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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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