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Question
A man of height 6 ft walks at a uniform speed of 9 ft/sec from a lamp fixed at 15 ft height. The length of his shadow is increasing at the rate of
Options
15 ft/sec
9 ft/sec
6 ft/sec
none of these
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Solution
6 ft/sec
Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y ft be the length of his shadow CE.
\[\text { Since the triangles ABE and CDE are similar }, \]
\[\frac{AB}{CD} = \frac{AE}{CE}\]
\[\Rightarrow \frac{15}{6} = \frac{x + y}{y}\]
\[ \Rightarrow \frac{x}{y} = \frac{15}{6} - 1\]
\[ \Rightarrow \frac{x}{y} = \frac{3}{2}\]
\[ \Rightarrow y = \frac{2}{3}x\]
\[ \Rightarrow \frac{dy}{dt} = \frac{2}{3}\left( \frac{dx}{dt} \right)\]
\[ \Rightarrow \frac{dy}{dt} = \frac{2}{3} \times 9\]
\[ \Rightarrow \frac{dy}{dt} = 6 \text { ft }/\sec\]
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