A moves 3km. east from his starting point,he then travels 5km north,from that point he moves 8kms to the east. how far is A from his starting point?

• Hope the ans is 13. A moves 3 km east and then 5 km towards north it doesn't mean from the point he left.(when we give a diagrammatic representation it ll be like 3'0 clk) .It is given in the ques that FROM THAT point so from 5 goes 8km E. so 3+5+(8-3)=13
• 12 years agoHelpfull: Yes(6) No(10)
• TOTAL KMS TRAVELLED TOWARDS EAST =3+8=11KMS
  TOTAL KMS TRAVELLED TOWARDS NORTH =5KMS
  ACCORDING TO HYPOTENUSE RULE,
  (HYPOTENUSE)2=SUM OF THE SQUARES OF TWO SIDES
  SQRT(121+25)=12.12KMS
  
• 8 years agoHelpfull: Yes(1) No(0)
• 11 kms.......
• 6 years agoHelpfull: Yes(1) No(0)
• (sqrt(3^2+(2.5)^2))+sqrt(8^2+(2.5)^2))
• 10 years agoHelpfull: Yes(0) No(0)
• TOTAL KMS TRAVELLED TOWARDS EAST =3+8=11KMS
  TOTAL KMS TRAVELLED TOWARDS NORTH =5KMS
  ACCORDING TO HYPOTENUSE RULE,
  (HYPOTENUSE)^2=SUM OF THE SQUARES OF TWO SIDES
  SQRT(121+25)=12.12KMS
• 8 years agoHelpfull: Yes(0) No(0)
• displacement from original position is 13. ((11)^2+(5^2)sq root)
• 8 years agoHelpfull: Yes(0) No(0)
• N
  W E
  S
  3 kms east,5kms north,8kms east
  
  
  fig. B;----------------- A
  | 8 :
  5 | : 5
  | :
  C----------;.................... E
  3 D 8
  now in fig we can see we can calculae the distance from A to C by applying pithagoras
  theorem.
  coz AB = 8 km
  so DE =8 km
  and BD = 5 km
  so AE = 5 km
  now we can calculate CE by adding CD+DE
  now AC^2 = AE^2+CE^2 CE=CD+DE
  AC^2=5^2+11^2
  AC^2=25+121
  AC^=146
  AC~12.08
• 5 years agoHelpfull: Yes(0) No(0)
• A is 13 km away from his starting point
  diagramatically:
  8
  ------------
  | |
  5 | |5
  ------ ----------
  3 8
  
  12
  
  therefore,
  by pythogrous theorem
  144+25=169
  which is sqrt(169)=13
• 5 years agoHelpfull: Yes(0) No(0)