A girl of height $90\text{ cm}$ is walking away from the base of a lamp-post at a speed of $1.2\text{ m/s}$. If the lamp is $3.6\text{ m}$ above the ground, find the length of her shadow after $4$ seconds.
Step-by-Step Solution
Key Concept: Distance walked $= 1.2 \times 4 = 4.8\text{ m}$. Let shadow $= x$. Similar triangles: $\dfrac{3.6}{0.9} = \dfrac{4.8 + x}{x} \Rightarrow 4 = \dfrac{4.8 + x}{x} \Rightarrow 4x = 4.8 + x \Rightarrow 3x = 4.8 \Rightarrow x = 1.6\text{ m}$.
Distance walked $= 1.2 \times 4 = 4.8\text{ m}$. Height of girl $= 0.9\text{ m}$. [0.5 Mark]
By similar triangles, $\dfrac{3.6}{0.9} = \dfrac{4.8 + x}{x} \Rightarrow 4x = 4.8 + x$. [1.0 Mark]
$3x = 4.8 \Rightarrow x = 1.6\text{ m}$. The length of shadow is $1.6\text{ m}$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Calculating distance $4.8\text{ m}$: 0.5 Mark
Setting up similar triangle ratio $4x = 4.8 + x$: 1.0 Mark
Solving shadow length $x = 1.6\text{ m}$: 0.5 Mark
Correct Answer: