A 5m 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 1.6m to towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
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Case I: In right angled $\triangle\text{LWD},$
$DW^2 = DL^2 - LW^2$
$\Rightarrow DW^2 = 5^2 - 4^2$
$\Rightarrow DW^2 = 25 - 16$
$\Rightarrow DW^2 = 9$
$\Rightarrow DW = 3m$
Case II: $RW = DW - DR$
$\Rightarrow RW = 3 - 1.6$
$\Rightarrow RW = 1.4m$
In right angled triangle RWE,
$EW^2 = RE^2 - RW^2$
$\Rightarrow EW^2 = 5^2 - 1.4^2$
$\Rightarrow EW^2 = 25 - 1.96$
$\Rightarrow EW^2 = 23.04$
$\Rightarrow\text{EW}=\sqrt{23.04}$
⇒ EW = 4.8m
$\therefore$ The distance by which the ladder shifted upward = EL = 4.8m - 4m = 08m
Hence, the ladder would slied upword on wall by 0.8m.
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