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## Geometry problem

Consider an alley with two ladders placed crosswise.
Kinda like this sucky diagram below:

$$\begin{array}{cccccccc} |&* & & & & & &| \\ |& &* & & & & &| \\ |& & &* & & &&/| \\ |& & & &* &/& &| \\ |& & &/& &* & &| \\ |&/& & & & &* &| \end{array}$$
( Ugh, I hope you get the idea)

One ladder has a length of 2 m. The other is 3 m long.
They cross each other at a height of 1 m.
What is the width of the alley?

It's not hard to see there's a unique solution to this question.
Good luck
 Recognitions: Gold Member Science Advisor Staff Emeritus $$\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)}$$ This might look pretty, but it's both messy and inelegant...there's got to be a nicer way. PS : That gives me w = about 1.231m
 The solution to these 'crossed ladder' problems, except in trivial cases, is a quartic. They are almost always messy: spoiler - highlight or Ctrl-A to view w^8 - 22w^6 + 163w^4 - 454w^2 + 385 = 0 w is about 1.2312 m. Best found (IMO) by successive approximation, ie. guessing.

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## Geometry problem

 Quote by Gokul43201 $$\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)}$$
You're right. (so is Ceptimus), but how did you figure that?
On the left side I see the height of the 3m ladder plus the height of the 2m ladder and on the right side the product. Why are these equal?
 Recognitions: Homework Help Science Advisor I don't know how they did things, but there are lots of similar triangles to pick from. It seems like that should be enough to work things out.
 could someone explain this one for me? this intrigues me
 Recognitions: Homework Help Science Advisor Me too. C'mon Gokul43201!
 Galileo, this is sort of a hint and i am not sure whether Gokul did it this way... Let the ladders be L1 and L2. Let H_L1 be the height at which ladder L1 meets the wall Let H_L2 be the height at which ladder L2 meets the wall can u show that, 1/H_L1 + 1/H_L2 = 1 -- AI
 Recognitions: Homework Help Science Advisor Here's a hint: highlight to see Use trigonometry (cos, sin, etc.......). It's not a difficult problem as you will see, the solution is quite "simple."

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