The coordinates of the vertices of the original figure are
A(5, 0), B(0, 5), C(-5, 0), and D(0, -5)
Therefore OB = 5, and OA = 5
Since AOB is a right-angled triangle with hypotenuse AB,
then
[tex]AB^2=5^2+5^2=50[/tex]
Hence,
[tex]AB=5\sqrt[]{2}[/tex]
From the graph, we can see that
BD = 10 =AC
and AC is perpendicular to BD
Therefore the figure is a square,
Which means that
[tex]AB=BC=CD=DA=5\sqrt[]{2}[/tex][tex]\begin{gathered} (x,y)\rightarrow(0.8x,0.8y) \\ \text{ is the same as} \\ (x,y)\rightarrow0.8(x,y) \end{gathered}[/tex]
Hence the image of the figure under the transformation is also a square with the length of a side equal to the product of 0.8 and the length of side of a side of the original figure
That is
[tex]\text{ length of each side of image =0.8}\times5\sqrt[]{2}=4\sqrt[]{2}[/tex]
Therefore
[tex]\begin{gathered} A^{\prime}B^{\prime}\text{ = 4}\sqrt[]{2} \\ B^{\prime}C^{\prime}\text{ = 4}\sqrt[]{2} \\ C^{\prime}D^{\prime}\text{ = 4}\sqrt[]{2} \\ D^{\prime}A^{\prime}\text{ = 4}\sqrt[]{2} \end{gathered}[/tex]
