Chapter 8, Problem 11CT

To determine

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

Answer to Problem 11CT

Perimeter is $17.66\text{ units}\text{.}$ and area is $15\text{ sq units}$ .

Explanation of Solution

Given:

  $J\left(-1,3\right),K\left(5,3\right)\text{ and }L\left(2,-2\right).$

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 that the figure has the midpoint of $\overline{JK}$ at point $\left(2,3\right)$ with a broken line connecting it to $L$ .

It needs to finding the are later.

The distances of the segments are:

  $\begin{array}{l}JK=\left|5-\left(-1\right)\right|.\\ \Rightarrow \left|5+1\right|.\\ \Rightarrow 6.\end{array}$

Similarly, the distance segment of $KL$ is:

  $\begin{array}{l}KL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}.\\ \Rightarrow \sqrt{{\left(2-5\right)}^2+{\left(-2-3\right)}^2}\\ \Rightarrow \sqrt{9+25}=\sqrt{34}=\mathrm{5.83.}\end{array}$

  $\begin{array}{l}LJ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}.\\ \Rightarrow \sqrt{{\left(2-\left(-1\right)\right)}^2+{\left(-2-3\right)}^2}.\\ \Rightarrow \sqrt{9+25}=\sqrt{34}.\\ \Rightarrow \mathrm{5.83.}\end{array}$

Therefore, the perimeter of the polygon is:

  $\begin{array}{l}JK+KL+LJ\\ \Rightarrow 6+5.83+5.83\\ \Rightarrow 17.66\text{ units}\end{array}$

As for the area of polygon, it can us e the distance of $JK$ as its base and height as the distance of the broken line a from the mid-point of $JK$ to $L\left(-2,-3=-5\right)$ .

  $\begin{array}{l}A=\frac{1}{2}bh.\\ \Rightarrow \frac{1}{2}\left(6\right)\left(5\right)=15\text{ sq units}\text{.}\end{array}$

The area is $15\text{ sq units}$ .

Hence, perimeter is $17.66\text{ units}\text{.}$ and area is $15\text{ sq units}$