Question

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of he ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Answer 1

In fig $\Delta DLW$ & $\Delta ERW$ is a wall DL and RE are two position of ladder of lengths $5m$.

Refer image.

In right angled $\Delta LWD$,

$D{W}^2+L{W}^2=D{L}^2$ (By Pythagoras)

$D{W}^2=D{L}^2-L{W}^2$

$\Rightarrow D{W}^2=5^2-4^2$

$=25-16=9$

$\Rightarrow DW=3$

Now, $RW=DW-DR$

$=3-1.6=1.4m$

In right angled triangle RWE,

$E{W}^2+R{W}^2=R{E}^2$ (BY Pythagoras)

$E{W}^2=R{E}^2-R{W}^2$

$=5^2-{1.4}^2=25-1.96$

$=23.04$

$EW=\sqrt{23.04}=4.8m$

$\therefore$ The distance by which the ladder shifted upward=

$EL=EW-LW=4.8m-4m=0.8m$.