Question

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.

Answer 1

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 metres.

Now, BD = 1.2 m $\times$ 4 = 4.8 m.

Note that in Δ ABE and Δ CDE,

∠ B = ∠ D (Each is of ${90}^{\circ }$ because lamp-post as well as the girl are standing vertical to the ground)

and ∠ E = ∠ E (Same angle)

So, $△$ ABE ~ $△$ CDE (AA similarity criterion)

Therefore, $\frac{BE}{DE}$ = $\frac{AB}{CD}$

i.e., $\frac{(4.8+x)}{x}$ = $\frac{3.6}{0.9}$

(90 cm = $\frac{90}{100}$m = 0.9 m)

i.e., 4.8 + x = 4x

i.e., 3x = 4.8

i.e., x = 1.6

So, the shadow of the girl after walking for 4 seconds is 1.6 m long.