Chapter 14.1, Problem 15WE

To determine

To find: The area and perimeter of each figure.

Answer to Problem 15WE

The area of square ABCD is 4 and perimeter is 8. Also, the area of other figure is $4\sqrt{2}$ and perimeter is $4\left(1+\sqrt{2}\right)$ .

Explanation of Solution

Given information: The vertices of square are $A\left(2,3\right)$ , $B\left(4,3\right)$ , $C\left(4,5\right)$ and $D\left(2,5\right)$ . The transformation is $T:\left(x,y\right)\to \left(x+y,y\right)$ .

Calculation:

The transformation of the point $A\left(2,3\right)$ is:

  $T:A\left(2,3\right)\to A^{\prime }\left(5,3\right)$

The transformation of the point $B\left(4,3\right)$ is:

  $T:B\left(4,3\right)\to B^{\prime }\left(7,3\right)$

The transformation of the point $C\left(4,5\right)$ is:

  $T:C\left(4,5\right)\to C^{\prime }\left(9,5\right)$

The transformation of the point $D\left(2,5\right)$ is:

  $T:D\left(2,5\right)\to D^{\prime }\left(7,5\right)$

  

Figure (1)

The distance between points $A\left(2,3\right)$ and $B\left(4,3\right)$ is:

  $\begin{array}{c}AB=\sqrt{{\left(4-2\right)}^2+{\left(3-3\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

The distance between points $B\left(4,3\right)$ and $C\left(4,5\right)$ is:

  $\begin{array}{c}BC=\sqrt{{\left(4-4\right)}^2+{\left(5-3\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

The distance between points $C\left(4,5\right)$ and $D\left(2,5\right)$ is:

  $\begin{array}{c}CD=\sqrt{{\left(2-4\right)}^2+{\left(5-5\right)}^2}\\ =\sqrt{{\left(-2\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

The distance between points $A\left(2,3\right)$ and $D\left(2,5\right)$ is:

  $\begin{array}{c}AD=\sqrt{{\left(2-2\right)}^2+{\left(3-5\right)}^2}\\ =\sqrt{{\left(-2\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

Area of square ABCD can be calculated as:

  $\begin{array}{c}\text{Area}=\left(2\times 2\right)\\ =4\end{array}$

The perimeter of square ABCD can be calculated as:

  $\begin{array}{c}\text{Perimeter}=4a\\ =4\times 2\\ =8\end{array}$

The distance between points $A^{\prime }$ and $B^{\prime }$ is:

  $\begin{array}{c}{A}^{\prime }{B}^{\prime }=\sqrt{{\left(7-5\right)}^2+{\left(3-3\right)}^2}\\ =\sqrt{2^2+0}\\ =\sqrt{4}\\ =2\end{array}$

The distance between points $B^{\prime }$ and $C^{\prime }$ is:

  $\begin{array}{c}{B}^{\prime }{C}^{\prime }=\sqrt{{\left(9-7\right)}^2+{\left(5-3\right)}^2}\\ =\sqrt{2^2+2^2}\\ =\sqrt{4+4}\\ =2\sqrt{2}\end{array}$

The distance between points $C^{\prime }$ and $D^{\prime }$ is:

  $\begin{array}{c}{C}^{\prime }{D}^{\prime }=\sqrt{{\left(7-9\right)}^2+{\left(5-5\right)}^2}\\ =\sqrt{{\left(-2\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

The distance between points $A^{\prime }$ and $D^{\prime }$ is:

  $\begin{array}{c}{A}^{\prime }{D}^{\prime }=\sqrt{{\left(7-5\right)}^2+{\left(5-3\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+2^2}\\ =\sqrt{4+4}\\ =2\sqrt{2}\end{array}$

The area of image $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is calculated as,

  $\begin{array}{c}\text{Area}=\text{length}\times \text{breadth}\\ =2\times 2\sqrt{2}\\ =4\sqrt{2}\end{array}$

The perimeter of $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is:

  $\begin{array}{c}\text{Perimeter}=2+2\sqrt{2}+2+2\sqrt{2}\\ =4+4\sqrt{2}\\ =4\left(1+\sqrt{2}\right)\end{array}$

Therefore, the area of square ABCD is 4 and perimeter is 8. Also, the area of other figure is $4\sqrt{2}$ and perimeter is $4\left(1+\sqrt{2}\right)$ .