A plane mirror 50 cm long, is hung on a vetical wall of a room, with its lower edge 50 cm above the ground. A man stands 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 pointsof the visible region perpendicular to the mirror ) of the floor visible to him dueto refliection from the mirror is (x)/(26)m . Find the value of x

A plane mirror 50 cm long, is hung on a vetical wall of a room, with i

1.	A plane mirror 50 cm long, is hung on a vetical wall of a room, with its lower edge 50 cm above the ground. A man stands 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 pointsof the visible region perpendicular to the mirror ) of the floor visible to him dueto refliection from the mirror is (x)/(26)m . Find the value of x
Answer» Solution :From figure, `DeltaEMP and DeltaPOS` are similar, So , `(EM)/(PO)=(MP)/(OS) rArr OS=1xx(2)/(0.8) m=2.5 m. Sumilarly, DeltaENQ and DeltaQOR` are similar. So, `(EN)/(QO)=(QN)/(RO) rArrRO=(0.5xx2)/((1.8-0.5)) m. =(10)/(13)m`. Hence, LENGTH of wall visible in mirror `=OS-OR=((5)/(2)-(10)/(13))m =(45)/(26),x=45`

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