A 5 miter long ladder is placed learning toward a vertical wall such that it reaches the wall at a point 4 m high. If the ladder is moved 1.6 m towards the wall,then find the distance by which the top of the ladder would slide upwards on the wall
Answers
Answered by
6
The point C is the base of wall
Originally, let the top of ladder reach the wall at the point A.
In triangle ABC, angle C =90
By Pythagoras theorem, we get
BC²+AC²=AB ²
BC = 3
Now, the foot of ladder is moved 1.6 m towards the wall
Let D be new position of foot ladder and E be the new position of its top
By Pythagoras theorem
EC²+ DC²= DE ²
EC = 4.8
4.8m- 4m = 0.8
I hope this will help you
By Devansh
Originally, let the top of ladder reach the wall at the point A.
In triangle ABC, angle C =90
By Pythagoras theorem, we get
BC²+AC²=AB ²
BC = 3
Now, the foot of ladder is moved 1.6 m towards the wall
Let D be new position of foot ladder and E be the new position of its top
By Pythagoras theorem
EC²+ DC²= DE ²
EC = 4.8
4.8m- 4m = 0.8
I hope this will help you
By Devansh
Devanshk:
AB = 5 m AC= 4m
Answered by
3
The answer is 0.8 m........
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