Ladders!!!

[S@NSIS]

Hi,
Can anyone help me solve this problem which my colleague has just given to me!!!

There are two ladders in an alleyway resting against opposite walls. (the feet of either ladder are firmly in the opposite corner). One ladder is 40m long, the other 60m long. The height at which they intersect is 15m.

How wide is the alley????

Any help will be appreciated

S@NSIS

[S@NSIS]

It looks like this...

|\ 40        /| 60
| \         / | 
|  \       /  |
|   \     /   |
|    \   /    |
|     \ /     |
|     /|\     |
|    / | \    |
|   /  |  \   |
|  /   |   \  |
| /    |15  \ |
|/     |     \|
|--------------
  <-- (x?) -->

Well, sort of!!!!!

[[Digital-X-Treme]]

Is there any information on the angles that the ladders make with the floor?

Laterz

[S@NSIS]

No, there isn't! I wish there was though as that would make it easier!!!

S@NSIS

[Bjwbell]

24 feet is the answer.
I used a graphing calculator to graph both lines and find out where they intersected.

[Guv]

BjwBell: I made a scale drawing using an alley 24 meters wide. It does not seem to be even approximately correct.

Everybody: I constructed a fairly accurate scale drawing using graph paper and the width looks like approximately 34 meters. While making the drawing, it looks as though the data given forces a unique solution.

If you draw a line parallel to the ground 15 meters above the ground, you get a bunch of similar triangles. It is possible to set up a lot of simultaneous equations in a lot of unknowns using ratios based on the similar triangles and the theorem of Pythagorus. A brute force solution based on this approach seems possible. I am too lazy to pursue this further.

I suspect that there is some simple idea which I am overlooking.

The information given might describe a special case. My scale drawing seems to indicate that the ladders meet at right angles, but I do not see an obvious proof of this. I would hesitate to make this assumption based merely on my scale drawing.

[Bjwbell]

I was wrong. the equation i used was
y=(15/(60/x)+15/(40/x))
and
y=15


60/x gives the slope of the first line and 40/x is the other lines slope.

15/(60/x) is how far from the first wall the first line is fifteen feet from the ground.

15/(40/x) is how far from the oposite wall the second line is fifteen feet from the ground

the equations i should have used are

y=(15/(60/x)+15/(40/x))
and
y=x

[DavidHooper]

Just a few thoughts:
*think of it on xy axes.
*I have the 60m ladder running bottom right to top left and vice versa for the 40m ladder. doesn;t really matter, but hey.
*there are four unknowns: the width of the alley, the x coordinate where the ladders intercept, the height of the alley (ie where the 60m cuts the y axis) and the y coordinate where the 40m ladder cuts the other side of the alley.

*Call the height of the alley c.
*Call the width of the alley q.
*Call the x coord of the interception p.
*Call the y coord where 40m cuts other side d.

*Now, check this out: you can find four equations from this setup and with four unknowns then (after a fair amount of work) the answer will come out.

*One equation can be found using pythagoras theorem with the 60m ladder - 60m hypotenuse, unknown height, unknown width. The equation is 60^2=q^2+c^2.

*Equation number two is found, again, using pythagoras with the 40m ladder. The equation is 40^2=q^2+d^2

*Equation three is found from the equation of the 40m ladder line. y=(d/q)*x so, specifically, using values, 15=dp/q.

*Equation four is, in the same vein, equation of the 60m ladder. y=-(c/q)*x+c. (c is the intercept). specifically, using values, 15=c-cp/q.

*Voila, 4 equations, 4 unknowns. soluble problem.

*Too much work for me to substitute all these equations into each other! does someone want to use excel solver function on it?

/dh/

[DavidHooper]

The width of the alley (q) is 33.73117848m.
The height of the alley (c) is 49.62063682m.
The ladder interception (p) occurs at x=23.53445974m.
The 40m ladders cuts at y=21.4990139m

[S@NSIS]

Thanks for all your help guys

[Active]

Originally posted by Bjwbell
60/x gives the slope of the first line and 40/x is the other lines slope.

Did you assume the wrong measurements ?
60m is the length of the ladder / and not the height | at which it meets the opposite wall.

[Bjwbell]

Yep.