Chapter 8, Problem 10CT

To determine

To find: the perimeter and area of the polygon using given vertices and explain the solution.

Answer to Problem 10CT

Perimeter is $22.32\text{ units}\text{.}$ and area is $30\text{ sq units}\text{.}$

Explanation of Solution

Given:

  $P\left(-3,4\right),Q\left(1,4\right),R\left(-3,-2\right)\text{ and S}\left(3,-2\right).$

Concept used:

Distance formula:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Area of triangle:

  $A=\frac{1}{2}bh$ .

Area of rectangle:

  $A=l\times b$ .

Perimeter of the any quadrilateral:

Sum of length of all side.

Calculation:

To find the perimeter of the polygon, plot the vertices on a coordinate plane. identify the segments of the polygon and find the distance between the points of the segments.

Plot the points on a coordinate plane first then sketch the graph of the polygon to know which distances of segments are needed to be identified.

Note in the figure that there is the presence of the point at $\left(1,-2\right)$ with a broken line connecting to $\left(1,4\right)$ .

  

The distances of the segments are:

  $\begin{array}{l}PQ=\left|1-\left(-3\right)\right|.\\ \Rightarrow \left|1+3\right|.\\ \Rightarrow 4.\end{array}$

Similarly, the distance segment of $RP$ is:

  $\begin{array}{l}RP=\left|-2-4\right|.\\ \Rightarrow \left|-6\right|.\\ \Rightarrow 6.\end{array}$

Similarly, length of the $SR$ is:

  $\begin{array}{l}SR=\left|3-\left(-3\right)\right|.\\ \Rightarrow \left|3+3\right|.\\ \Rightarrow 6.\end{array}$

  $\begin{array}{l}QS=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}.\\ \Rightarrow \sqrt{{\left(3-1\right)}^2+{\left(-2-4\right)}^2}.\\ \Rightarrow \sqrt{4+36}=\sqrt{40}.\\ \Rightarrow 2\sqrt{10}\approx \mathrm{6.32.}\end{array}$

Therefore, the perimeter of the polygon is:

  $\begin{array}{l}PQ+RP+SR+QS\\ \Rightarrow 3+6+6+6+\mathrm{6.32.}\\ \Rightarrow 22.32\text{ units}\end{array}$

As for the area of polygon, it forms two shapes are rectangle and a triangle.

Find both are and add them together.

For the rectangle:

  $\begin{array}{l}A=l\times b\\ \Rightarrow \left(RP\right)\left(PQ\right).\\ \Rightarrow \left(6\right)\left(4\right)=24\text{ sq units}\text{.}\end{array}$

For the triangle it can use the broken line, which is congruent to $RP$ ,as its height and the distance between point $\left(1,-2\right)$ to $S\left(3,-2\right)$ .

  $\begin{array}{l}A=\frac{1}{2}bh.\\ \Rightarrow \frac{1}{2}\left(6\right)\left(2\right)=6\text{ sq units}\text{.}\end{array}$

Add the two area together to find the area of the entire polygon therefore, adding two areas:

  $24+6=30\text{ sq units}\text{.}$

Hence, perimeter is $22.32\text{ units}\text{.}$ and area is $30\text{ sq units}\text{.}$