The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?
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Given:

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground.

To do:

We have to find the height its tip reaches when the ladder is shifted in such a way that its foot is 8 m away from the wall.

Solution:

Let the length of the ladder be $AC = DE = x\ m$.

In $∆ABC$,

By Pythagoras theorem

$AC2 = AB^2 + BC^2$

$x^2 = 8^2 + 6^2$

$x^2=64+36=100$

In $∆DBE$,

By Pythagoras theorem,

$DE^2 = DB^2 + BE^2$

$x^2 = DB^2 + 8^2$

$DB^2=x^2-64$

$DB^2 = 100 -64$   (since $x^2=100$)

$DB^2 = 36$

$DB = \sqrt{36}\ m$

$DB = 6\ m$

Therefore, the tip of the ladder reaches to a height of $6\ m$. 

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Updated on: 10-Oct-2022

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