MENSURATION 15 triangle is cquel to half a rectangle of the same base and altitude, we have at once area ABC = ah. Example. Let a = 40 chains and h = 14 52 chains, then area = i x 40 x 14 52 = 290 4 square chains. (ft) When two sides a and c and the included angle Bare given. From fig. 3 = sinB , and therefore 7t = csinB ; hence area ^ah = ^ac sin B ; or Iog2area loga + logc + LsinB-10. Example. Let a - 40 , c = 30, and B = 30, then area = Jacsin B = |x40x30x^ = 300. (7) When the three sides a, b, c are given. From 8 2 __ h = - Vs(s - a)(s - b)(s - c) , and therefore area = ah - a x Vs(s-)(s-&)(s-c)= Vs(s - a)(s - 6)(s - c) ; or log area = J {logs + log(s - a) + log(s - fc) + log(s - c)} . Since 2s = + b + c , we have area of triangle = | N/2(a%*+ 6V + c 2 a 2 ) - (a 4 + 6 4 Example 1. Let = 13, i = 14, andc=15, then a -4(18 + 14 + 15) -21, -a = 21-13 = 8,
- -i = 21-14 = 7, ands-c = 21-15 = 6 ;
therefore area = V21 x 8 x 7 x 6 = 84 . Example 2. Let = 255, & = 238, and c = 221, then logs = log 357= 2-5526682 log (s - a) = log 102= 2-0086002 log(s-6)-logll9- 2-0755470 Iog(s-c)=logl36= 2-1335389 therefore hence lo area .$(87703543) -A -3851 771; area = 24276. (5) When any two angles B and C and the adjacent side a are given. Since c sinC < _ftsinC a sin A sin A and therefore (by ft} , -r, 2 sinBsinC , A iorv /r> , r- area = iacsmB = -, where A = 180 -(B + C), 2 sin A or Iog2 area = 2 loga + L sin B + L sin C + L cosec A - 30 . Since all the angles of a triangle are given when any two are given, we can find the area of a triangle when any two angles and any one side are given. Thus, when A, B, and c are given, we know C also, and the problem reduces to a case of the preceding. (e) When the three medians a, ft, y are given. If a, b, c be the three sides of a triangle, and a, ft, y the three medians, i.e., the lines drawn from the angles to the middle points of the opposite sides, then by well-known geometrical propositions wo have 4(o 2 + ft- + 7 2 ) = 3(ft 2 + b- + c") , 1 6(a 2 ft- + ft-y 2 + 7 2 a 2 ) = 9(a 2 6 2 + &V + cPa 2 ) , and 16(a 4 + /3 4 + 7 4 ) = 9(a 4 + Z 4 + c 4 ) . Now ( 9, 7) area of triangle = ^ /2(a-b 2 + b -c* + c-a-) - (a 4 + b 4 + c 4 ) , therefore = J V2(a-- + ft-y- + y-a-) - (a 4 + ft 4 + y 4 ) . D. Parallelograms. 10. The opposite sides and angles of a parallelogram equal, three independent data, one of which at least is a length, are necessary and sufficient to determine it completely. In the parallelogram A BC D (fig. 4) let BC = DA = , AB = CD = 6, AC = c, AE = A, the angle ABC = a and being D Since the diagonal AC divides the parallelogram into two equi valent triangles, we obtain (a) area of A BCD = 2 area of triangle ABC (ft) area of ABCD = 2 area ABC or logarea=log + lo (7) a ( 9, - 10 ; ) area of ABCD = 2 area ABC = 2( ABO + CBO) 2{iBO.AOsinAOB + JBO.COsinCX)B}-2{iBO. = iBD. AC sin $ = $cdsin ft , a6sina; ACsinj8} 11. If the parallelogram be equiangular (a rectangle), c = d, and area = ^c 2 sin/3. If it be equilateral (a rhombus), /3 = 90, and area ^cd. If it be both equiangular and equilateral (a square), c = d and /3 = 90, and area =ic 2 as before ( 5, /3). E. Trapeziums. 12. To determine a trapezium completely four data are neces sary and sufficient. In the trapezium ABCD (fig. 5) let BC = a, CD = b, DA = c, AB = d, A D B E F C Fig. 5. and AE perpendicular to BC = /i, and draw AF parallel to CD, then (a) area ABCD = area ABC + area ADC or the area is equal to half the sum of the parallel sides multiplied by the perpendicular between them. Again, area of ABF = BF x AE ( 9^a) = i(a J - L c)7t , also area of ABF = Vs(s - AB)(* - BF)(s - FA) , where hence h = Vs(s - AB)(s - BF)(s - FA), therefore (13) areaof ABCD-4(+e)fc-- ctt c - AB)(* - BF)(s - FA) since -a + 6-fc + d)(a + b-c- d)(a + b-c + d}(a -b-c + d), AB = rf, BF = a-c, andFA = CD = i. Thus we can find the area of a trapezium in terms of its sides. 13. If c = 0, ABCD becomes a triangle, and its area = jV( - ft + b + d)(a + b- d)(a + b + d)(a -b + d) . Again, if c = a, then also b = d, and ABCD becomes a parallelo gram, and its area takes the indeterminate form-, as it should do, since four sides do not completely determine a parallelogram. F. Quadrilaterals Generally. 14. A quadrilateral is completely determined when five inde pendent data are given. We consider the following cases. (a) When any diagonal and the perpendiculars on it from the opposite vertices are given. The quadrilateral ABCD (fig. 6) =ABD + BCD or the area is equal to half the product of the diagonal and the sum of the perpendiculars. If the diagonal BD fall Bi without the figure, as in the concave quadrilateral ABCD / -fif (fig. 7), then it is clear that / area ABCD = BD(AE-CF). (ft) When the diagonals and their included angle arc given. In the quadrilateral ABCD (fig. 8, p. 16) let BD = /, AC = /t, and angle DEA = a, then ABCD =ABD+ BCD ina ( 10, 7) or the area is equal to the product of the diagonals and the sine of their contained angle. The same result holds when one of the dia gonals falls without the quadrilateral, as in fig. 7, as the reader can easily verify. (7) When the four sides and the angle bc-
Fig. tween the diagonals arc B given. If a, b, c, d be the sides and a the angle between the diagonals it can easily be shown that
area of quadrilateral = |(a 2 - V 2 + c 2 - d 2 ) tan o .