Question

if a cubic cell with 2-cm sides doubles the dimensions of its sides, what change occurs to the ratio of its surface area to volume?
a. the ratio is halved.
b. the ratio is doubled.
c. the ratio is quadrupled.
d. the ratio is quartered.

Explanation:

Step1: Calculate the surface - area - to - volume ratio for the original cube

The surface area formula of a cube \(S = 6s^{2}\), and the volume formula \(V=s^{3}\), where \(s\) is the side length of the cube.
For \(s = 2\mathrm{cm}\), \(S_1=6\times(2)^{2}=24\mathrm{cm}^{2}\), \(V_1=(2)^{3}=8\mathrm{cm}^{3}\).
The ratio \(r_1=\frac{S_1}{V_1}=\frac{6s^{2}}{s^{3}}=\frac{6}{s}\), substituting \(s = 2\), we get \(r_1=\frac{6}{2}=3\).

Step2: Calculate the surface - area - to - volume ratio for the new cube

When \(s\) doubles, \(s = 4\mathrm{cm}\).
\(S_2=6\times(4)^{2}=96\mathrm{cm}^{2}\), \(V_2=(4)^{3}=64\mathrm{cm}^{3}\).
Using the ratio formula \(r_2=\frac{S_2}{V_2}=\frac{6s^{2}}{s^{3}}=\frac{6}{s}\), substituting \(s = 4\), we get \(r_2=\frac{6}{4}=1.5\).

Step3: Analyze the change in the ratio

We find the relationship between \(r_1\) and \(r_2\). Since \(r_1 = 3\) and \(r_2=1.5\), and \(r_2=\frac{1}{2}r_1\).

Answer:

A. The ratio is halved.