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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

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.

Answers (1)

Answer: [0.8 m]

Here,   AC = 5m, BC = 4 m, AD = 1.6 m

Let       CE = x m

In \Delta ABC use Pythagoras theorem

AC^{2}=AB^{2}+BC^{2}

\left ( 5 \right )^{2}=\left ( AB \right )^{2}+\left ( 4 \right )^{2}

25-16=AB^{2}

9=AB^{2}

AB=\sqrt{9}

AB=3\; m

DB=AB-AD

=3-1.6=1.4\; m

Similarly, in triangle EBD use Pythagoras theorem

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

\left ( 5 \right )^{2}=EB^{2}+\left ( 1.4 \right )^{2}

23.04=EB^{2}

EB=\sqrt{23.04}

EB=4.8

EB=EB-BC

4.8-4=0.8\; m

Hence the top of the ladder would slide up words on the wall at distance is 0.8 m.

Posted by

infoexpert23

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