Question

A ladder 5 m long leans against a vertical wall. The bottom of a ladder is 3m
from the wall. If the bottom of a ladder is pull 1m farther from the wall, how much
does the top of the ladder slide down the wall ?​

Answer 1

Step-by-step explanation:

Answer

Let ym be the height of the wall at which the ladder touches. Also, let the foot of the ladder be xm away from the wall.

Then, by Pythagoras theorem, we have:

x

2

+y

2

=25 [Length of the ladder =5m]

⇒y=

25−x

2

Then, the rate of change of height (y) with respect to time (t) is given by,

dt

dy

=

25−x

2

−x

⋅

dt

dx

It is given that

dt

dx

=2cm/s

∴

dt

dy

=

25−x

2

−2x

Now, when x=4m, we have:

dt

dy

=

25−4

2

−2×4

=−

3

8

Hence, the height of the ladder on the wall is decreasing at the rate of

3

8

cm/s.

Answer 2

Step-by-step explanation:

The wall, ladder and floor will make a right angle triangle whose hypotenuse would be the ladder.

Applying Pythagoras in the triangle we will get the third side that is the height from the ground on the wall where the ladder top touch which would be 4m (1)

Now as the ladder is taken 1m more far from the wall so now it is at a distance of 4m from wall but Ladder is now too 5m long so again apply Pythagoras theorem,

This time the third side will come = √(5^2 - 4^2) = √(25-16) = √9 = 3m (2)

So the amount of ladder slide down is (1) - (2)

= 4-3 = 1m

Hence 1m is the answer

First make the figure to understand well