Question

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 6m towards the wall then find the distance by which the top of the ladder would slide upwards on the wall.

Answer 1

Let AC be the ladder of length 5 m and BC = 4m be the height of the wall, which ladder is placed, if the foot of the ladder is moved 1.6m towards the wall i.e., AD = 16m, then the in right angled $\Delta$ ABC
$A{C}^2=A{B}^2+B{C}^2$ [by Pythagoras theorem]
$\Rightarrow$ $(5{)}^2=(AB{)}^2+(4{)}^2$
$\Rightarrow$ $A{B}^2=25-16=9\Rightarrow AB=3m$
$\therefore$ $DB=AB-AD=3-1.6=1.4m$



In right angled $\Delta EBD$, $E{D}^2=E{B}^2+B{D}^2$ [by Pythagoras theorem]
$\Rightarrow$ $(5{)}^2=(EB{)}^2+(1.4{)}^2$ $[\because BD=1.4m]$
$\Rightarrow$ $25=(EB{)}^2+1.96$
$\Rightarrow$ $(EB{)}^2=25-1.96=23.04$
$\Rightarrow$ $EB=\sqrt{23.04}=4.8$
Now, $EC=EB-BC=4.8-4=0.8$
Hence, the top of the ladder would slide upwards on the wall at distance 0.8m.