# 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.

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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$=DA=DC-AC=4.8-4=0.8\ m$.

Updated on 10-Oct-2022 13:28:12