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

25.6 units

Step-by-step 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=\sqrt{4+64}

d=\sqrt{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=\sqrt{64+64}

d=\sqrt{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.

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