1) Calculati aria triunghiului ABC, daca:
[tex]\displaystyle a) ~~~AB=5~cm,~~BC=6~cm~\text{ si }~m(\ \textless \ B)=30^o. \\ \\
A = \frac{AB\times BC\times \sin B}{2} = \frac{5\times 6\times \sin 30^o}{2} = \frac{30\times \frac{1}{2} }{2} =\frac{15 }{2} = \boxed{7,5~cm^2} [/tex]
[tex]\displaystyle b)~AB=AC=BC=12~dm\\ \\
A = \frac{l^2 \sqrt{3} }{2} =\frac{12^2 \sqrt{3} }{2} =\frac{144 \sqrt{3} }{2} =\boxed{72\sqrt{3}~dm^2}
[/tex]
[tex]\displaystyle c)~AB=4\sqrt{6}~m,~AC=BC~\text{ si }~m(\ \textless \ C)=90^o. \\
\text{Este un triunghi dreptunghic isoscel cu ipotenuza AB} \\ \\
A = \frac{AB^2}{4} = \frac{AB^2}{4} =\frac{(4\sqrt{6})^2}{4} =\frac{16 \times 6}{4}=4 \times 6 = \boxed{24~m^2}[/tex]
[tex]\displaystyle d)~AB=6~m,~AC=9~m,~BC=11~m \\
\text{Aplicam formula lui Heron: }\\
P=AB+AC+BC=6+9+11 = 26~m \\ \\
p= \frac{P}{2} = \frac{26}{2} =13 ~m \\ \\
A = \sqrt{p(p-AB)(p-AC)(p-BC)}= \\
= \sqrt{13(13-6)(13-9)(13-11)}= \\
= \sqrt{13 \times 7 \times 4 \times 2}= \sqrt{728} = \boxed{2\sqrt{182}~m^2} ~\approx~ \boxed{26,98~m^2 } \\ [/tex]
