d(H,N)=HN
NM⊥(EFG)⇒NM⊥HM⊂(EFG)⇔ΔNHM dreptunghic in M
⇒HN=√(NM²+HM²)= √(NM²+HG²+GM²)=√8²+12²+6²= √(100+144)=√244=2√61cm, cerinta
NM⊥(EFG)
EF⊂(EFG)
MF⊥EF
⇒(T3p)NF⊥EF⇔ΔNEF dreptunghic in F⇒Aria ΔNEF=NF*EF/2
NF=√(NM²+FM²)=√(8²+6²)=10
EF=12(ipoteza)
Aria ΔNEF=NF*EF/2=19*12/2=60cm², cerinta
