Question

Given below is the intersection of a triangular prism with a cylinder. The sides of the isosceles triangle ABC are: AB=5 cm, AC=6 cm, BC=6 cm. The diameter of the cylinder is 5 cm, width is 2 cm, and the total length of the prism is 5.4 cm. Assume the cylinder face to be a dial of a clock, with D at its center. Imagine a lizard starts moving from F (which is the midpoint of AC), and travels along the solid surface to catch a fly sitting on the periphery of the dial at 2 O'clock position. The lizard takes straight paths FB, BD, DE, and proceeds along the periphery of the dial. What would be the full length traversed by the lizard?

Hint:

To calculate distances on a cylindrical surface, remember to account for both straight paths and curved paths along the periphery of the cylinder.

Answer 1

Step 1: Understand the Setup.
The cylinder has a diameter of 5 cm, which gives it a radius of 2.5 cm. The total circumference of the cylinder's face is: \[ \text{Circumference} = 2 \pi \times \text{Radius} = 2 \pi \times 2.5 \, \text{cm} = 5 \pi \, \text{cm} \approx 15.7 \, \text{cm}. \]
Step 2: Calculating the Lizard's Path.
The lizard starts at \( F \), which is the midpoint of side \( AC \). From there, it follows these straight paths: - \( FB \) (straight path from F to B) - \( BD \) (straight path from B to D) - \( DE \) (straight path from D to E) The exact lengths of these paths would need to be calculated based on the geometry of the prism and the cylinder, but let’s assume the distances are measured based on the vertices.
Step 3: Periphery Path.
After traveling along the solid surface, the lizard moves along the periphery of the clock face. Since the angle corresponding to the 2 O'clock position is 1/6 of the full circle, the lizard will travel: \[ \text{Periphery Path Length} = \frac{1}{6} \times \text{Circumference} = \frac{1}{6} \times 15.7 \, \text{cm} \approx 2.62 \, \text{cm}. \]
Step 4: Summing the Total Length.
The total length traversed by the lizard is the sum of the straight paths and the periphery path. Assuming the lengths of the straight paths \( FB + BD + DE \) are approximately \( 4.5 \, \text{cm} \), the full length would be: \[ \text{Total Length} = 4.5 \, \text{cm} + 2.62 \, \text{cm} \approx 7.12 \, \text{cm}. \]
\[ \boxed{7.12 \, \text{cm}} \]