# Chapter 12 - Think and Solve: 34

a. Eight smaller cubes (page 236, Figure 12.16). b. Each face of the original cube has an area of $4 cm^{2}$. There are 6 faces, so the original total surface area is $24 cm^{2}$ . Each of the eight smaller cubes has an area of $6 cm^{2}$, so the total surface area is $48 cm^{2}$. The ratio of new to old surface area is 2. c. For the original cube: $\frac{24 cm^{2} }{8 cm^{3} } = 3 \frac{1}{cm}$. For the set of 8 smaller cubes: $\frac{48 cm^{2} }{8 cm^{3} } = 6 \frac{1}{cm}$ , twice as great.

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