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Fie punctele A(8;16), B(12;12)si C (0;4) Stabiliti natura triunghiului ABC si calculati perimetrul triunghiului

Răspuns :

[tex]AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{(12-8)^2+(12-16)^2}=\\=\sqrt{4^2+4^2}=\sqrt{2\cdot4^2}=4\sqrt2\\AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}=\sqrt{(0-8)^2+(4-16)^2}=\\=\sqrt{8^2+12^2}=\sqrt{4^2(2^2+3^2)}=4\sqrt{13}\\BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}=\sqrt{(0-12)^2+(4-12)^2}=\\=\sqrt{12^2+8^2}=4\sqrt{13}\\AC=BC\Rightarrow \triangle ABC\text{ isoscel }\\P_{\triangle ABC}=AB+AC+BC=4\sqrt2+2\cdot4\sqrt{13}=4\sqrt2+8\sqrt{13}[/tex]