Ex. 2. What is the area of a square, whose side is 70.25 two-pole chains?

Ans. 124A. 1R. 1P.

Ex. 3. What is the area of a rhombus, whose side is 60 perches, and its height 45 perches? Ans. 16A. 3R. 20P.

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?

LEMMA.

PL. 8 fig. 9.

Ans. 1A. 3R. 8P.

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, DĞ.

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 doth 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.

2

Produce CA to H, till AH=EB or GB; so will HC be equal to half the sum of the sides, viz. to AB + ¦ AC + 1 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 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+ 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×HK ×HC=FC× FA× HA×HK; let HK be taken from both, and multiply each side by CH; then the square of CH by the square of FD=FCX X FAXHAXCH.

2

It is plain, by the foregoing problem, that AB × DE+ BCX DG + ACX 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× 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. QE. D.

3

Cor. 1. If all the sides be equal, the rule will become a×a×a×aaa3, for the equilateral triangle whose side is a.

2

4

Hence the following problem will be evident.

PROB. 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.

* The demonstration of this is plain from the foregoing Lemma, and the nature of Logarithms.

3

or, 4.663 Acres.

Or, 15.96×5.32×3.68× 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?

Ans. 29A. 7P.

3. What quantity of land is contained in a triangle, the three sides of which are 20, 30, and 40 four-pole chains?

Ans. 29A. 7.579P.

4. How many acres are in a triangle, whose sides are 49, 50.25, 25.69 four-pole chains? Ans. 61A. 1R. 39.68P.

PROB. IX.

Two sides of a plane triangle and their included angle given, to find

the area.

RULE.*

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.

Or, As radius

To the sine of any angle of a triangle So is the product of the sides including the angle: To double the area of the triangle.

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?

* Demonstration. This follows from rule 2. prob. 5. and from the nature of Logarithms, because a triangle is half a parallelogram of the same base and height.

Or, thus. Pl. 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 pro portion 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.