Correct question :-

A girl of height 90 CM is walking away from the base of a lamp post at a speed of 1.2 metre if the length is 3.6 metre above the ground find the length of her shadow after 4 seconds.

Solution :-

According to the GIVEN information draw a rough figure

[ Refer to attachment ]

Height of the lamp pole BC = 3.6 m

Height of the girl DE = 90 cm = 0.9 m [ Because 1 cm = 1/100 m ]

Speed of the girl = 1.2 m/s

Time taken to walk = 4 seconds

We know that

Speed = Distance / Time

⇒ Distance = Speed * Time

So, Distance walked by the girl from the lamp pole = 1.2 * 4 = 4.8 m = EB

Length of the shadow of the girl = AE

Consider ΔDEA and ΔCAB

DE ⊥ AB, CB ⊥ AB [ HEIGHTS are considered to be perpendicular to the ground ]

⇒ ∠DEA = ∠CBE = 90°

∠A is common

Third angles are also equal

Therefore by AAA similarity ΔDEA ~ ΔCAB

If 2 TRIANGLES are similar corresponding sides are in proportion

⇒ AE/AB = DE/CB

⇒ AE/(AE + EB) = DE/CB

⇒ AE/(AE + 4.8) = 0.9/3.6

⇒ AE/(AE + 4.8) = 9/36

⇒ AE/(AE + 4.8) = 1/4

⇒ 4AE = AE + 4.8

⇒ 4AE - AE = 4.8

⇒ 3AE = 4.8

⇒ AE = 1.6

Hence the length of the shadow of girl is 1.6 m.