FAD=EAD by construction, and AFD

AED, being each a right angle, and of consequence ADF= ADE; wherefore AD:DE:: AD: DE: and since AD bears the same proportion to DF, that it doth to DE, DF=DE, and the triangle AFD=AED. The same way DE=DG, and the triangle DEB= DGB, and FD=DE=

therefore D will be the centre of a circle that will pass through E, F, G.

In the same way if A and C were bisected, the same point D would be had, therefore a line from D to C will bisect C, and thus the triangles DFC, DGC will be also equal.

Produce CA to H, till AH-EB or GB; so will HC be equal to half the sum of the sides viz. to į AB + AC + į BC; for FC, FA, EB, are severally equal to CG, AE, BG; and all these together are equal to the sum of the sides of the triangle; therefore FC + FA + EB or CH, are equal to half the sum of the sides.

FC=CH-AB, for AF=AE and HA=EB; therefore HF= AB; and AF=CH-BC; for CF=CG, and AH=GB; therefore BC= HA + FC, and AH=CH-AC.

Continue DC, till it meets a perpendicular drawn upon H in K; and from K draw the perpendicular KI, and join AK.

Because the angles AHK and AIK are two right ones, the angles HIA and K together, are equal to two right'; since the angles of the two triangles contain four right : in the same way FDE + FAE=(2 right angles =) FAE + IAH, let FAE be taken from both, then FDE=IAH, and of course FAE=K: the quadrilateral figures AFDE, and KHAI, are therefore similar, and have the sides about the equal angles proportional ; and it is plain the triangles CFD and CHK are also proportional : hence,

FD : HA::FA: HK
FD : FC:: HK : HC

Wherefore by multiplying the extremes, and means in both, it will be the square of FD X HK x HC FC ~ FA HA YHK; let HK be taken from both, and multiply each side by CH;

then the square of CH by the square of FD = FC ~ FA X HAXCH.

It is plain, by the foregoing problem, that * AB x DE, + į BC * DG +į

AČ ~ FD = the area of the triangle ; or that half the sum of the sides, viz. CH * FD = the triangle ; wherefore the square of CH by the square of FD = FC X

FA ~ HA CH, that is, the half sum multiplied continually into the differences between the half sum and each side, will be the square of the area of the triangle, and its root the area. Q. E. D.

Hence the following problem will be evident.

PROBLEM VIII.

Three sides of a plane triangle given to find

the area.

RULE. From half the sum of the three sides subtract each side severally ; take the logarithms of half the sum and three remainders, anŭ half their total will be the logarithm of the area : or, take the square root of the continued product of the half sum and three remainders for the area.

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EXAMPLES.
Plate VIII. fig. 9.

1. In the triangle ABC, are

AB= 10.64
Given, AC= 12.28

four pole chains ;
CB=

goo

required the area?

Sum 31.92

Half sum 15 96

5.32 Remainders 3.68

6.96

Log. 1.20303

0.72591
0.56585
0.84261

2)3.33740

Answer, Sqr. Ch. 46.63 Log. 1.66870

or, 4.663 Acres.

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