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

`AB= √((-2-(-1))^2+(1-4)^2)= √(1+9)= √(10) \approx 3.2\\\\ BC= √((2-(-2))^2+(1-1)^2)= √(16)=4\\\\ AC= √((2-(-1))^2+(1-4)^2)= √(9+9)= √(18) \approx 4.2 \\\\ P_(ABC)=AB+BC+AC=3.2+4+4.2=11.4`

The perimeter of ∆ABC = 11.4 units

Answer 2

The perimeter is 11.4 units

Step-by-step explanation

Perimeter is the distance around the figure.

We use the distance formula,

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

to determine the length of all the sides and add them.

`|AB|=√((-2--1)^2+(1-4)^2)`

`|AB|=√((-1)^2+(-3)^2)`

`|AB|=√(1+9)`

`|AB|=√(10) \approx 3.162`

`|AC|=√((2--1)^2+(1-4)^2)`

`|AC|=√((2+1)^2+(1-4)^2)`

`|AC|=√((3)^2+(-3)^2)`

`|AC|=√(9+9)`

`|AC|=√(18) \approx 4.24`

`|BC|=√((2--2)^2+(1-1)^2)`

`|BC|=√((2+2)^2+(1-1)^2)`

`|BC|=√((4)^2+(0)^2)`

`|BC|=√(14)=4`

Therefore perimeter=
`|AB|+|BC|+|AC|`

=
`4.00+3.16+4.24`

=
`11.40` units