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.  Answers:	A) 1.8	B) 1.9	C) 1.6	D) 1.7	E) 56
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= 
<gadget id="calculator">8/5</gadget>
<output>8/5 = around 1.6</output>
1.6   Therefore, the desired height is 1.6 meters.   ANSWER:1.6
<result>C</result>
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.