$A$ girl of height $90\, cm$ is walking away from the base of a lamp-post at a speed of $1.2\, m/s$. If the lamp is $3.6\, m$ above the ground,find the length of her shadow after $4\, seconds$.

(N/A) Let $AB$ denote the lamp-post and $CD$ the girl after walking for $4\, seconds$ away from the lamp-post.
From the figure,you can see that $DE$ is the shadow of the girl. Let $DE$ be $x$ meters.
Now,$BD = 1.2\, m/s \times 4\, s = 4.8\, m$.
Note that in $\Delta ABE$ and $\Delta CDE$:
$\angle B = \angle D = 90^{\circ}$ (Each is $90^{\circ}$ because the lamp-post and the girl are standing vertical to the ground).
$\angle E = \angle E$ (Common angle).
So,$\Delta ABE \sim \Delta CDE$ ($AA$ similarity criterion).
Therefore,$\frac{BE}{DE} = \frac{AB}{CD}$.
Given $AB = 3.6\, m$ and $CD = 90\, cm = 0.9\, m$.
$\frac{4.8 + x}{x} = \frac{3.6}{0.9}$.
$\frac{4.8 + x}{x} = 4$.
$4.8 + x = 4x$.
$3x = 4.8$.
$x = 1.6\, m$.
So,the length of the shadow of the girl after walking for $4\, seconds$ is $1.6\, m$.