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Ex. 4. What is the area of a rhomboides whose less angle is 40° and the including sides 80 and 25 four-pole chains? Ans. 128A. 2R. 9P. Ex. 5. What is the area of a triangle whose base is 12 and its perpendicular height 6 two-pole chains : Ans. 1A. 3R. 8P. 4 = 3-2.2+ *-

LEMMA.

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If from half the sum of the sides of any plane triangle 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. Bisect any two of the angles, as A and B, with the lines AD, 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 : DF; and since AD bears the same proportion to DF that it does to DE, DF=DE, and the triangle AFD= AED. The same way DE=DG, and the triangle DEB= DGB, and FD=DE=DG ; 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 4AB+}AC+}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 AHGB ; therefore BC=HA-H FC, and AC=CH-AH. 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 HAI and K together are equal to two right; since the angles of the two triangles contain four right: in the same way FDE-FFAE= (two right angles =) FA E--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 FDX HKx HC=FCx FAx HAx HK: let HK be taken from both, and multiply each side by CH; then the square of CH X by the square of FD=FCX FA X HAX CH. *

It is plain by the foregoing problem, that ABX DE-H4BC × DG+4ACX FD = the area of the triangle; or that half the sum of the sides, viz. CHX FD = the triangle; wherefore, the square of CH x by the square of FD=FCX FA × HAx 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.

Cor 1. If all the sides be equal, the rule will become M3 a × # a × 4 a × 4 a-#aa V3, for the equilateral triangle whose side is a.

Hence the following problem will be evident.

PROBLEM VIII.
The 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, and 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.

EXAMPLES.
PL. 8. fig. 9.

1. In the triangle ABC are

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AB = 10.64
Given & AC-12.28 ! four-pole chains; required the area.

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The demonstration of this is plain from the foregoing lemma, and the nature of logarithms.

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* Demonstration. This follows from rule 2, prob. 5, and from the na-

ture of logarithms, because a triangle is half a parallelogram of the same

base and height.

Or thus, PL. 11, 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 sine of
the given angle; which proportion gives the operation as in the rule above.

ExAMPLEs.
PL. 5. 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 contents be? Given angle 30° sine 9.698970 64 log. 1.806180

Containing sides } 40.5 log. 1.607455

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4A. 8P. 2. Required the area of a triangle, two sides of which are 49.2 and 40.8 perches, and their contained angle 1444 degrees. Ans. 3A. 2R. 22P. 3. What quantity of ground is enclosed in an equilateral triangle, each side of which is 100 perches, either angle being 60 degrees : - Ans. 27A. 10P.

PROBLEM X.

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.

* Demonstration. The trapezoid ABCD (pl. 14, fig. 8) is equivalent to the rectangle contained by its altitude and half the sum of the parallel sides BC and AD. For draw CE parallel to AB (prob. 8), bisect ED in F, and draw FG parallel to AB, meeting the production of BC in G. Because BC is equal to AE, BC and AD are together equal to AE and AD, or to twice AE with ED, or to twice AE and twice EF, that is, to twice AF; consequently, AF=}(BC+AD). Wherefore, the rectangle contained by the altitude of the trapezoid and half the sum of its parallel sides is equivalent to the rhomboid BF: but the rhomboid EG is equivalent to the triangle ECD (theo. 12, cor. 2); add to each the rhomboid BE, and the rhomboid BF is equivalent to the trapezoid ABCD. Note.—On this proposition is founded the method of offsets, which enters so largely into the practice of land surveying. In measuring a field of a very irregular shape, the principal points only are connected by straight lines, forming sides of the component triangles, and the distance of each remarkable flexure of the extreme boundary is taken from these rectilineal traces. The exterior border of the polygon is therefore considered as a collection of trapezoids, which are measured by multiplying the mean of each pair of offsets or perpendiculars into their base or intermediate distance, which is one of the other sides, because the parallel sides are perpendicular to it. * Demonstration. For the trapezium ABDC = the triangles ABC+ Apc=oto-oxac, a B.D.

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.

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2. In the trapezoid ABCD the parallel sides are, AD 20 perches, BC 32, and their perpendicular distance, AB .26; required the contents. Ans. 4A. 36P.

PROBLEM XI.
To find the contents of a trapezium.
RULE 1.*

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.

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