A sheet of paper is successively folded along its length into 3 different shapes of right circular cylinder, a hollow shaft with a square base and a prism with an equilateral triangle as base. The shape enclosing the largest volume of air is (assume both ends in each shape are open)

Options

A. Square base
B. Triangular base
C. Circular base
D. All three enclose equal volume

Answer

To determine which shape encloses the largest volume of air when a sheet of paper is folded into three different shapes (a right circular cylinder, a hollow shaft with a square base, and a prism with an equilateral triangle as the base), we will calculate the volumes of each shape based on a constant perimeter.

### Step-by-step Solution:

1. **Assume a Perimeter**:
Let's assume the perimeter of the sheet of paper is \( P = 44 \) units. This will be the same for all three shapes.

2. **Volume of the Right Circular Cylinder**:
- The circumference of the circular base is equal to the perimeter:
\[ 2\pi r = 44 \]
- Solving for \( r \):
\[ r = \frac{44}{2\pi} = \frac{22}{\pi} \]
- The volume \( V \) of the cylinder is given by:
\[ V = \pi r^2 h \]
- Substituting \( r \):
\[ V = \pi \left(\frac{22}{\pi}\right)^2 h = \pi \cdot \frac{484}{\pi^2} \cdot h = \frac{484}{\pi} h \]
- Approximating \( \pi \approx 3.14 \):
\[ V \approx \frac{484}{3.14} h \approx 154 h \]

3. **Volume of the Hollow Shaft with Square Base**:
- The perimeter of the square base is:
\[ 4a = 44 \quad \Rightarrow \quad a = 11 \]
- The volume \( V \) of the hollow shaft is:
\[ V = a^2 h = 11^2 h = 121 h \]

4. **Volume of the Prism with Equilateral Triangle Base**:
- The perimeter of the equilateral triangle is:
\[ 3a = 44 \quad \Rightarrow \quad a = \frac{44}{3} \]
- The area \( A \) of the equilateral triangle is:
\[ A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left(\frac{44}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{1936}{9} = \frac{484\sqrt{3}}{9} \]
- The volume \( V \) of the prism is:
\[ V = A \cdot h = \frac{484\sqrt{3}}{9} h \]

5. **Comparing Volumes**:
- Volume of the cylinder: \( V_c = 154h \)
- Volume of the square base: \( V_s = 121h \)
- Volume of the triangular base: \( V_t = \frac{484\sqrt{3}}{9}h \)

To compare \( V_t \) with \( V_c \):
- Calculate \( \sqrt{3} \approx 1.732 \):
\[ V_t \approx \frac{484 \cdot 1.732}{9}h \approx \frac{838.088}{9}h \approx 93.1h \]

6. **Conclusion**:
- The volumes are approximately:
- Cylinder: \( 154h \)
- Square base: \( 121h \)
- Triangle base: \( 93.1h \)
- The largest volume is from the right circular cylinder.

### Final Answer:
The shape enclosing the largest volume of air is the **right circular cylinder**.