+ Į 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 + B or CH, are equal to half the sum of the sides. FC=CH-AB, for AF=AE, and HA :-EB; there. fore HF-AB; and AF=CH-BO; 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 angle 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 Wherefore by multiplying the extremes, and means in both, it will be the square of FDxHKXHC=FC *FAXHAXHK ; let HK be taken from both, and multiply each side by CH; then the square of CHX It is plain, by the foregoing problem, that į ABx DE, * { BC * DG * AC FD = the area of the triangle; or that half the sum of the sides, viz. CHX FD = the triangle; wherefore the square of CH by the square of FD=FC * FAXHAXCH, 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. 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 the half 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. 2)3.33740 Answer, Sqr. Ch. 46.63 log. 1.66870 or, 4.663 Acres. Or, 15.96x5.32x3.68x6.96=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? Answer 29A. YP. PROB. IX. Two sides of a plane triangle and their included an gle 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. 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? 2. Required the area of a triangle, two sides of which 19.2 and 10.8 perches, and their contained angle 14+1 degrees ? Answer, 3A. 2R. 22P. 3. What quantity of ground is inclosed in an equilateral triangle, each side of which is 100 perches, either angle being 60 degrees ? Answer, 27A. 10P. Let AH be perpendicular to AB and equal to AC and HE, FCG, parallel to AB then making AH (=A C) 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 |