question_answer

A \[\text{5 m}\] Long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point \[\text{4 m}\] height If the foot of the ladder is moved \[\text{1}\text{.6 m}\] towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is:

A) \[\text{0}\text{.6 m}\]

B) \[\text{0}\text{.2 m}\]

C) \[\text{0}\text{.4 m}\]

D) \[\text{0}\text{.8 m}\]

Correct Answer: D

Solution :

[d] Let AC be the ladder of length \[5\,m\]and \[\text{BC}=\text{4 m}\]be the height of the wall, on which ladder is placed.

If the foot of the ladder is moved \[1.6\,m\] towards the wall, i.e., \[AD=1.6\,m,\]then the ladder will slide upward, i.e., \[CE=x\,m.\]

In right angled \[\Delta ABC,\]

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

[By Pythagoras theorem]

\[{{(5)}^{2}}={{(AB)}^{2}}+{{(4)}^{2}}\,\,\,\,\Rightarrow \,\,A{{B}^{2}}=25-16=9\]

\[AB=3\,m\]

\[DB=AB-AD=3-1.6=1.4\,\,m\]

In right angled \[\Delta EBD,\]

\[E{{D}^{2}}=E{{B}^{2}}+B{{D}^{2}}\]   [By Pythagoras theorem]

\[{{(5)}^{2}}={{(EB)}^{2}}+{{(1.4)}^{2}}\] \[[BD=14\,m]\]

\[25={{(EB)}^{2}}+1.96\]

\[{{(EB)}^{2}}=24-196=23.04\]

\[EB=\sqrt{23.04}=4.8\]

Now,     \[EC=EB-BC=4.8-4=0.8\]

Hence, the top of the ladder would slide upwards on the wall by a distance of\[0.\text{8 m}\].