A solid block of dense foam shaped as a perfect cube is split into two equal rectangular slabs by a single planar cut made parallel to its bottom face. Calculate the percentage increase in the total surface area of the two newly formed pieces combined compared to the surface area of the original cube.

The correct option is 33.33%.

Step 1: Surface area of the original cube

Let the side length of the solid foam cube be $s$.

A cube consists of 6 identical square faces, with each face having an area of $s^2$.

The total surface area of the original cube ($A_{\text{original}}$) is given by:

$A_{\text{original}}=6s^2$

Step 2: Combined surface area of the two newly formed slabs

Making a single planar cut parallel to the bottom face divides the cube into two equal rectangular slabs.

This cut exposes 2 new interior square faces, each having an area of $s^2$.

Therefore, the total surface area of the two newly formed pieces combined ($A_{\text{new}}$) is the sum of the original surface area and the area of the 2 new faces:

$A_{\text{new}}=6s^2+2s^2=8s^2$

Step 3: Calculate the increase in total surface area

The increase in total surface area is:

$\text{Increase in Area}=8s^2-6s^2=2s^2$

Step 4: Calculate the percentage increase

The percentage increase in surface area is calculated as:

$\text{Percentage Increase}=\left(\frac{\text{Increase in Area}}{\text{Original Surface Area}}\right)\times 100\%$

Substituting the values into the formula:

$\text{Percentage Increase}=\left(\frac{2s^2}{6s^2}\right)\times 100\%$

$\text{Percentage Increase}=\frac{1}{3}\times 100\%=33.33\%$

Thus, the percentage increase in the total surface area of the two newly formed pieces combined compared to the original cube is 33.33%.