Find the perimeter of the figure shown. Round your answer to the nearest tenth.

To calculate the perimeter fo the figure you have to add all of its sides, so the first steo is to determine the length of said sides:
Side AD
This side is parallel to the x-axis.
To determine the length of this side you have to calculate the difference between the x-coordinates of points D and A.
[tex]AD=x_D-x_A=4-(-4)=4+4=8[/tex]Side CD
You'll have to use the pythagoras theorem to calculate this since its not parallel to either axis:
The theorem states that
[tex]a^2+b^2=c^2[/tex]The sum of squares of the base and height of a triangle is equal to the square of its hypotenuse.
Side CD will be the hypothenuse of this triangle, and the base and height will be given by the difference x and y coordinates of points D and C, so:
[tex]\begin{gathered} CD^2=(x_C-x_D)^2+(y_C-y_D)^2 \\ CD^2=(2-4)^2+(4-(-3))^2 \\ CD^2=53 \\ CD=\sqrt[]{53}\cong7.28 \end{gathered}[/tex]Side BC
You have to apply the same procedure as with side CD:
[tex]\begin{gathered} BC^2=(x_C-x_B)^2+(y_C-y_B)^2 \\ BC^2=(2-(-3))^2+(4-2)^2 \\ BC^2=40 \\ BC=\sqrt[]{40}=2\sqrt[]{10}\cong6.32 \end{gathered}[/tex]Side AB
You have to use the theorem to calculate this side length too:
[tex]\begin{gathered} AB^2=(x_B-x_A)^2+(y_B-y_A)^2 \\ AB^2=(-3-(-4))^2+(2-(-3))^2 \\ AB^2=26 \\ AB=\sqrt[]{26}\cong5.099 \end{gathered}[/tex]Using the calculated side lengths you can calculate the perimeter:
[tex]P=AD+CD+BC+AB=8+\sqrt[]{53}+2\sqrt[]{10}+\sqrt[]{26}=26.703\cong26.7[/tex]The perimeter of the figure is 26.7 units