A ladder 25m long reaches a window of building 20m above the ground. Then the distance of the foot of the ladder from the building is?
A) 35m b)45m c)30m d)15m

• let it forms a right angled triangle and follow c^2=a^2+b^2
  then 25^2=20^2+b^2
  b^2=625-400
  b=15
• 11 years agoHelpfull: Yes(1) No(0)
• this question is solve using Pythagoras theorem of right angle tringle
  so let y is distance from building
  25^2=20^2+y^2
  y^2=625-400=225
  y=15
  d.15 is right ans
• 11 years agoHelpfull: Yes(1) No(0)
• the distance of window from the ground is 20 meter.
  and the length of ladder is 25 meter.
  
  so the laddermake the right hand triangle from the wall so from the pythagoras theorem
  
  
  (the distance of the foot of ladder from the building)^2 =
  (length of ladder)^2 -(distance of window from ground)^2
  
  
  so (the distance of the foot of ladder from the building)^2 = (25)^2 - (20)^2
  = 625 - 400
  = 225 so the distance of the foot of ladder from the building = 15 meter
  
• 11 years agoHelpfull: Yes(0) No(0)
• ((heights)^2)+((base)^2))=((hypotenuse)^2)
  ((base)^2))=((hypotenuse)^2)-((heights)^2)
  ((base)^2))=(25^2)-(20^2)
  ((base)^2))=625-400=225
  base =sqrt(225)
  base =15m is the distance of the foot of the ladder from the building.
• 11 years agoHelpfull: Yes(0) No(0)
• 25^2=20^2+x^2 625-400=x^2 x=15
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• it is pithagirus theorem format s0 15m is the ans
• 11 years agoHelpfull: Yes(0) No(0)