SOLUTION: I have been working on this problem and can't seem to figure it out. I was wondering if someone could help me? Please and Thank you!! I would deeply appreciate it!
Two vertical
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Question 272872: I have been working on this problem and can't seem to figure it out. I was wondering if someone could help me? Please and Thank you!! I would deeply appreciate it!
Two vertical poles have heights 6 ft and 12 ft. A rope is stretched from the top of each pole to the bottom of the other. How far above the ground do the ropes cross? (hint: The lengths y and z do not affect the answer.)
Found 2 solutions by richwmiller, Alan3354:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
One way to solve is to plot it on a coordinate plane.
one pole is 12 and the other is 6
let's put the 12 ft at (0,0) and the 6 ft pole 6 away at 6,0.
the top of the 12ft pole will be at 0,12
the top of the 6 ft pole will be at 6,6
the lines will be from 0,12 to 6,0
and from 0,0 to 6,0
so find the equations of the lines going through 0,12 and 6,0 and from 0,0 to 6,0
and find where they cross
y=mx+b and (y-y)/(x-x)=m
12-0/0-6=12/-2=-2=m
12=-2*0+b
12=b
y=-2x+12
2nd line
6-0/6-0=m
0=1*0+b
0=b
y=1*x+0
y=x
so now we solve the system of equations
y=-2x+12 and y=x
-2x+12=x
12=3x
4=x
4=y
They cross at 4,4 so they cross 4 ft above the ground.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Two vertical poles have heights 6 ft and 12 ft. A rope is stretched from the top of each pole to the bottom of the other. How far above the ground do the ropes cross?
--------------------
Strangely enough, this is similar to parallel resistors, and parallel flow problems.
They cross at the product over the sum
= 6*12/(6+12) = 72/18
= 4 feet
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