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

Let's solve the problem step by step.

### Step 1: Area of the Rectangular Sheet
The area \( A \) of the rectangular sheet is given by:
\[
A = \text{length} \times \text{width} = 54 \, \text{cm} \times 4 \, \text{cm} = 216 \, \text{cm}^2
\]

### Step 2: Surface Area of the Cube
The surface area of a cube with side length \( s \) is:
\[
\text{Surface Area of Cube} = 6s^2
\]
We know that the surface area of the cube is equal to the area of the rectangular sheet, so:
\[
6s^2 = 216 \, \text{cm}^2
\]
Solving for \( s^2 \):
\[
s^2 = \frac{216 \, \text{cm}^2}{6} = 36 \, \text{cm}^2
\]
Thus, \( s = \sqrt{36} \, \text{cm} = 6 \, \text{cm} \).

### Step 3: Volume of the Cube
The volume \( V_{\text{cube}} \) of the cube is:
\[
V_{\text{cube}} = s^3 = 6^3 \, \text{cm}^3 = 216 \, \text{cm}^3
\]

### Step 4: Volume of the Cylinder
The sheet is rolled along its longer edge to form a cylindrical tube. The circumference of the base of the cylinder (which is equal to the longer edge of the sheet) is:
\[
\text{Circumference} = 54 \, \text{cm}
\]
The radius \( r \) of the cylinder's base can be found using the formula for the circumference of a circle:
\[
2\pi r = 54 \, \text{cm}
\]
\[
r = \frac{54 \, \text{cm}}{2\pi} = \frac{27}{\pi} \, \text{cm}
\]
The height \( h \) of the cylinder is equal to the width of the rectangular sheet:
\[
h = 4 \, \text{cm}
\]
The volume \( V_{\text{cylinder}} \) of the cylinder is given by:
\[
V_{\text{cylinder}} = \pi r^2 h = \pi \left(\frac{27}{\pi}\right)^2 \times 4
\]
Simplifying:
\[
V_{\text{cylinder}} = \frac{729}{\pi} \times 4 = \frac{2916}{\pi} \, \text{cm}^3
\]

### Step 5: Ratio of Volumes
The ratio of the volume of the cylinder to the volume of the cube is:
\[
\text{Ratio} = \frac{V_{\text{cylinder}}}{V_{\text{cube}}} = \frac{\frac{2916}{\pi}}{216} = \frac{2916}{216\pi} = \frac{13.5}{\pi}
\]
This simplifies further to:
\[
\text{Ratio} = \frac{4.5}{\pi} \approx \frac{4.5}{3.14} \approx 1.43
\]

So, the ratio of the volume of the cylindrical tube to the volume of the cube is approximately \( \frac{4.5}{\pi} \).

Would you like more details on any part of this solution? Here are some follow-up questions you might find interesting:

1. How does the ratio change if the sheet's dimensions are doubled?
2. What is the relationship between the side of the cube and the height of the cylinder if their volumes are equal?
3. How would the ratio change if the cylinder was formed by joining the shorter edges of the sheet instead?
4. Can the surface area of another shape be equal to that of the cube with the same volume?
5. How does the volume of the cylinder compare to that of a cone with the same base and height?

**Tip:** When comparing volumes or surface areas of different shapes, it's often useful to express them in terms of the same variable or geometric property.

Explore the solution to a geometry problem involving a cylindrical tube and a cube. Understand how to calculate volumes and ratios using basic geometric formulas. Suitable for students in Grades 7-9.

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