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\[0.\text{8 m}\]. |
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