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

Find the perimeter of the figure shown. Round your answer to the nearest tenth.-example-1

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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.


AD=x_D-x_A=4-(-4)=4+4=8

Side CD

You'll have to use the pythagoras theorem to calculate this since its not parallel to either axis:

The theorem states that


a^2+b^2=c^2

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:


\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}

Side BC

You have to apply the same procedure as with side CD:


\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}

Side AB

You have to use the theorem to calculate this side length too:


\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}

Using the calculated side lengths you can calculate the perimeter:


P=AD+CD+BC+AB=8+\sqrt[]{53}+2\sqrt[]{10}+\sqrt[]{26}=26.703\cong26.7

The perimeter of the figure is 26.7 units

User GarryOne
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