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As shown in the given figure, 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.
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BD = speed $\times$ time = $1.2 \times 4 = 4.8 \text{ m}$ (1 Mark)
$\triangle ABE \sim \triangle CDE$ (1/2 Mark)
$\frac{BE}{DE} = \frac{AB}{CD}$ (1/2 Mark)
$\frac{4.8 + DE}{DE} = \frac{3.6}{0.9}$ (1 Mark)
$\Rightarrow DE = 1.6 \text{ m}$ (1/2 Mark)
$\therefore$ Shadow of the girl after walking for 4 seconds is 1.6 m long
$\triangle ABE \sim \triangle CDE$ (1/2 Mark)
$\frac{BE}{DE} = \frac{AB}{CD}$ (1/2 Mark)
$\frac{4.8 + DE}{DE} = \frac{3.6}{0.9}$ (1 Mark)
$\Rightarrow DE = 1.6 \text{ m}$ (1/2 Mark)
$\therefore$ Shadow of the girl after walking for 4 seconds is 1.6 m long