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** **Two walls are at a distance of 5m from each other. A ladder, kept resting on one wall; touch the wall at a point 6m above the ground. When moves towards the second wall, keeping the foot of the ladder fixed, touches the second wall 5m above the ground. Find the distance of the foot of the ladder from the first wall.

Please find below the solution to the asked query:

From given information we form our diagram , As :

Here AB = 5 m ( Distance between two walls )

and

AD = 5 m ( As ladder touches the second wall 5 m above the ground )

BC = 6 m ( As ladder touches the first wall 5 m above the ground )

Let length of ladder = EC = ED =

*x*

And

Distance of the foot of the ladder from the first wall = BE =

*y*, So

Distance of the foot of the ladder from the first wall = AE = 5 -

*y*

Now we apply Pythagoras theorem in AED and get

ED

^{2}= AD

^{2}+ AE

^{2}, Substitute all values and get

*x*

^{2}= 5

^{2}+ ( 5 -

*y*)

^{2},

*x*

^{2}= 25 + 25 +

*y*

^{2}

*-*10

*y*

*x*

^{2}= 50 +

*y*

^{2}

*-*10

*y*----- ( 1 )

Now we apply Pythagoras theorem in BEC and get

EC

^{2}= BC

^{2}+ BE

^{2}, Substitute all values and get

*x*

^{2}= 6

^{2}+

*y*

^{2},

*x*

^{2}= 36 +

*y*

^{2}, Substitute that value in equation 1 and get

36 +

*y*

^{2}= 50 +

*y*

^{2}

*-*10

*y*

10

*y*= 14

*y*= 1.4

So,

**Distance of the foot of the ladder from the first wall = 1.4 m ( Ans )**

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