AB=BC=CD=DA=9√2≈12,72 m
Perimetrul=4x9√2=36√2
muchia =324√6:36√2=9√3 ≈ 15,58 m
pachetele paralelipipedice vor fi asezate cu dimensiunea de 3 pe muchia prismei
12,72 : 4 ≈3
12,72:6≈2
15,58 :3≈ 5
deci vor intra 3x2x5=30 pachete
diagonala bazei=√[(9√2)²+(9√2)²]=√(162+162)=√324=18 m
1/2 diagonala = 18/2=9
distanta de la F la AC=√[9²+(9√3)²]=√324=18 m