Question

Calcule o perímetro do triangulo de vértice A(-6,-1),B(4,2),C(3,-1)

Answer 1

Boa tarde

A(-6,-1),B(4,2),C(3,-1)

AB = √109
BC = √10
AC = √81 = 9

perimetro P = √109 + √10 + 9 = 22.6026 

Answer 2

Distância de A(-6,-1) até B(4,2):

$d_{AB}= \sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \\ \\ d_{AB}= \sqrt{(4+6)^2+(2+1)^2} \\ \\ d_{AB}= \sqrt{10^2+3^2} \\ \\ d_{AB}= \sqrt{100+9} \\ \\ d_{AB}= \sqrt{109}$

Distância de B(4,2) até C(3,-1):

$d_{BC}= \sqrt{(x_C-x_B)^2+(y_C-y_B)^2} \\ \\ d_{BC}= \sqrt{(3-4)^2+(-1-2)^2} \\ \\ d_{BC}= \sqrt{(-1)^2+(-3)^2} \\ \\ d_{BC}= \sqrt{1+9} \\ \\ d_{BC}= \sqrt{10}$

 

Distância de C(3,-1) até A(-6,-1):

$d_{CA}= \sqrt{(x_A-x_C)^2+(y_A-y_C)^2} \\ \\ d_{CA}= \sqrt{(-6-3)^2+(-1+1)^2} \\ \\ d_{CA}= \sqrt{(-9)^2+(0)^2} \\ \\ d_{CA}= \sqrt{81+0} \\ \\ d_{CA}= \sqrt{81}\\ \\ d_{CA}=9$

 

Perímetro: 

$P=d_{AB}+d_{BC}+d_{CA} \\ \\ P=\sqrt{109} + \sqrt{10} +9 \\ \\ P=10,44+3,16+9 \\ \\ P= 22,60$

Bons estudos!