a)
[tex]\it AB^2=(x_B-x_A)^2+(y_B-y_A)^2=(-4-2)^2+(-5-3)^2=36+64=100\\ \\ BC^2=(x_C-x_B)^2+(y_C-y_B)^2=(-6+4)^2+(9+5)^2=4+196=200\\ \\ AC^2=(x_C-x_A)^2+(y_C-y_A)^2=(-6-2)^2+(9-3)^2=64+36=100[/tex]
Vom arăta că triunghiul este dreptunghic, folosind
reciproca teoremei lui Pitagora.
[tex]\it AB^2+AC^2=100+100=200=BC^2 \Rightarrow \Delta ABC\ -\ dreptunghic,\ \hat A=90^o[/tex]
[tex]\it b)\\ \\ AB^2=100 \Rightarrow AB=\sqrt{100}=10\ cm \\ \\ AC^2=100 \Rightarrow AC=\sqrt{100}=10\ cm\\ \\ BC^2=200 \Rightarrow BC=\sqrt{200}=\sqrt{100\cdot2}=10\sqrt2\ cm\\ \\ \\ \mathcal{P}=AB+AC+BC=10+10+10\sqrt2=20+10\sqrt2\ cm\\ \\ \mathcal{A}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AB\cdot AC}{2}=\dfrac{10\cdot10}{2}=50\ cm^2[/tex]
