EXAMPLES. Let BC be a fence whose length is 24C. 40L. by which it is required to lay out a triangular piece of ground, whose content shall be 4A. 1R. 10P. Required the perpendicular. Half the Base=24.8 perches, A. R. P. and 4. 1. 10. 690, divided by 24.8 gives 27.82 the perpendicular. 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 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. Bisest 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: DF:: 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=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 AB, + { 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 + B 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 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 FD: FC: HK: HC Wherefore by multiplying the extremes, and means in both, it will be the square of FDxHKXHC=FC >FA×HA×HK; 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 AB× DE, BC × DG x AC × FD the area of the × × 2 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. 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. |