Q4 of 62 Page 186

A girl is in the beach with her father. She spots a swimmer drowning. She shouts to her father who is 50 m due west of her. Her father is 10 m nearer to a boat than the girl. If her father uses the boat to reach the swimmer, he has to travel a distance 126 m from that boat. At the same time, the girl spots a man riding a water craft who is 98 m away from the boat. The man on the water craft is due east of the swimmer. How far must the man travel to rescue the swimmer? (Hint: see figure).



Here,


In the above figure,


B is the girl.


A is her father.


F is the boat.


D is the Drowning Man.


E is the man driving water craft.


We need to find the distance ED.


Now,


In ΔFAB and ΔFED


∠AFB = ∠EFD (vertically opposite angle)


∠FAB = ∠FED by alternate angles (AB∥DE)


∴ ΔFAB ∼ ΔFED


(∵ ΔFAB ∼ ΔFED)



⇒ x = 45 m


(∵ ΔFAB ∼ ΔFED)



⇒ ED = 140 m (∵ x = 45 m)


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