Question

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Answer 1

Solution:

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Given:


Condition 1 : A ladder resting against a vertical wall has its foot on the ground at a

distance of 6ft,. from the wall.


Let AB of the following diagram be the vertical wall.


Let AC be the ladder resting on the side of it.

   
   A
    l\
    l  \
E l---\ . D
    l      \
    l        \
    l--------\
   B        C
    l---6ft--l


Condition 2 :  A man climbs two - thirds of the ladder.


Let D be the point on the ladder (two - third) of the wall.

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To find :


What will be his distance from the wall now?


=> ED =?

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According to the given data,


we can say that,.


In ΔABC & Δ AED


=> ED ll BC ,.


=> ∠AED = ∠ABC,. (90° as the floor and vertical wall are perpendicular to each  other,.)


=> ∠A =∠A (same vertices of triangle)


=> AE = AB - $\frac{2}{3} AB$


=> AE = $\frac{1}{3}AB$


=> 3 AE = AB


=> $\frac{AB}{AE} = 3$


=> $\frac{AE}{AB} = \frac{1}{3}$ .....(i)


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According to similarity of triangles,.


We can say that,.


=> $\frac{AE}{AB} = \frac{AD}{AC} = \frac{ED}{BC}$   (the-1)


=> $\frac{AE}{AB} = \frac{ED}{BC}$


=> $\frac{1}{3} = \frac{ED}{6}$


=> $3ED = 6$


=> ED =2 ft,.


∴ His distance from the wall will be 2ft.


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                                      Hope it Helps !!