A fifteen feet long ladder is placed across a street such that while its base is at one edge of the street its top rests against the opposite wall at a height of nine feet. Similarly, another ladder, twenty feet long, is placed resting across the other side-wall so that the two ladders cross each other. A spider wishing to cross the street, climbs up one ladder till it gets to the meeting point; thereafter, it climbs down the other.

How long will the spider take to accomplish the crossing assuming that he covers a foot in ten seconds?

The width of the street can be calculated using the Pythagorean theorem:
a² = b² + c²
Where a the street width, b is the height of the first wall (9 feet), and c is the height of the first ladder (15 feet), the street width is calculated to be equal to 12 feet. The height of the second wall is then calculated using the theorem's equation, and is found to be 16 feet high.

 The length of the first ladder from the street to where it crosses the second ladder is equal to sin(tan^-1(12/16))*16, i.e., approximately 9.6. feet

The length of the second ladder from the street to where it crosses the first ladder is equal to sin(tan^-1(12/9))*9, i.e., 7.2.

Therefore the time required to cross the street would be approximately 168 seconds or 2 minutes and 48 seconds.

Edited on July 12, 2006, 8:48 pm

Posted by Dej Mar on 2006-07-12 15:57:20