Question

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

Answer 1

Answer:

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.