[tex]1)BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}=\sqrt{100}=10\\
AC=\sqrt{6,4^{2}+4,8^2}=\sqrt{64}=8\\
AB=\sqrt{3,6^2+4,8^2}=\sqrt{36}=6\\
Deoarece\ AB^2+AC^2=BC^2=\ \textgreater \ reciproca\ teoremei\ lui\ Pitagora=\ \textgreater \ \\
\triangle ABC \ dreptunghic\ in\ A\\
2)A_{\triangle ABC}= \frac{AC\cdot AB}{2}= \frac{48}{2}=24\\
3)P_{\triangle ABC}= AB+BC+AC=6+10+8=24.[/tex]
