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 :

\frac{45}{26} m

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\frac{35}{13} m

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100 cm

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90 cm

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Solution

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

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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Q. A plane mirror 50 cm long, is hung on a vertical wall of a room,with its lower edge 50 cm above the ground.A man stands infront of the mirror at a distance 2 m away from the mirror. If his eyes are at a height 1.8 m above the ground,then the length (distance between the extreme points of the visible region perpendicular to the mirror) of the floor visible to him due to reflection from the mirror is

\frac{x}{26}

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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 :

The correct option is A 4526 m From the figure we can see that : Δ ABD≃ΔEFD ⇒ 1.81=BDFD ⇒ 1.81=2+FDFD ⇒FD=52 m Also, Δ ABC≃ΔGFC ⇒ 1.80.5=2+FCFC ⇒ 18FC=10+5FC ⇒ FC=1013 m Length of floor visible : CD = FD − FC CD=52−1013 CD=4526 m

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 :