Question

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

Answer 1

Finding the distance by ladder would slide upwards on the wall:

Let the length of the ladder is $\text{AC=5 m}$

Let the height of the wall on which ladder is placed $\text{BC= 4 m}$

Let $\text{CE=x, BD=y}$

To find the distance we use Pythagoras Theorem

Pythagoras Theorem : ${\text{H}}^2{\text{=B}}^2{\text{+P}}^2$

In right- angled $△\text{ABC}$

${\text{AC}}^2{\text{=BC}}^2{\text{+AB}}^2$ (here $\text{AB=AD+DB}$ )

$$\begin{gathered} 5^2=4^2+(1.6+{\text{y)}}^2 \\ {\text{25=16+(1.6+y)}}^2 \\ {\text{25-16=(1.6+y)}}^2 \\ {\text{9=(1.6+y)}}^2 \\ (\text{1.6+y)=}\sqrt{9} \\ \text{1.6+y=3} \\ \text{y=3-1.6} \\ \text{y=1.4 m} \end{gathered}$$

So $\text{BD=y=1.4 m}$

Now in right-angled $△\text{DBE}$

${\text{DE}}^2{\text{=BE}}^2{\text{+BD}}^2$ (here $\text{BE=BC+CE}$)

$$\begin{gathered} 5^2=(4+{\text{x)}}^2{\text{+(1.4)}}^2 \\ {\text{25=(4+x)}}^2\text{+1.96} \\ 25-1.96=(4+{\text{x)}}^2 \\ {\text{23.04=(4+x)}}^2 \\ \text{(4+x)=}\sqrt{23.04} \\ 4+\text{x=4.8} \\ \text{x=4.8-4=0.8 m} \end{gathered}$$

Hence, top of the ladder would slide upwards on the wall by a distance of $0.8\text{m}$.