Question

a 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at point 4 m high if the the foot of the foot ladder is moved 1.6 m find out the distance between by which the top of the ladder would slide upwards on the wall​

Answer 1

Answer:

In fig ΔDLW & ΔERW is a wall DL and RE are two position of ladder of lengths 5m.

Refer image.

In right angled ΔLWD,

DW2+LW2=DL2   (By Pythagoras)

DW2=DL2−LW2

⇒DW2=52−42

=25−16=9

⇒DW=3

Now, RW=DW−DR

=3−1.6=1.4m

In right angled triangle RWE,

EW2+RW2=RE2  (BY Pythagoras)

EW2=RE2−RW2 

=52−1.42=25−1.96

=23.04

EW=23.04=4.8m

∴ The distance by which the ladder shifted upward= 

EL=EW−LW=4.8m−4m=0.8m.

Step-by-step explanation:

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