Two vertical poles 2 meter and 8 meter high stand apart on a horizontal plane. find the height in meters of the point of intersection of the lines joining the top of each pole to the bottom of the other pole.
a.1.8 b.Cannot be determined without knowing the distance between bottom of the pole c.1.6 d.5.6

• Let A be the point of the top of the 2m pole.
  Let B be the point of the foot of the 2m pole.
  Let C be the point of the top of the 8m pole.
  Let D be the point of the foot of the 8m pole.
  Let E be the point of intersection of the line segments AD and BC.
  From E drop a perpendicular to the ground.
  Let F be the point on the ground where perpendicular from E meets the ground.
  Let X be the distance from E to F.
  Let P be the distance from B to F.
  Let Q be the distance from F to D.
  Triangles BEF and BCD are similar so we get Equation 1:
  X/P = 8/(P+Q)
  Triangles DEF and DAB are similar so we get:
  X/Q = 2/(P+Q)
  Multiplying both sides by 4 we get Equation 2:
  (4X)/Q = 8/(P+Q)
  Combining Equations 1 and 2 we get:
  X/P = (4X)/Q
  Simplifying this we get:
  1/P = 4/Q
  And then we get:
  Q = 4P
  Substituting Q=4P into Equation 1 we get:
  X/P = 8/(P+4P)
  Simplifying this we get:
  X/P = 8/(5P)
  X = (8P)/(5P)
  X = 8/5
  X = 1.6
  Therefore, the desired height is 1.6 meters.
• 8 years agoHelpfull: Yes(58) No(0)
• c) 1.6
  using trigonometry
  
• 8 years agoHelpfull: Yes(4) No(3)
• Simply use the formula = m*n/m+n
  [where m and n are height of the poles]
• 7 years agoHelpfull: Yes(2) No(1)
• 1.6 m
  Draw figure and apply property of similar triangle
• 8 years agoHelpfull: Yes(0) No(1)