Question

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

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

Note that in $\Delta ABE$ and $\Delta CDE$,

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

and $\angle E=\angle E$ [Common angle]

So, $\Delta ABE\sim \Delta CDE$ [AA similarity criterion]

$\therefore \frac{BE}{DE}=\frac{AB}{CD}$

$\Rightarrow \frac{(4.8+x)}{x}=\frac{3.6}{0.9}$

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

$\Rightarrow 4.8+x=4x$

$\Rightarrow 3x=4.8$

$\Rightarrow x=1.6$

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