There is a 6 metres wide alley. Both walls of the alley are perpendicular to the ground. Two ladders, one 10 metres long, the other 12 metres, are propped up from opposite corners to the adjacent wall, forming an X shape. All four feet of each ladder are firmly touching either the corner or the wall. The two ladders are also touching each other at the intersection of the X shape.
What is the distance from the point of intersection from the ground?
Drawing a graph of how I interpret this problem, I get two lines (for 0 < x < 6):
1) y=-(4/3)x + 8
and
2) y=((sqrt108)/6)x
Substituting 2) into 1), then the result back into 2), I get a height of (8*sqrt(108)) / (8 + sqrt(108)) or approx 4.5m
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Posted by jude
on 2003-04-21 09:05:33 |
Interesting problem
