Shortest diagonal distance

Question

We have an empty cardboard box of dimensions breadth = 3 m, length = 4 m, and height = 2 m. An ant and a fly are sitting at one of the 8 corners inside the box and wish to go get some food at the corner diagonally across (the farthest point in the box for them). What is the minimum distance the fly must travel before it gets to the food? How about the ant?

Answer 1

Suppose the ant and the fly are at the an inside bottom corner of the box.  The length of the diagonal to the opposite corner of the bottom is ^√(3² + 4²) = 5 m.  So the shortest distance for the ant to get to the food is go to the opposite corner at 5 m and crawl up the height of of 2 m, or a total distance of 7m.

 

The fly can fly directly from the corner where it is to the opposite corner.  The shortest distance for him is

√((3² + 4²) + 2²) = √(29) ≈ 5.4 m.

Answer 2

Hey Jw.

 

Step one of this problem would be figuring out the diagonal length across the bottom panel of the box. If you have a breadth of 3m and a length of 4m, then your diagonal will be:

 

3^2 + 4^2 = d^2

 

d = 5

 

Your ant will have to climb down the height of the wall then make the straight line across the diagonal. Since you know the height of the wall is 2 and the diagonal is 5, the ant will travel 2m down then 5m across, which is 7m.

 

Your fly has the luxury of making a straight line from the top corner to the bottom corner across the box. His flight path can be determined by constructing a right triangle where the height is 2m and the length is 5m (the length of your diagonal). The fly's straight line through the air will have a distance of:

 

2^2 + 5^2 = d^2

 

d = √29 m

 

Hope this helps.