Question

A vertical lamp-post, 6 m high, stands at a distance of 2 m from a wall, 4 m high. A 1.5-m-tall man starts to walk away from the wall on the other side of the wall, in line with the lamp-post.The maximum distance to which the man can walk remaining in the shadow is

• A. $\frac{5}{2}m$
• B. $\frac{3}{2}m$
• C. $4m$
• D. none of these

Answer 1

The correct option is A $\frac{5}{2}m$

From similar $△ABQ$ and $△CDQ$, we get
$\frac{BR}{DR}=\frac{AB}{CD}=\frac{4}{\frac{3}{2}}=\frac{8}{3}$ ...(1)
Now, from similar $△PQR$ and $△ABR$, we get
$\frac{PQ}{AB}=\frac{QR}{BR}\Rightarrow \frac{6}{4}=\frac{BQ+BR}{BR}=\frac{2+BR}{BR}$
$\Rightarrow BR=4$ ...(2)
Substituting (2) in (1), we get
$\frac{4}{DR}=\frac{8}{3}\Rightarrow DR=\frac{3}{2}$ ...(3)
From (2) and (3)
$BD=BR-DR=4-\frac{3}{2}=\frac{5}{2}$