ericamcfaddentx
ericamcfaddentx ericamcfaddentx
  • 01-08-2018
  • Mathematics
contestada

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

To the nearest tenth find the perimeter of ABC with vertices A14 B21 and C21 show your work class=

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Mustela Mustela
  • 01-08-2018

we have to firstly apply the distance formula to find the length of sides of the triangle

distance formula = [tex] \sqrt{(x_{2} -x_{1})^2 +(y_{2}- y_{1})^2} [/tex]

So length AB = [tex] \sqrt{(-2-(-1))^2+(1-4)^2}= \sqrt{1 +9}=\sqrt{10} [/tex]

BC= [tex] \sqrt{(2-(-2))^2+(1-1)^2}= \sqrt{16+0}=4 [/tex]

AC = [tex] \sqrt{(2-(-1))^2+ (1-4)^2}= \sqrt{9+9}=\sqrt{18}=3\sqrt{2} [/tex]

Now perimeter = AB+BC+AC = [tex] \sqrt{10}+4+3\sqrt{2} [/tex]

Plug in calculator

perimeter= 11.4 units

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