b,
ΔABF=dreptunghic in F.Cunoastem AB=15cm si AF=12cm. Aplicam T Pitagora:
BF²=AB²-AF²⇒BF²=15²-12²=225-144=81=9²
BF=9cm
In ΔABD, AF=naltime. Conform T inaltimii;
AF²=BF×FD ⇒ FD=AF²/BF
FD=12²/9 ⇒ FD=144/9 ⇒ FD=16
Dar BD=AF+FD
Atunci:
BD=9+16=25cm.
c.
AF⊥BD; AP⊥(ABD) ⇒conform Teoremei celor 3 perpendiculare:
PF⊥BD.
ΔAFP dreptunghic in A. Aplicam T Pitagora:
PF²=AP²+AF²
PF²=16²+12² ⇒ PF²=256+144=400
PF=20cm
Aria ΔBPD=S=BD×PF/2
S=25×20/2=250cm²
