Question

A plane mirror $50\text{cm}$ long hung on a vertical wall of a room with its lower edge $50\text{cm}$ above the ground. A man is standing in front of the mirror at $2\text{m}$ distance. If his eyes are $1.8\text{m}$ above the ground, then the length of the floor visible to him in the mirror is :

• A. $\frac{45}{26}\text{m}$
• B. $90\text{cm}$
• C. $100\text{cm}$
• D. $\frac{35}{13}\text{m}$

Answer 1

The correct option is A $\frac{45}{26}\text{m}$


From the figure we can see that :
$\Delta \text{ABD}\simeq \Delta \text{EFD}$
$\Rightarrow \frac{1.8}{1}=\frac{\text{BD}}{\text{FD}}$
$\Rightarrow \frac{1.8}{1}=\frac{2+\text{FD}}{\text{FD}}$
$\Rightarrow \text{FD}=\frac{5}{2}\text{m}$
Also,
$\Delta \text{ABC}\simeq \Delta \text{GFC}$
$\Rightarrow \frac{1.8}{0.5}=\frac{2+\text{FC}}{\text{FC}}$
$\Rightarrow 18\text{FC}=10+\text{5FC}$
$\Rightarrow \text{FC}=\frac{10}{13}\text{m}$

Length of floor visible :
$\text{CD = FD}-\text{FC}$
$\text{CD}=\frac{5}{2}-\frac{10}{13}$
$\text{CD}=\frac{45}{26}\text{m}$