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 trick here is to recognize that the two ladders create multiple of 3X4X5 right triangles with the street and walls. The street is now 12 ft wide. The 20' ladders then rests at 16' on the wall.
The street is now 12 ft wide. The 20' ladders then rests at 16' on the wall.
It it shares the same base angles, the intersecting angle must be 90 degress, making the intersecting triangle of the same form 3X4X5, in real measurements 7.2 X 9.6 X 12.
Therefore, the spider must travel 7.2+9.6 = 16.8 feet in 168 seconds or 2 minutes 48 seconds.
Edited on September 18, 2023, 2:12 am