A child was looking for his father. He went 90 metres in the East before turning to his
right. He went 20 metres before turning to his right again to look for his father at his
uncle’s place 30 metres from this point. His father was not there. From here he went 100
metres to the North before meeting his father in a street. How far did the son meet his
father from the starting point?

• first he went 90 meters to east,
  then he went 20 meters to south,
  next he went 30 meters to west,
  finally he went 100 meters to north then he found his father,
  distance b/w the starting point and ending point is =sqrt((100-20)^2+(90-30)^2)
  =100 meters
• 10 years agoHelpfull: Yes(17) No(0)
• sqrt{(90-30)^2+(100-20)^2}=100m
• 10 years agoHelpfull: Yes(10) No(1)
• sqrt(80^2+60^2)=100
• 10 years agoHelpfull: Yes(3) No(1)
• Ans :100m
  The required distance or displacement is the hypotenuse of a right triangle
  having its base and height 60m and 80m respectively.
  Therefore by pythagoras theorem,the required length = length of hypotenuse
  = squareroot(60^2 + 80^2) = 100
• 10 years agoHelpfull: Yes(2) No(0)
• ans= 100m from starting pt. Draw acc. to qn. n u will find a right angle triangle,whose base is 60 and height is 80 so hyp.=100.
• 10 years agoHelpfull: Yes(1) No(1)
• ans 100m
  make a pictorial representation of the question and you will that the distance that is to be is simply the hypotenuse of the small rt angle triangle formed
  so,
  distance^2=60^2+80^2
  distance=100
• 10 years agoHelpfull: Yes(1) No(1)
• 100 m
  sqrt(80^2 +60^2)
  
• 10 years agoHelpfull: Yes(1) No(0)
• Ans : 240m
  
• 10 years agoHelpfull: Yes(0) No(5)