Chapter 9.6, Problem 30PPS

(a)

To determine

To graph: The WXYZ and find its perimeter.

Answer to Problem 30PPS

The graph of WXYZ is shown in figure (1) and its perimeter is $23.1\text{units}$

Explanation of Solution

The coordinates of the polygon are:

  $\begin{array}{c}W\left(6,2\right)\\ X\left(3,7\right)\\ Y\left(-1,4\right)\\ Z\left(4,-2\right)\end{array}$

The graph of $\text{quad}WXYZ$ is shown below.

  

Figure (1)

The perimeter is calculated as,

  $\begin{array}{c}P=WX+XY+YZ+ZW\\ =\left[\begin{array}{l}\left(\sqrt{{\left(6-3\right)}^2+{\left(2-7\right)}^2}\right)+\left(\sqrt{{\left(3+1\right)}^2+{\left(7-4\right)}^2}\right)+\\ \left(\sqrt{{\left(-1-4\right)}^2+{\left(4+2\right)}^2}\right)+\left(\sqrt{{\left(4-6\right)}^2+{\left(-2-2\right)}^2}\right)\end{array}\right]\\ =\sqrt{34}+\sqrt{25}+\sqrt{61}+\sqrt{20}\\ =5.8+5+7.8+4.5\\ =23.1\text{units}\end{array}$

Therefore, the graph is shown in figure (1) and the perimeter is $23.1\text{units}$ .

(b)

To determine

To graph: The image WXYZ.

Answer to Problem 30PPS

The graph of image of WXYZ is shown in figure (2).

Explanation of Solution

The scale factor is $\frac{1}{2}$ .

The new coordinate are $\left(x,y\right)\to \left(kx,ky\right)$ . So,

  $\begin{array}{c}W\left(6,2\right)\to W^{\prime }\left(3,1\right)\\ X\left(3,7\right)\to X^{\prime }\left(1.5,3.5\right)\\ Y\left(-1,4\right)\to Y^{\prime }\left(-0.5,2\right)\\ Z\left(4,-2\right)\to Z^{\prime }\left(2,-1\right)\end{array}$

The graph of the new quadrilateral with old quadrilateral is shown below.

  

Figure (2)

Therefore, the image of the polygon is shown in figure (2).

(c)

To determine

To find: The perimeter of the dilated image and compare with the preimage.

Answer to Problem 30PPS

The perimeter of the dilated image is $11.6\text{unit}$ and it is $\frac{1}{2}$ times the original image.

Explanation of Solution

The perimeter of dilated image is calculated as,

  $\begin{array}{c}{P}^{\prime }=W^{\prime }{X}^{\prime }+X^{\prime }{Y}^{\prime }+Y^{\prime }{Z}^{\prime }+W^{\prime }{Z}^{\prime }\\ =\left[\begin{array}{l}\left(\sqrt{{\left(3-1.5\right)}^2+{\left(1-3.5\right)}^2}\right)+\left(\sqrt{{\left(1.5+0.5\right)}^2+{\left(3.5-2\right)}^2}\right)+\\ \left(\sqrt{{\left(-0.5-2\right)}^2+{\left(2+1\right)}^2}\right)+\left(\sqrt{{\left(2-3\right)}^2+{\left(-1-1\right)}^2}\right)\end{array}\right]\\ =\sqrt{8.5}+\sqrt{6.25}+\sqrt{15.25}+\sqrt{5}\\ =2.9+2.5+3.9+2.2\\ =11.6\text{units}\end{array}$

On comparison, the perimeter of the dilated image is $\frac{1}{2}$ times the original quadrilateral.

Therefore, the perimeter of the dilated image is $11.6\text{unit}$ and it is $\frac{1}{2}$ times the original image.