Question

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.

Answer 1

Let AC be the ladder of length 5 m and BC = 4 m be the height of the wall, which ladder is placed.

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

Then the ladder is slide upward i.e., CE = x m.

In right angled ∆ABC,

AC² = AB² + BC²  ...[By pythagoras theorem]

⇒ (5)² = (AB)² + (4)²

⇒ AB² = 25 – 16 = 9

⇒ AB = 3 m

Now, DB = AB – AD

= 3 – 1.6

= 1.4 m

In right angled ∆EBD,

ED² = EB² + BD²   ...[By pythagoras theorem]

⇒ (5)² = (EB)² + (14)²  ...[BD = 1.4]

⇒ 25 = (EB)² + 1.96

⇒ (EB)² = 25 – 1.96 = 23.04

⇒ EB = `sqrt(23.04)` = 4.8

Now, EC = EB – BC

= 4.8 – 4

= 0.8

Hence, the top of the ladder would slide upwards on the wall at distance is 0.8 m.