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, the find the distance by which the top of the ladder would slide upwards on the wall.

Answer 1

Answer:

The distance by which the top of the ladder would slide upwards on the wall is 0.8 meter.

Step-by-step explanation:

Given : 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.

To find : The distance by which the top of the ladder would slide upwards on the wall ?

Solution :

We have sketch a rough image of the situation,

Let the length of the ladder AC=5 m

The height of the wall BC=4 m

If the foot of the ladder is moved 1.6 m towards the wall i.e, AD = 1.6 m

Refer the attached figure below.

Now, Consider ΔABC and ΔEBD be right angle triangle.

Applying Pythagoras theorem in ΔABC,

$AC^2=AB^2+BC^2\\5^2=AB^2+4^2\\25=AB^2+16\\AB=\sqrt{25-16}\\AB=\sqrt{9}\\AB=3$

So, AB= 3 m

We know, DB=AB-AD

DB=3-1.6

DB=1.4 m

Applying Pythagoras theorem in ΔEBD,

$ED^2=EB^2+BD^2\\5^2=EB^2+1.4^2\\25=EB^2+1.96\\EB=\sqrt{25-1.96}\\EB=\sqrt{23.04}\\EB=4.8$

So, EB= 4.8 m

We know, EC=EB-BC

EC=4.8-4

EC=0.8 m

Therefore, The distance by which the top of the ladder would slide upwards on the wall is 0.8 meter.