. A fly is there 1 foot below the ceiling right across a wall length is 30m at equal distance from both the ends. There is a spider 1 foot above floor right across the long wall equidistant from both the ends. If the width of the room is 12m and 12m, what distance is to be travelled by the spider to catch the fly, if it takes the shortest path?

• Answer is:
  30 - 2 foot
• 11 years agoHelpfull: Yes(8) No(9)
• 10 feets....
  
  this is the room..
  |
  12 | 30
  |------------
  /
  12 /
  /
  fly is 1 feet below the roof and spider is 1 feet bove the floor
  hence. also they are at same line coz given in question tht they are equidistant from both ends...hence 12-2=10
• 11 years agoHelpfull: Yes(7) No(7)
• plz explain
• 11 years agoHelpfull: Yes(2) No(1)
• plz explain
  
• 11 years agoHelpfull: Yes(1) No(2)
• 1+11+12 = 24 (take fly 1 down)
• 11 years agoHelpfull: Yes(1) No(1)
• 31.75 m
  Reason - The room given in this case is a 3-d structure hence we have to consider the length,breadth also along with the height. Its given in the ques that both spider and the fly are 1 foot above the floor and 1 foot below the ceiling respectively but its not given that both are exactly in line, hence they can be considered anywhere in the room and in any 3-d structure diagonal is the only shortest distance between two points. Hence we will construct a right angled triangle whose base will be the breadth of the room i.e 12m and height will be 3000-(30+30)= 2940cm (coz 1 feet=30cm and 30m = 3000cm) which in turn = 29.40m Now to calculate the length of diagonal we can apply Pythagoras Theorem which comes out to be 31.75m (Ans)
• 11 years agoHelpfull: Yes(0) No(1)
• 1 meter = 3 feet.
  15 m
  _______
  | |1 feet
  |_______|_________
  | |
  |_______|
  
  So to it follow the above square to reach the fly quickly.
  Therefore, (2*15m) + (2*1feet) = 30.99 m
  ANS: 30.99m
• 10 years agoHelpfull: Yes(0) No(1)
• as equal distance at all in the room........
  11.39m to take the fly.
• 10 years agoHelpfull: Yes(0) No(0)