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**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 the 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:

Let the length of the ladder = AC = 5 m

Let the height of the wall on which ladder is placed = BC = 4m.

From right angled ∆EBD,

Using the Pythagoras Theorem,

ED^{2} = EB^{2} + BD^{2}

(5)^{2} = (EB)^{2} + (14)^{2} { BD = 1.4}

25 = (EB)^{2} + 1.96

(EB)^{2} = 25 –1.96 = 23.04

EB = √23.04 = 4.8

Now, we have,

EC = EB – BC = 4.8 – 4 = 0.8

Hence, the top of the ladder would slide upwards on the wall by a distance of 0.8 m.

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