Question

A cube with edges of length $b$ is divided into $8$ equal smaller cubes. Calculate the difference between the combined surface area of the $8$ smaller cubes and the surface area of the original cube.

• A. $\frac{3}{2}{b}^2$
• B. $\frac{3}{4}{b}^2$
• C. $\frac{9}{2}{b}^2$
• D. $6b^2$
• E. $12b^2$

Answer 1

Answer: (D)

$6b^2$

Surface area of original cube is $6b^2$
Volume of cube is $b^3$
Volume of smaller cube is $\frac{b^3}{8}$, which implies side length of smaller cube is $\frac{b}{2}$
Surface area of $8$ smaller cubes combined is $8\times 6\times \frac{b}{2^2}=12b^2$
Difference is $12b^2-6b^2=6b^2$