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A \( 5 \mathrm{~m} \) long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point \( 4 \mathrm{~m} \) high. If the foot of the ladder is moved \( 1.6 \mathrm{~m} \) towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Given:
A $5\ m$ long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point $4\ m$ high. The foot of the ladder is moved $1.6\ m$ towards the wall.
To do:
We have to find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
The length of the ladder$=$Hypotneuse $AB=5\ m$
Base $BC =?$
Height of the wall $=$ altitude $AC=4\ m$
Using Pythagoras theorem, in $\vartriangle ABC$
$\Rightarrow BC^2=5^2-4^2$
$\Rightarrow BC^2=25-16$
$\Rightarrow BC=\sqrt{9}$
$\Rightarrow BC=3\ m$
When base is reduced by $1.6\ m$
Base $EC=BC-BE=3-1.6=1.4\ m$
Hypotneuse $=$ Height of ladder$DE=5\ m$
Altitude$DC=?$
$\Rightarrow DC^2=5^2-1.4^2$
$\Rightarrow DC^2=25-1.96$
$\Rightarrow DC=\sqrt{23.04}$
$\Rightarrow DC=4.8\ m$
$\therefore$ Distance by which the top of the ladder would slide upwards on the wall$=DA=DC-AC=4.8-4=0.8\ m$.