Question

Se um triangulo tem vertices nos pontos
A=(1,-3)
B=(-2,0)
C=(9,5)
Entao o triangulo tem perimetro igual a:

Answer 1

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

$\text{med}\left(\overline{AB} \right )=\sqrt{\left(x_{B}-x_{A} \right )^{2}+\left(y_{B}-y_{A} \right )^{2}}\\ \\ =\sqrt{\left(-2-1 \right )^{2}+\left[0-\left(-3 \right ) \right ]^{2}}\\ \\ =\sqrt{\left(-3 \right )^{2}+\left(3 \right )^{2}}\\ \\ =\sqrt{9+9}\\ \\ =\sqrt{9 \cdot 2}\\ \\ =3\sqrt{2} \text{ u.c.}$


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

$\text{med}\left(\overline{BC} \right )=\sqrt{\left(x_{C}-x_{B} \right )^{2}+\left(y_{C}-y_{B} \right )^{2}}\\ \\ =\sqrt{\left[9-\left(-2 \right) \right ]^{2}+\left(5-0 \right )^{2}}\\ \\ =\sqrt{\left(11 \right )^{2}+\left(5 \right)^{2}}\\ \\ =\sqrt{121+25}\\ \\ =\sqrt{146} \text{ u.c.}$


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

$\text{med}\left(\overline{CA} \right )=\sqrt{\left(x_{A}-x_{C} \right )^{2}+\left(y_{A}-y_{C} \right )^{2}}\\ \\ =\sqrt{\left(1-9 \right )^{2}+\left(-3-5 \right )^{2}}\\ \\ =\sqrt{\left(-8 \right )^{2}+\left(-8 \right )^{2}}\\ \\ =\sqrt{64+64}\\ \\ =\sqrt{64 \cdot 2}\\ \\ =8\sqrt{2} \text{ u.c.}$


O perímetro do triângulo $ABC$ é dado por

$\text{P}_{\Delta ABC}=\text{med}\left(\overline{AB} \right )+\text{med}\left(\overline{BC} \right )+\text{med}\left(\overline{CA} \right )\\ \\ \text{P}_{\Delta ABC}=3\sqrt{2}+\sqrt{146}+8\sqrt{2}\\ \\ \text{P}_{\Delta ABC}=\left(3+8 \right )\sqrt{2}+\sqrt{146}\\ \\ \boxed{\text{P}_{\Delta ABC}=11\sqrt{2}+\sqrt{146} \text{ u.c.}}$