Question

os pontos A=(3, 6, -6), B=(0, 5, -3), C=(1, 2, 0) são os vértices de um triângulo ABC podemos afirmar que a área do triângulo ABC em cm², é aproximadamente.

Answer 1

$Sabemos\ que: \\\\ A=(3,6,-6) \ \| \ B=(0,5,-3) \ \| \ C=(1,2,0) \\ d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \\\\ Calculando\ d_{AB}: \\\\ d_{AB}= \sqrt{(0-3)^2+(5-6)^2+(-3-(-6))^2}\\ d_{AB}= \sqrt{(-3)^2+(-1)^2+(3)^2}\\ d_{AB}= \sqrt{9+1+9}= \sqrt{19}\ cm\\\\ Calculando\ d_{AC}:\\\\ d_{AC}= \sqrt{(1-3)^2+(2-6)^2+(0-(-6))^2}\\ d_{AC}=\sqrt{(-2)^2+(-4)^2+(6)^2}\\ d_{AC}=\sqrt{4+16+36}=\sqrt{56}\\ d_{AC}=2\sqrt{14}\ cm$

$Calculando\ d_{BC}:\\\\ d_{BC}=\sqrt{(1-0)^2+(2-5)^2+(0-(-3))^2}\\ d_{BC}=\sqrt{(1)^2+(-3)^2+(3)^2}\\ d_{BC}=\sqrt{1+9+9}=\sqrt{19}\ cm\\\\ Logo:\\\\ d_{AB}=\sqrt{19}\ cm\\ d_{AC}=2\sqrt{14}\ cm \ --\ \textgreater \ \ is\acute osceles\\ d_{BC}=\sqrt{19}\ cm\\\\ Calculando\ a\ altura:\\\\ l^2=(\frac{b}{2})^2+h^2\\ h^2=(\sqrt{19})^2-(\frac{2\sqrt{14}}{2})^2\\h^2=19-14\\h^2=5\ --\ \textgreater \ \ h=\sqrt{5}\ cm\\\\Calculando\ a\ \acute area:\\\\ A=\frac{b.h}{2}\ --\ \textgreater \ \ A=\frac{2\sqrt{14}.\sqrt{5}}{2}\\\\ \| \ A=\sqrt{70}\ cm^2\ \|$