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A $ 6 \mathrm{~m} $ long ladder leans on a wall to reach the height of $ 3.6 \mathrm{~m} $ on the wall. Find the distance of the lower end of the ladder from the wall.
Given:
A \( 6 \mathrm{~m} \) long ladder leans on a wall to reach the height of \( 3.6 \mathrm{~m} \) on the wall.
To do:
We have to find the distance of the lower end of the ladder from the wall.
Solution:
Let $AB$ be the ladder and $AC$ be the wall.
$BC$ is be the distance between the wall and the foot of the ladder.
Therefore,
$AB=6\ m$
$AC=3.6\ m$
In $\vartriangle ABC$, using Pythagoras theorem,
$AB^2=AC^2+BC^2$
$\Rightarrow ( 6)^2=( 3.6)^2+BC^2$
$\Rightarrow 36=12.96+BC^2$
$\Rightarrow BC^2=36-12.96$
$\Rightarrow BC^2=23.04$
$\Rightarrow BC=\sqrt{23.04}$
$\Rightarrow BC=4.8\ m$
The distance between the wall and the foot of the ladder is $4.8\ m$.
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