The area of the rectangle formed by the intersection of the plane and cube is [tex]4\sqrt{2}cm^2[/tex]
The diagram of the cube and the cutting plane is shown in the figure.
Since the cube has a volume of [tex]8cm^3[/tex], the sides are [tex]2cm[/tex] each.
From the diagram, the red rectangle is the rectangle formed from the intersection of the plane and the cube. It has an area given by
[tex]A_R=\text{side of cube}\times \text{diagonal}=2d[/tex]
and, using Pythagoras' theorem,
[tex]d=\sqrt{2^2+2^2}\\d=\sqrt{8}=2\sqrt{2}\text{ cm}[/tex]
Substituting the value of [tex]d[/tex] into the formula for [tex]A_R[/tex]
[tex]A_R=2\times d\\=2\times 2\sqrt{2}\\=4\sqrt{2}[/tex]
The area of the rectangle formed by the intersection of the plane and cube is [tex]4\sqrt{2}cm^2[/tex]
Learn more about areas formed from cubes here: https://brainly.com/question/25636539
