Question

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

Answer 1

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

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

So length AB =
`√((-2-(-1))^2+(1-4)^2)= √(1 +9)=√(10)`

BC=
`√((2-(-2))^2+(1-1)^2)= √(16+0)=4`

AC =
`√((2-(-1))^2+ (1-4)^2)= √(9+9)=√(18)=3√(2)`

Now perimeter = AB+BC+AC =
`√(10)+4+3√(2)`

Plug in calculator

perimeter= 11.4 units