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Moving Object, Stationary Plane Mirror

A plane mirro...

Question

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

100 cm

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\frac{35}{13} m

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90 cm

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\frac{45}{26} m

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Solution

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

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

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

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