Răspuns:
Explicație pas cu pas:
a) ΔAEB dreptunghic isoscel
AB=80 m => AB=AE=80 m
Aria ΔAEB=[tex]\frac{c1*c2}{2}[/tex]=[tex]\frac{AB*AE}{2}[/tex]=[tex]\frac{80*80}{2}[/tex]=40*80=3200 m²
ABCD patrat, BCF si AEB Δ isoscele => AB=BF=80 m
Aria ΔEBF= Aria ΔAEF - Aria ΔAEB
Aria ΔEBF= 6400 - 3200
A ΔAEF= [tex]\frac{80*160}{2}[/tex]=80*80=6400 m²
Aria ΔEBF= 3200
Aria ΔAEB= 3200 => Aria ΔEBF=Aria ΔAEB
b) ABCD patrat, ΔAEB=ΔBCF=ΔCDG=ΔADH => ΔEBF=ΔCGF=ΔDGH=ΔEAH
Aria ΔEBF=3200 m² => Aria totala azonei de relaxare = 3200*4= 12800 m²
c) EFGH patrat => EG diagonala
triunghiurile ΔBCF=ΔCDG=ΔDAH=ΔAEB => intersectia diagonalelor din ABCD = intersectia digonalelor din EFGH => O∈AG => E, O, G coliniare