intercepts between two triangles...trig help?

Question

This is kind of hard to explain without a picture but here goes...

Two poles measuring 3 feet and 6 feet stand vertically in a field. Wires are attached from the top of each pole to the bottom of the other. A third wire is then run vertically from teh meeting of the first two wires to the ground. What is the length of this wire?

It looks like it would be 2.5 feet. But I really have no idea how to figure this out.

I\
I \
6 I \
I \ I 3ft
I x\ I
I / x \ I
----------------------Not a great pic I know...It's pretty much two triangles....and i am trying to find out the length of the x line that connects the two. Any help would be appreciated...

Answer 1

Using similar triangles:

let z be the height of that vertical wire.
x be the distance from the base of that 3rd wire to the base of 3 ft pole
y is the distance from the base of that 3rd wire to the base of 6 ft pole.

then

3/z = (x+y) / y ... & ... 6/z = (x+y) / x

then 3y/z = x + y .. & .. 6x/z = x + y

3y = 6x means y = 2x

then
3/z = (x+y) / y becomes 3/z = 3x/2x = 3/2

thus z = 2 feet.

d:

Answer 2

Let the distance between the poles be d.
Let the length of the third wire be x.
Let the distance form the base of the 3 ft pole to the point where the third wire touches the ground be y.
The distance form the base of the 6 ft pole to the point where the third wire touches the ground is d-y.
By simliar triangles, write the ratios
x/y=6/d and x/(d-y)=3/d
6y=xd and 3d-3y=xd
So 6y+3y=3d or 3y=d
Since x=6y/d=6y/(3y)=2
Answer 2 ft.

Answer 3

Draw a horizontal line. On this horizontal line erct a perpendicular at point A and extend it to B so that AB = 3.
Now go to point C on the horizontal line and erect a perpendicular at point C and extending it to point D so that CD = 6. DrawAD and Bc and label their point of intersection as E. Now drop a perpendicular from point E to the horizotal line at point F. That's your picture.

Note that AB || EF || BC because they are all perpendicular to the same horizontal line. From this you can easily prove that
6/EF = AD/AE since triangle ACD similar to triangle AFE.

Also triangle CED similar triangle triangle BEA and they have a scale factor of 6/3 = 2/1. Thus DE = 2AE.
Thus AD/AE= 3AE/AE = 3.
6/EF = 3 --> EF = 2