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.

#### Solution

“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.

In the right ∆ABC,

BC^{2} = AC^{2} – AB^{2}

= 5^{2} – 4^{2}

= 25 – 16

= 9

BC = `sqrt(9)` = 3 m

When the foot of the ladder moved 1.6 m toward the wall.

The distance between the foot of the ladder to the ground is

BE = 3 – 1.6 m = 1.4 m

Let the distance moved upward on the wall be “h” m

The ladder touch the wall at (4 + h) M

In the right ∆BED,

ED^{2} = AB^{2} + BE^{2}

5^{2} = (4 + h)^{2} + (1.4)^{2}

25 – 1.96 = (4 + h)^{2 }

∴ 4 + h = `sqrt(23.04)`

4 + h = 4.8 m

h = 4.8 – 4

= 0.8 m

Distance moved upward on the wall = 0.8 m