91Ó°ÊÓ

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 $T:\left(x,y\right)→\left(x+y,y\right)$.

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

$T:A\left(2,3\right)→A\text{'}\left(5,3\right)$

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

$T:B\left(4,3\right)→B\text{'}\left(7,3\right)$

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

$T:C\left(4,5\right)→C\text{'}\left(9,5\right)$

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

$T:D\left(2,5\right)→D\text{'}\left(7,5\right)$

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:

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

The perimeter of square ABCD can be calculated as:

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

The distance between points $A\text{'}$ and $B\text{'}$ is:

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

The distance between points $B\text{'}$ and $C\text{'}$ is:

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

The distance between points $C\text{'}$ and $D\text{'}$ is:

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

The distance between points $A\text{'}$ and $D\text{'}$ is:

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

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

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

The perimeter of $A\text{'}B\text{'}C\text{'}D\text{'}$ 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 $4\left(1+\sqrt{2}\right)$.