A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the 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 teh wall.

Explanation

Let AB be a ladder of length 5 m and BE = 4 m. when the foot of ladder A is moved 1.6 m towards the wall , then ladder would be in the position CD.
Here AC = 1.6 m
In `triangleABE`
`triangle AE^(2)+ BE^(2)= AB^(2)`
`Rightarrow " " AE^(2)+4^(2)= 5^(2)`
`Rightarrow " " AE^(2)= 25 - 16 =9`
AE = 3m
CE = AE -AC = 3- 1.6 = 1.4 m
Now CD = AB = 5 m
` In triangleCDE`
`CE^(2)+ED^(2)=CD^(2)`
` Rightarrow " " ED^(2) = CD^(2)-CE^(2)= 5^(2)- (1.4)^(2)= 2.5 - 1.96 = 23.04`
ED = 4.8cm
BD = ED- BE = 4.8 - 4 = 0.8 m