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

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.

Doubt by Ananya

Solution : 

Let OB'=x & AA'=y

In Rt. ∆ AOB

(AB)²=(OA)²+(OB)² [By Pythagoras Theorem]
(5)²=(4)²+(OB)²
25 = 16+(OB)²
25-16=(OB)²
9 = (OB)²
OB=√9
OB=3m
OB'+B'B=3
x+1.6=3
x=3-1.6
x=1.4m

Now, In Rt. ∆AOB'
(A'B')²=(OA')²+(OB')²
(5)² = (OA')²+x²
25 = (OA')²+ (1.4)² [∵ x=1.4 m]
25 = (OA')²+ 1.96
25-1.96 = (OA')²
23.04 = (OA')²
OA'= √(23.04)
OA'= 4.8
OA + AA' = 4.8
4 + y = 4.8
y = 4.8-4
y = 0.8 m

Therefore, the top of the ladder would slide up by 0.8 m on the wall.