Question

# A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of he 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.

Solution

## In fig $$\Delta DLW$$ & $$\Delta ERW$$ is a wall DL and RE are two position of ladder of lengths $$5m$$.Refer image.In right angled $$\Delta LWD$$,$$DW^2+LW^2=DL^2$$   (By Pythagoras)$$DW^2=DL^2-LW^2$$$$\Rightarrow DW^2=5^2-4^2$$$$=25-16=9$$$$\Rightarrow DW=3$$Now, $$RW=DW-DR$$$$=3-1.6=1.4m$$In right angled triangle RWE,$$EW^2+RW^2=RE^2$$  (BY Pythagoras)$$EW^2=RE^2-RW^2$$ $$=5^2-1.4^2=25-1.96$$$$=23.04$$$$EW=\sqrt{23.04}=4.8m$$$$\therefore$$ The distance by which the ladder shifted upward= $$EL=EW-LW=4.8m-4m=0.8m$$.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More