Question

A cube with volume 8 $\text{cm}^3$ is cut in half by a plane which contains two edges of the cube. What is the area of the rectangle formed by the intersection of the plane and cube? express your answer in simplest radical form.

Answer 1

The area of the rectangle formed by the intersection of the plane and cube is $4\sqrt{2}cm^2$

The diagram of the cube and the cutting plane is shown in the figure.

Since the cube has a volume of $8cm^3$, the sides are $2cm$ 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

$A_R=\text{side of cube}\times \text{diagonal}=2d$

and, using Pythagoras' theorem,

$d=\sqrt{2^2+2^2}\\d=\sqrt{8}=2\sqrt{2}\text{ cm}$

Substituting the value of $d$ into the formula for $A_R$

$A_R=2\times d\\=2\times 2\sqrt{2}\\=4\sqrt{2}$

The area of the rectangle formed by the intersection of the plane and cube is $4\sqrt{2}cm^2$

Learn more about areas formed from cubes here: brainly.com/question/25636539