A moves 3 kms east from his starting point . He then travels 5 kms north. From that point he moves 8 kms to the east.How far is A from his starting point?

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shobana

  • Jul 19th, 2007
 

A is 11 kms far from his starting point.

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madhucd

  • Oct 12th, 2007
 

ans is 12.43...

diagramatically, u wil get a right angled triangle with sides 5 n 8.. using pythogoras th. 3rd side will b 9.43..

9.43+3= 12.43

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sriharsha.ab

  • Jul 18th, 2008
 

                                          _____8 kms east_________________>
                                         / 
                                          |
                                          |
                               5 kms     north    
                                          |
                                          |
_3kms east___________>


&


_3kms east___________>_____8 kms east_________________>



just to 11 kms far from starting point.

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swarnakarm

  • Jun 1st, 2009
 

THE ANSWER IS 12.08 as 
                   ______________8kms
                  |
                  |5kms
                  |
                  |
---------------
3kms
<---------------------11kms------->

so applying pythegorus theorem we get  90 degree b/w 11kms and 5kms.
so result is 12.083..

tiru5796

  • Jun 8th, 2009
 

                                                         .c___________________________.d
                                                         |
                                                         |
                                                         |
                                                         |
                                                         |
                                                         |
                                                         |
.___________________________.|

a                                                     b

Now draw a straight line from {a to d}. measure that approximatly 12.08
or aplly pythagurus formula.

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pradeepagrahari

  • Nov 7th, 2010
 

       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

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p.anand.3190

  • Nov 8th, 2010
 

12.08
           ---------8-------------
          |
          5      
          | 
  ---3---
           ---------8-------------
          |                            |
          5                           5
          |                            |
----------------11-------------
 Now take  pythogoras thoram and find the value. Take 11 as base and 5 as altitude.Find the hypotenuse

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pranoti

  • Jan 30th, 2011
 

Draw the diagram and join the two end points. We obtain a triangle base=11, height=5. Apply Pythagoras theorem.

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