a) Aplicam teorema lui Pitagora in ΔABC dreptunghic in A: BC²=AB²+AC²⇒BC=10
A=AB*AC/2=4,8
b)Aplicam teorema lui Pitagora in ΔABC dreptunghic in A:BC²=AB²+AC²⇒
AC=15
A=AB*AC/2=150
c)A=AB*AC/2⇔864=36*AC/2⇒AC=46
Aplicam teorema lui Pitagora in ΔABC dreptunghic in A:BC²=AB²+AC²⇒
BC=2√853
d)A=AB*AC/2=BC*AD/2⇒AD=1,44 m
