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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= 

$$EL=EW-LW=4.8m-4m=0.8m$$.

1790118_1551554_ans_d1b00e63ac024b77b166999f87e26320.png

Mathematics

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