Question

What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?

Answer 1

25.6 units

Explanation:

From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

`d=\sqrt{(x_2-x_1)^(2) +(y_2-y_1)^(2)}`

where

`(x_1,y_1)` are the coordinates of the first point

`(x_2,y_2)` are the coordinates of the second point

- For AB:

`d=\sqrt{[1-(-5)]^(2)+(4-4)^2}`

`d=\sqrt{(1+5)^(2)+(0)^2}`

`d=\sqrt{(6)^(2)}`

`d=6`

- For BC:

`d=\sqrt{(3-1)^(2) +(-4-4)^(2)}`

`d=\sqrt{(2)^(2) +(-8)^(2)}`

`d=√(4+64)`

`d=√(68)`

`d=8.24`

- For AC:

`d=\sqrt{[3-(-5)]^(2) +(-4-4)^(2)}`

`d=\sqrt{(3+5)^(2) +(-8)^(2)}`

`d=\sqrt{(8)^(2) +64}`

`d=√(64+64)`

`d=√(128)`

`d=11.31`

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

`p=AB+BC+AC`

`p=6+8.24+11.31`

`p=25.55`

`p=25.6`

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.