Discussion
A rectangular paper sheet of dimensions 54 cm × 4 cm is taken. The two longer edges of the sheet are joined together to create a cylindrical tube. A cube whose surface area is equal to the area of the sheet is also taken.
Then, the ratio of the volume of the cylindrical tube to the volume of the cube is
Answer: A
The ratio is \(1/\pi\).
Joining the two 54 cm edges makes a tube 54 cm long whose circumference is the 4 cm edge, so its radius is \(r = 4/2\pi = 2/\pi\) cm:
\[ \begin{aligned} V_{tube} &= \pi r^{2} h \\ &= \pi \cdot \frac{4}{\pi^{2}} \cdot 54 = \frac{216}{\pi} \end{aligned} \]
The sheet's area is \(54 \times 4 = 216\ \text{cm}^{2}\). A cube with that surface area has \(6a^{2} = 216\), so \(a = 6\) cm and \(V_{cube} = 216\ \text{cm}^{3}\). The ratio is \(\frac{216/\pi}{216} = \frac{1}{\pi}\).
The trap is to roll the sheet the other way (circumference 54 cm, length 4 cm), which gives a far larger tube; the stem joins the longer edges, so the circumference is the short side. The other ratios do not follow from either reading.
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