Question

Determine o perímetro e a área de cada um dos polígonos cujas coordenadas são

Polígono 1: A(-3,2), B(5,2) C(5,-4) e D(-3, -4).
Polígono 2: E(2,3), F(-4,3), G(-4, -3) e H(2, -3).

Answer 1

Resposta:

Boa noite, Karol! Tudo bem?

Explicação passo-a-passo:

Vamos, primeiramente, encontrar a distancia desses lados dos polígonos através da raiz quadradas de seus pontos.

Polígono I

$d_{AB}=\sqrt{(5-(-3))^{2} +(2-2)^{2} }=\sqrt{(8)^{2} +0^{2} }=\sqrt{64}=8u.c\\\\\\ d_{BC}=\sqrt{(5-5)^{2} +(-4-2)^{2} }=\sqrt{0^{2} +(-6)^{2} }=\sqrt{36 }=6u.c \\\\\\d_{CD}=\sqrt{(-3-5)^{2} +(-4-(-4))^{2} }=\sqrt{(-8)^{2} +0^{2} } =\sqrt{64} =8u.c \\\\\\d_{AD}=\sqrt{(-3-(-3))^{2} +(-4-2)^{2} }=\sqrt{0^{2} +(-6)^{2} }=\sqrt{36}=6u.c\\\\\\Perimetro:\\\\2*6+2*8=12+16=28u.c\\\\\\Area:\\\\6*8=48u.c^2$

Polígono II

$d_{EF}=\sqrt{(-4-2)^{2} +(3-3)^{2} }=\sqrt{(-6)^{2} +0^{2} }=\sqrt{36} = 6u.c \\\\\\d_{FG}=\sqrt{(-4-(-4))^{2} +(-3-3)^{2} }=\sqrt{0^{2} +(-6)^{2} } =\sqrt{36} = 6u.c \\\\\\d_{GH}=\sqrt{(2-(-4))^{2} +(-3-(-3))^{2} }=\sqrt{(6)^{2} +0^{2} }=\sqrt{36} = 6u.c\\\\\\d_{EH}=\sqrt{(2-2)^{2} +(-3-3)^{2} }=\sqrt{0+(-6)^{2} }=\sqrt{36} = 6u.c\\\\\\Perimetro:\\\\6*4=24u.c\\\\\\Area:\\\\6*6=36u.c^2$

Prof Alexandre

Bom Conselho/PE