Question

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 towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is

(a) 0.6m (b) 0.2m (c) 0.4m (d) 0.8m

Answer 1

Explanation:

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

Refer image.

In right angled ΔLWD,

DW

2

+LW

2

=DL

2

(By Pythagoras)

DW

2

=DL

2

−LW

2

⇒DW

2

=5

2

−4

2

=25−16=9

⇒DW=3

Now, RW=DW−DR

=3−1.6=1.4m

In right angled triangle RWE,

EW

2

+RW

2

=RE

2

(BY Pythagoras)

EW

2

=RE

2

−RW

2

=5

2

−1.4

2

=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.

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Answer 2

Answer:

(d) 0.8m

Explanation:

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