a)
Zerlege das Viereck ABCD in 2 Dreiecke ACD und ABC.
A = AACD + AABC
AACD = 1/2 * CD*DA = 1/2*(195,7m+65,7m)*168,1m = 21970,67 m^2
AABC = 1/2*AC*hAC
AC = √(CD2 + DA2) = √((195,7m+65,7m)2 + (168,1m)2) = 310,79 m
hAC = sin (CAE) *AB
CAE = DAE - CAD
DAE = 116,5°
CAD = arcsin (CD/AC) = arcsin ((195,7m+65,7m)/310,79m) =57,25°
CAE = 116,5° - 57,25° = 59,25°
hAC = sin (CAE) *AB = sin (59,25°) * (60,6m+182,5m) = 208,91 m
AABC = 1/2*AC*hAC = 1/2 * 310,79 m * 208,91 m = 32463,83 m^2
A = 21970,67 m^2 + 32463,83 m^2 = 54434,5 m^2 = 544,34 Ar ≈ 544 Ar
b) Einheit ist immer m, daher weggelassen
ED = √(DA2 + AE2 - 2*DA*AE*cos DAE ) (Kosinussatz für Dreieck DAE)
ED = √(168,12 + 60,62 - 2*168,1*60,6*cos 116,5° ) = 202,54
sin EDA = (AE * sin DAE) / ED (Sinussatz für Dreieck DAE)
EDA = arcsin [(AE * sin DAE) / ED]
EDA = arcsin [60,6*sin 116,5°/202,54] = 15,53°
FDE = FDA - EDA = 90° - 15,53° = 74,47°
EF = √(FD2 + ED2 - 2*FD*ED*cos FDE ) (Kosinussatz für Dreieck FDE)
EF = √(195,72 + 202,542 - 2*195,7*202,54*cos 74,47°)
EF = 241,03
Der Weg durch den Wald ist also 241m lang.