Question 1034439
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A 7 ft ladder is leaning against a vertical wall. A point near the bottom of the ladder that is 1 ft from the ground and 1 ft from the wall. Find the exact or approximate distance from the top of the ladder to the ground.
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Unclear.
Read your post.
Try to understand.
The re-formulate and re-post.
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<U>Comment from student</U>: Hello, I know it sounds confusing, but if it may help, here's a link to the picture of the problem being asked: 
http://i66.tinypic.com/103gcq1.jpg Perhaps the image may be a little more helpful! 
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My response:


OK. I found an appropriate Figure from my archive (on the right),  
and I will use it as an illustration, 
although it is not in the scale (sorry).


So, BC represents the ladder, its length is 7 ft.

DE and FE are of 1 ft length each, according to the condition (! not in the scale !). &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

We need to find the length of AC.


Let "x" be the length of FC and "y" be the length of DB.
Consider the triangles EFC and BDE.
They are similar.


From the similarity, you have a proportion {{{abs(CF)/abs(FE)}}} = {{{abs(ED)/abs(DB)}}}, or {{{x/1}}} = {{{1/y}}},

so y = {{{1/x}}}.
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  <TD>

{{{drawing( 200, 250, 0.5, 4.5, 0.5, 5.5, 
            line( 1, 1, 4, 1), 
            line( 1, 1, 1, 5),
            line( 4, 1, 1, 5), 
       blue(line( 1, 2.714, 2.714, 2.714)), 
       blue(line( 2.714, 1, 2.714, 2.714)), 

            locate ( 0.9,  1,    A),
            locate ( 4.0,  1,    B),
            locate ( 0.9,  5.3,  C),
            locate ( 0.78, 2.95, F),
            locate ( 2.8,  2.95, E),
            locate ( 2.65, 1.0,  D)
)}}}

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</TABLE> 
Now you have the legs AC and AB of the length of (x+1) and (1 + 1/x) respectively.


So, you can write the Pythagorean equation


{{{(x+1)^2}}} + {{{(1 + 1/x)^2}}} = {{{7^2}}}.


Yes, it is non-linear equation and it is not quadratic. (It is quartic).


But you can find its approximate solution by solving it graphically.


Good luck!