Question

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!

Answer 1

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

`d =\sqrt{(x_(2) -x_(1))^2 +(y_(2) -y_(1))^2}`

`AB = √((1-4)^2 +(-2+1)^2)= √(10)`

`√(BC) = √((1-1)^2 + (2+2)^2)=4`
Perimeter is the sum of all sides.

`Perimeter = AB + BC + AC = √(10)+4 +√(18)`

`Perimeter= 11.4`