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?

(In reply to 7 legs for an 8 legged walk by Dej Mar)

Dej Mar, the angle of intersection is indeed 90 degrees, but the triangle formed with the street is not isosceles, as the ladders are not the same length and don't intersect exactly halfway across the street. Actually, the triangle formed by the two lower ends of the ladder and the street is also proportional to the two triangles formed by the ladders, walls and street, also this time 12 is the length of the hypotenuse. Patrick gives the correct measurements of the two legs: 7.2 and 9.6 feet. So the spider travels 16.8 feet, in 168 seconds.

Posted by Jyqm on 2006-07-12 16:55:03