AP EAMCET · Maths · Heights and Distances
A man of height \(2 \mathrm{~m}\) walks at a uniform speed of \(7 \mathrm{~m} / \mathrm{min}\) away from a lamp post of height \(9 \mathrm{~m}\). The rate \((\mathrm{m} / \mathrm{min})\) at which the length of his shadow increases is
- A 2
- B \(\frac{5}{2}\)
- C 3
- D \(\frac{7}{2}\)
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
Let \(A B\) be the lamp-post and \(P Q\) the man, \(C P\) be his shadow at time \(t\). Let \(A P={ }^` x, P C=y\), Also \(A B=9 \mathrm{~m}, P Q=2 \mathrm{~m}\) Now, \(\triangle C A B\) and \(\triangle C P Q\) are equiangular and hence similar.…
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