Question

4- Determine o perimetro do triângulo ABC
​

Answer 1

Explicação passo-a-passo:

$AB = d(AB)= \sqrt{(3 + 1)^{2} + (5 - 2)^{2} } = > AB= \sqrt{ {4}^{2} + {3}^{2} } = > AB= \sqrt{16 + 9} = >AB = \sqrt{25} = > AB = 5$

$AC = d( AC) = \sqrt{( - 3 - 3)^{2} + ( - 6 - 2)^{2} } = > AC = \sqrt{( - 6)^{2} + ( - 8)^{2} } = > AC = \sqrt{36 + 64} = > AC = \sqrt{100} = > AC = 10$

$BC = d(BC) = \sqrt{( - 3 + 1)^{2} + ( - 6 - 5)^{2} } = > BC = \sqrt{( - 2)^{2} + ( - 11)^{2} } = > BC = d(BC) = \sqrt{4 + 121} = > BC = \sqrt{125} = > BC = 5 \sqrt{5}$

$Perímetro = AB + AC + BC =5 + 10 + 5 \sqrt{5} = 15 + 5 \sqrt{5}$