Question

A vertical lamp-post PQ, $6m$ high, stands at a distance of $2m$ from a wall AB, $4m$ high. A $1.5m$ tall man(CD) starts to walk, 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. $5m$

Answer 1

Answer: (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)
$BD=BR-DR=4-\frac{3}{2}=\frac{5}{2}$