Question

Below is a schematic diagram of a room 2m long, 2m wide, and 3m high. An ant is at position A on the ground and a lizard is at position B on the ceiling. 
Calculate the shortest distance the lizard has to travel to eat the ant. Write the answer in the box provided. 

 

Hint:

When finding the shortest path on a 3D object like a room, unfold the walls into a 2D plane to turn a complex 3D problem into a simple 2D one. Then use the Pythagorean theorem to calculate the straight-line distance between the start and end points. Always compare possible unfoldings to find the true shortest path.

Answer 1

The shortest distance the lizard has to travel is found by "unfolding" the room into a two-dimensional net and calculating the straight-line distance (the hypotenuse of a right-angled triangle) between point A (Ant) and point B (Lizard). 
The room dimensions are: 
Length (L) = 2 m 
Width (W) = 2 m 
Height (H) = 3 m 
Since the ant (A) is on the ground and the lizard (B) is on the ceiling at the diagonally opposite corner, the path must involve crossing at least two faces. We need to check the three main ways to unfold the room to find the shortest distance. 
1. Path Across Two Adjacent Side Walls (Length and Width Walls) 
The shortest path will go up two adjacent vertical walls. 
Total horizontal distance = Length + Width = \( L + W = 2 + 2 = 4 \) m 
Vertical distance = Height = \( H = 3 \) m 
This forms a right-angled triangle with sides 4 m and 3 m: \[ d_1 = \sqrt{(L + W)^2 + H^2} = \sqrt{(2 + 2)^2 + 3^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m} \] 2. Path Across a Side Wall and the Ceiling/Floor 
This path goes up one wall and across the ceiling or floor. 
Length along one side = \( L + H = 2 + 3 = 5 \) m 
Other side = \( W = 2 \) m 
This forms another right-angled triangle: \[ d_2 = \sqrt{(L + H)^2 + W^2} = \sqrt{(2 + 3)^2 + 2^2} = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \text{ m} \] Conclusion 
Since \( 5<5.39 \), the shortest distance the lizard has to travel is: \[ \boxed{5 \text{ meters}} \]