Similar Triangles
Asked by Rishabh1 | 6th Jul, 2009, 07:53: PM
Expert Answer:
BD is the string, Dis the fly which is 3.6m away from point A. BC is the vertical dist of the tip of the rod to the water.
In right triangle BCD
(BC)2 + (CD)2 =(BD)2
(1.8)2 + (2.4)2 =(BD)2
9.00 =(BD)2
Therefore the length of the string BD is 3m which is nothing but 300cm.
Now if the string is pulled @ of 5 cm /sec then ,it is pulled 60 cm in 12 sec.which means length BD (string)now becomes 240 cm(300-60) i.e 2.4m.now we must find out dist AD, of which dist AC remains same.therefore to find CD now
In triangle BCD
(BC)2 + (CD)2 =(BD)2
(1.8)2 + (CD)2 =(2.4)2
(2.4)2 - (1.8)2 =(CD)2
5.76 -3.24 = (CD)2
2.52 = (CD)2
CD =1.587 or 1.6 m
therefore after 12 sec the fly is 2.8 m (1.6 +1.2) m away from Nazima
Answered by | 10th Jul, 2009, 02:04: PM
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