two terms become BC. DE: BC. sin B sin (B+c) : therefore the last BC.DE. sine.sin(B+C+D), BC.DE Sin BSinD sin B. sin D + sin c. sin (B+C+D). and this ex sin (B+C) pression by means of the formula for 4 arcs (art 30 ch. iii,) becomes Be. 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 (e+) +BC. DC sin c +BC. DE. Sin (c+D) +PC. DE . sin ». Cor. Taking away from this expression, the 1st, 2d, and 4th terms, which together make double the trapezium ABCD, there will remain Or, twice area of the triangle AEF AB. BC. sin B AP. EF. sin (B+C+D+E) AB . EF . Sin (B+C+D+E) DE. EF. Sin E {+DE. | +BP. EF. Sin (B+C+D+E) Now, writing for BP, CP, their respective values, BC. sin c sin (B+c) sin (B+c) the sum of the last two expressions, in the double areas of AEF, will become BC. EF and this, by means of the formula for 5 arcs (art. 30 ch. iii) Twice the area AB . BC . sin B +AB. CD. Sin (B+C) ་ +BC. DE. Sin (C+D) +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 n-1 n—2 above, and the number of terms → 1 when the number of sides of the polygon is n for the number of terms 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 AB . BC . sin B +CD. AD. şin D. The pentagon ABCDE may be divided into the trapezium ABCD, and the triangle ADE, whence Twice area of pentagon 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 +AB. DC. Sin (B+C) +DE. EF. Sin E And lastly, the heptagon may be divid- Twice area of heptagon AB. BC. Sin B +AB. CD. Sin (B+c) +BC. DE . Sin (C+D) +CD. DE sin D +EF. FG. sin F +EF. GA. Sin (F+G) +FG. GA. Sin G. The same method may obviously be extended to other poly gons, with great ease and simplicity. It It often happens, however, that only one side of a polygon can be measured, and the distant angles be determined by intersection; in this case the area may be found, independent of construction, by the following problem. PROBLEM I. Given the Length of One of the Sides of a Polygon, and the The same method manifestly applies to polygons of any number of sides: and all the terms except the last are so perfectly symmetrical, while that last term is of so obvious a form, that there cannot be the least difficulty in extending the formula to any polygon whatever. PROBLEM II. Given, in a Polygon, All the Sides and Angles, except three ; to find the unknown Parts. This problem may be divided into three general cases, as shown at the beginning of this chapter: but the analytical solution of all of them depends on the same principles; and these are analogous to those pursued in the analytical investigations of plane trigonometry. In polygonometry, as well as trigonometry, when three unknown quantities are to be found, it must be by means of three independent equations, involv ing the known and unknown parts. These equations may be deduced from either theorem 1, or 3, as may be most suited to the case in hand; and then the unknown parts may each be found by the usual rules of extermination. For an example, let it be supposed that in an irregular hexagon Abcdef, there are given all the sides except AB, BC, and all the angles except в; to determine those three quantities, The angle B is evidently equal to (2n D D F A B 4) right angles, Atc tote† F); n being the number of sides, and the angles being here supposed the interior ones. y A A + EF. COS AB EF + AF. COS ABaAF ; * . COS B + AF . COS BC AF + FE. COS BCAFE. +DE, COS BCADE + DC. COS BC1CD, In the first of the above equations, let the sum of all the terms after y. cos B, be denoted by c; and in the second the sum of all those which fall after x. cos в, by d; both sums being manifestly constituted of known terms: and let the known co-efficients of x and y be m and n respectively. Then will the preceding equations become x = ny + c ... y = mx + d. Whence Substituting for y, in the first of the two latter equations, its |
