2. For a pentagon, as ABCDE. Its area is obviously equal to the sum of the areas of the trapezium ABCD, and of the triangle ADE. Let the sides AB, DC, as before, meet when produced at P. Then, from the above, we have Twice area of the trapezium ABCD }= AB. BC . sin B B +AB. DC. sin (B+c) And, by the preceding corollary, Twice triangle DAE That is, twice triangle DAE }={ BC. sin c AP. DE. Sin (P+D) or sin (B+C+D) + DP. DE. sin D. AB DE. sin (B+C+D) + DC. DE. sin D +BP. DE. sin (B+C+D) BC. sin B sin (B+c) BC. DE. sin c. sin (B+C+D) sin (B+c) Now, BP = and CP = sin (B+ c)' two terms become =BC.DE · sin B. sin D + sin c. sin (B+C+D) sin (B+ c) :therefore the last + BC. DE. Sin B. sind sin (B+C) : and this expression, by means of the formula for 4 arcs (art. 30 ch. iii), becomes BC. DE. sin (c+D). Hence, collecting the terms, and arranging them in the order of the sides, they become Cor. Taking away from this expression, the 1st, 2d, and 4th terms, which together make double the trapezium ABCD, there will remain Twice area of DAE. } AB. DE. sin (B+C+D) +BC. DE. sin (C+D) DC. DE. sin D. 3. For a hexagon, as ABCDEF. The double area will be found, by supposing it divided into the pentagon ABCDE, and the triangle AEF. For, by the last rule, and its corollary, we have, B Twice area of ABCDE Twice area of) AEF Or, twice area the triangle of AEF. BP. EF. sin (B+C+D+E) +CP. EF . sin (D+E). Now, writing for BP, CP, their respective values, BC.sin c and BC. sin B sin (B+c)' the sum of the last two expressions, sin (B+ c) BC.EF. sin c : and this, by means of the formula for 5 arcs (art. 30 ch. iii) becomes BC. EF . sin (c+D+E). Hence, collecting and properly arranging the several terms as before, we shall obtain AB. BC. sin B Twice the area of the hexa- = +AB. CD. sin (B+C) +AB. DE. sin (B+C+D) +BC. DE. sin (C+D) +CD. EF. sin (D+E) +DE. EF. sin E. 4. In a similar manner may the area of a heptagon be determined, by finding the sum of the areas of the hexagon and the adjacent triangle; and thence the area of the octagon, nonagon, or of any other polygon, may be inferred; the law of continuation being sufficiently obvious from what is done above, and the number of terms=".2, when the number of sides of the polygon is n: for the number of terms is evidently the same as the number of ways in which n - 1 quantities can be taken, two and two; that is, (by the nature of Permutations) = 1.2-2 1 1 Scholium. B +AB. DC. sin (B+c) And, by the preceding corollary, Twice triangle DAE That is, twice triangle DAE }={ } BC. sin c Now, BP = = sin (B+c)' two terms become BC.DE . AP.DE.Sin (P+ D) or sin (B+C+D) + DP.DE. sin D. AB. DE. sin (B+C+D) + DC. DE. sin D +BP. DE. sin (B+C+D) sin (B+c) therefore the last BC. DE. sin c. sin (B + C + D) BC.DE.sin B. sind sin B. sin D + sin c. sin (B+C+D) sin (B+ c) sin (B+C) : and this expression, by means of the formula for 4 arcs (art. 30 ch. iii), becomes BC. DE. sin (C + D). Hence, collecting the terms, and arranging them in the order of the sides, they become Twice the area' of the penta = gon ABCDE AB. BC . sin B + AB. DC. sin (B+c) · AB. DE sin (B+C+D) + BC. DE sin (C+D) 3. For a hexagon, as ABCDef. double area will be found, by supposing it divided into the pentagon ABCDE, and the triangle AEF. For, by the last rule, and its corollary, we have, B 13.3. s B 3.3+C) 34 sin B+C+D' -43-27-sin.s+c+p+s) -3C.. sin c -3C.DE.sin (C+D) CD DE. sin D →CD . EF . sin (D+8) +DE. B. sin È. 4:r mer may the area of a heptagon be á Lining he sum of the areas of the hex ** de arrange: and thence the area of the v MOLT I ET her polygon, may be interes of figu are the municer of terms = rine fhe polygon is n: for the muc”, te time as the number of ways ***** Ken, two and two; that *** = Scholium. This curious theorem was first investigated by Simon Lhuillier, and published in 1789. Its principal advantage over the common method for finding the areas of irregular polygons is, that in this method there is no occasion to construct the figures, and of course the errors that may arise from such constructions are avoided. In the application of the theorem to practical purposes, the expressions above become simplified by dividing any proposed polygon into two parts by a diagonal, and computing the surface of each part separately. Thus, by dividing the trapezium ABCD into two triangles, by the diagonal AC, we shall have Twice area? H AB. BC. sin B +CD. AD. sin D. The pentagon ABCDE may be divided into the trapezium ABCD, and the triangle ADE, whence AB. BC . sin B + BC DC. sin c +DE. AE. sin E. Thus again, the hexagon may be divided into two trapeziums, by a diagonal drawn from A to D, which is to be the line excepted in the theorem; then will Twice area of AB. BC. sin B +DE. EF . sin E +DE. AF. Sin (E+F) B And lastly, the heptagon may be divided into a pentagon and a trapezium, the diagonal, as before, being the excepted line: so will the double area be expressed by 9 instead of 15 products, thus : AB . BC . sin B +AB. CD. sin (B+C) Ꮯ ت F +EF. GA. Sin (F+G) L+FG. GA. Sin G. The same method may obviously be extended to other polygons, with great ease and simplicity. |