A 5 m pole long ladder is placed leaning forward on a vertical wall , such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m toward the wall then find the distance by which the top of the ladder slide upwards on the wall.

Asked by vans75a | 31st Jul, 2015, 08:09: PM

Expert Answer:

 
Initially AD is the ladder of length 5 m which reaches the wall at the point D which is 4 m high from C. 
When we move it 1.6m towards the wall, now it's at point B in the figure. The topp of the ladder now reaches point E on wall. 
We have to find the distance DE. In the figure, DE=x m and BC=y m. AD=BE=5 m (length of the ladder)
In the right angled triangle ACD, we have
A D squared equals A C squared plus C D squared space rightwards double arrow 5 squared equals left parenthesis 1.6 space plus space y right parenthesis squared space plus space 4 squared rightwards double arrow left parenthesis 1.6 space plus space y right parenthesis squared equals 5 squared minus space 4 squared rightwards double arrow left parenthesis 1.6 space plus space y right parenthesis squared equals 9 rightwards double arrow 1.6 space plus space y equals 3 rightwards double arrow y equals 1.4 space m
Similarly, in the right angled triangle BCE, we have
B E squared equals B C squared plus C E squared rightwards double arrow 5 squared equals y squared plus open parentheses 4 plus x close parentheses squared rightwards double arrow 5 squared equals 1.4 squared plus open parentheses 4 plus x close parentheses squared rightwards double arrow open parentheses 4 plus x close parentheses squared equals 5 squared minus 1.4 squared rightwards double arrow open parentheses 4 plus x close parentheses squared equals 3.6 cross times 6.4 space space left parenthesis u sin g space a squared minus b squared equals open parentheses a plus b close parentheses cross times open parentheses a minus b close parentheses right parenthesis rightwards double arrow open parentheses 4 plus x close parentheses equals square root of 3.6 cross times 6.4 end root rightwards double arrow 4 plus x equals 4.8 rightwards double arrow x equals 4.8 minus 4 equals 0.8 space m
Hence, the top of the ladder slides upward by a distance of 0.8 m.

Answered by satyajit samal | 2nd Aug, 2015, 12:14: AM

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