Two poles of height 6 meters and 11 meters stand on a plane ground. If the distance between their feet is 12 meters then find the difference in the distance between their tops:

a. 12m
b. 5m
c. 13m
d. 11m

• 13 mtrs
  distance between their tops = sqrt( 12^2+(11-6)^2) = sqrt 169=13m
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• answer is 5m
• 10 years agoHelpfull: Yes(2) No(9)
• by rhs rule distance= (12^2+5^2)^2
  =13
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• 13 mtr
  pythagrs
  height 5 distantce 12 so diagonal is 13
  
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• ans is c 13m
  by using phytogorous theroem
  we know two sides of the triangle
  distance btn two poles=12m
  difference btn the heights of the poles=5m
  let the difference btn their tops be xm
  x^2=12^2+5^2
  =>144+25=169
  x=13m
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• distance between tops of poles=(12)^2+(11-6)^2=169=sqr.root of 169=13 meters
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• ans:c
  ans=sqrt(12^2 + 5^2)= 13m
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• 13 meter
  distance between their tops = sqrt( 12^2+(11-6)^2) = sqrt 169=13meter
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• c,h=sqrt(12^2+5^2)=sqrt(169)=13
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• explain the formula
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• Definitely it will be 13.
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• drawing the figure a/ q and labeling it one can get the answer as 13m
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• Hypotenuse
  Root of(12 square+5square)=13
  Option c
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• 13
  it will form a trainagle
  base 12
  n perpendicular will be 5 = 11-6
  so difference that is hypotenuse will be
  =square root of(144+25)
  =13
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• |
  12m | 5m(11-6)
  |---------------|
  6m | | 11m
  |____________|
  12 m
  hyp=sqrt(12^2 + 5^2)
  =sqrt(144+25)
  = sqrt(169)
  =13 meters
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