A 25 m long ladder is placed against a vertical wall such that the foot of the ladder is 7 m from the feet of wall. If the top of the ladder slides down by 4 m, by how much distance will the foot of the ladder slide?

Given:

A 25 m long ladder is placed against a vertical wall such that the foot of the ladder is 7 m from the feet of the wall.

The top of the ladder slides down by 4 m.

To do:

We have to find the distance the foot of the ladder slides. 

Solution:

The length of the ladder be $AB = DE = 25\ m$.

In $\triangle ABC$,

By Pythagoras theorem

$AB^2 = AC^2 + BC^2$

$25^2 = 7^2 + x^2$

$x^2=625-49=576$

$x=\sqrt{576}=24$

In $\triangle DEC$,

By Pythagoras theorem,

$DE^2 = DC^2 + CE^2$

$25^2 = DC^2 + (x-4)^2$

$DC^2=625-(24-4)^2$

$DC^2 = 625-400$   (since $20^2=400$)

$DC^2 = 225$

$DC = \sqrt{225}\ m$

$DC = 15\ m$

The distance slid by the foot of the ladder$=(15-7)\ m=8\ m$

Therefore, the foot of the ladder will slide by $8\ m$. 

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Updated on: 10-Oct-2022

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