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I have been trying to solve this problem for the last two days and I can't find a way to solve it I tried to look for similar problems but all the problems I find include the length of the ladders Any help would be appreciated.

The ladders intersect in the middle and they are in the middle of each building

1.How long is the alley's width?

2.The height of the buildings?

3.The lengths of the ladders?

h = 1.80m

building 1 4 floors in height

building 2 6 floors in height

I was not given more information and thanks for your time

enter image description here

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    "The ladders intersect in the middle" -- do you mean the intersection point is the midpoint of each ladder? Then you could mark some equal segments> "They are in the middle of each building" -- does that mean that one ladder ends at a height two floors up and the other at a height three floors up? (And we are I suppose to assume that the heights of one floor of a building is the same in each)? Then we can mark one vertical segment as ratio 3:2 of the other, one being 2a and one 3a. Aha! Now we have a lot more information. Work on it and get back to me.. – 2017-02-08
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    No, the ladders cannot meet at their midpoints -- similar triangles. And it can't be the middle of the alley either. This is a very poorly phrased question. – 2017-02-08
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    [This is a super well known problem](https://en.wikipedia.org/wiki/Crossed_ladders_problem) – 2017-02-08
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    If you estimate one floor of a building is approximately 10 feet = 3 meters you can do this. Otherwise we are short of info. – 2017-02-08
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    Thanks for the new information I'll keep trying – 2017-02-08
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    @BrevanEllefsen It's not a duplicate of the quoted problem -- in this question, we are not given the ladder lengths; instead we have to find them. – 2017-02-08

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It's a trick question. You can get the heights of the buildings (assuming each story is the same height in both buildings), but you cannot deduce the width of the alley (and therefore can't deduce the lengths of the ladders).enter image description here Using similar triangles, you get $$\frac yh=\frac{x+y}A\qquad\text{and}\qquad \frac xh=\frac{x+y}B=\frac{x+y}{\frac64 A}.$$ Rearrange the first equation to $$y=\frac{(x+y)h}A$$ and the second to $$\frac{(x+y)h}A=\frac32 x,$$ so that $y=\frac32 x$, from which you can deduce $A=\frac53 h$ and $B=\frac52 h$. But that's as far as you can get. Since we have only height information, there are infinitely many possibilities for $x$ and $y$ (but they all satisfy $y=\frac32 x$).

You can visualize this: Try pulling the two buildings farther apart from each other, keeping the ladder-to-wall contact points unchanged. The point where the ladders cross remains at height $h$.

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    `The ladders intersect in the middle` Where does the above use that given premise (which, honestly, is not obvious to me what must mean)? – 2017-02-08
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    @dxiv I agree the problem is not precisely posed. I would rephrase that premise as "the point where the ladders intersect is 1.80 m above the ground". My solution proceeds on the assumption that $h$ is known. – 2017-02-08