Question

Calcule o perímetro do triângulo ABC, sendo A(1,0), B(3,7) e C(-2,4)

Answer 1

a) medida do segmento $\overline{AB}$:

$\text{med}\left(\overline{AB} \right )=\sqrt{\left(x_{B}-x_{A} \right )^{2}+\left(y_{B}-y_{A} \right )^{2}}\\ \\ =\sqrt{\left(3-1 \right )^{2}+\left(7-0 \right )^{2}}\\ \\ =\sqrt{\left(2 \right )^{2}+\left(7\right )^{2}}\\ \\ =\sqrt{4+49}\\ \\ =\sqrt{53} \text{ u.c.}$


b) medida do segmento $\overline{BC}$:

$\text{med}\left(\overline{BC} \right )=\sqrt{\left(x_{C}-x_{B} \right )^{2}+\left(y_{C}-y_{B} \right )^{2}}\\ \\ =\sqrt{\left(-2-3 \right)^{2}+\left(4-7 \right )^{2}}\\ \\ =\sqrt{\left(-5 \right )^{2}+\left(-3 \right)^{2}}\\ \\ =\sqrt{25+9}\\ \\ =\sqrt{34} \text{ u.c.}$


c) medida do segmento $\overline{CA}$:

$\text{med}\left(\overline{CA} \right )=\sqrt{\left(x_{A}-x_{C} \right )^{2}+\left(y_{A}-y_{C} \right )^{2}}\\ \\ =\sqrt{\left[1-\left(-2 \right ) \right ]^{2}+\left(0-4 \right )^{2}}\\ \\ =\sqrt{\left(3 \right )^{2}+\left(-4 \right )^{2}}\\ \\ =\sqrt{9+16}\\ \\ =\sqrt{25}\\ \\ =5 \text{ u.c.}$


O perímetro do triângulo $ABC$ é dado por

$\text{P}_{\Delta ABC}=\text{med}\left(\overline{AB} \right )+\text{med}\left(\overline{BC} \right )+\text{med}\left(\overline{CA} \right )\\ \\ \boxed{\text{P}_{\Delta ABC}=\sqrt{53}+\sqrt{34}+5 \text{ u.c.}}$