Respuesta :
[tex]AB= \sqrt{(-2-(-1))^2+(1-4)^2}= \sqrt{1+9}= \sqrt{10} \approx 3.2\\\\
BC= \sqrt{(2-(-2))^2+(1-1)^2}= \sqrt{16}=4\\\\
AC= \sqrt{(2-(-1))^2+(1-4)^2}= \sqrt{9+9}= \sqrt{18} \approx 4.2 \\\\
P_{ABC}=AB+BC+AC=3.2+4+4.2=11.4[/tex]
The perimeter of ∆ABC = 11.4 units
The perimeter of ∆ABC = 11.4 units
ANSWER
The perimeter is 11.4 units
EXPLANATION
Perimeter is the distance around the figure.
We use the distance formula,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
to determine the length of all the sides and add them.
[tex]|AB|=\sqrt{(-2--1)^2+(1-4)^2}[/tex]
[tex]|AB|=\sqrt{(-1)^2+(-3)^2}[/tex]
[tex]|AB|=\sqrt{1+9}[/tex]
[tex]|AB|=\sqrt{10} \approx 3.162[/tex]
[tex]|AC|=\sqrt{(2--1)^2+(1-4)^2}[/tex]
[tex]|AC|=\sqrt{(2+1)^2+(1-4)^2}[/tex]
[tex]|AC|=\sqrt{(3)^2+(-3)^2}[/tex]
[tex]|AC|=\sqrt{9+9}[/tex]
[tex]|AC|=\sqrt{18} \approx 4.24[/tex]
[tex]|BC|=\sqrt{(2--2)^2+(1-1)^2}[/tex]
[tex]|BC|=\sqrt{(2+2)^2+(1-1)^2}[/tex]
[tex]|BC|=\sqrt{(4)^2+(0)^2}[/tex]
[tex]|BC|=\sqrt{14}=4[/tex]
Therefore perimeter=[tex]|AB|+|BC|+|AC|[/tex]
=[tex]4.00+3.16+4.24[/tex]
=[tex]11.40[/tex] units
