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

Answer:
`\bold{4+3\sqrt2+√(10)}`

Explanation:

Perimeter is the sum of the lengths of the sides. Use the distance formula to find the lengths of each of the sides, then find their sum.

`d_(AB)=√((x_B-x_A)^2+(y_B-y_A)^2)`

`=√((-2+1)^2+(1-4)^2)`

`=√((-1)^2+(-3)^2)`

`=√(1+9)`

`=√(10)`

`d_(AC)=√((x_C-x_A)^2+(y_C-y_A)^2)`

`=√((2+1)^2+(1-4)^2)`

`=√((3)^2+(-3)^2)`

`=√(9+9)`

`=√(18)`

`=3√(2)`

`d_(BC)=√((x_C-x_B)^2+(y_C-y_B)^2)`

`=√((2+2)^2+(1-1)^2)`

`=√((4)^2+(0)^2)`

`=√(16+0)`

`=√(16)`

`=4`