karias1980
karias1980 karias1980
  • 07-08-2018
  • Mathematics
contestada

To the nearest tenth, find the perimeter of ∆ABC with vertices A(-1,4), B(-2,1) and C(2,1).

*Please show all the work!

Respuesta :

Riia
Riia Riia
  • 08-08-2018

First we need to find the length of each side of the triangle , and for that we need to use the distance formula, that is
[tex] d =\sqrt{(x_{2} -x_{1})^2 +(y_{2} -y_{1})^2} [/tex]
[tex] AB = \sqrt{(1-4)^2 +(-2+1)^2}= \sqrt{10} [/tex]
[tex] \sqrt{BC} = \sqrt{(1-1)^2 + (2+2)^2}=4 [/tex]
Perimeter is the sum of all sides.
[tex] Perimeter = AB + BC + AC = \sqrt{10}+4 +\sqrt{18} [/tex]
[tex] Perimeter= 11.4 [/tex]

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