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A not quite as tough as jothaxe's problem problem

Q

Quaggoth

Senior member
You have a 6' ladder and a 10' ladder. Each are leaning across an alley. There is a plumbline dropped from where they intersect, and that line is 3' long. How wide is the alley?


OOPS! It is supposed to be a 10' ladder. sorry. I originally posted the problem with an 8' ladder. My bad.
 
Hmm, not sure of a definition for a plumbline, but basically, there is a line dropped from the intersection straight to the ground and it is 3' long. Basically it's just saying that the intersection of the 2 ladders is 3' from the ground.
 
| |
|' |
| ' |
| ' '|
| '' |
| '| ' |
|' | '|


is this how it is supposed to look?

the two vertical lines in the middle being the plumb? And the other long vertical lines being the wall alleys. The short vertical lines are a bad interpretation of the ladders-one going one way and the other going the other.


forget it...didn't look like i wanted it to look

let me know

sig.
 
if you think about it logically, a 12.6 foot alley would not be able to accomodate a 6 foot ladder leaning across it.

think about it. The ladder wouldn't even make it halfway.

sig.
 
Please state HOW you solved the problem too please. A guy walked up to me when I was working at Best Buy (A LONG time ago) and gave me the problem. He told me if I could answer him in less than 30 seconds he could probably get me into mensa (Something I have no interest in). I gave him my answer in about 5 seconds and he just walked away. I still don't know if my answer was right, which is part of the reason I chose to post it here.


sigity..... That just looks wierd. Just draw 2 walls and a floor, then put an X in it. the plumbline drops from the center of the X, and each bar in the X is a ladder. your choice as to which one is which length.
 
Look at this diagram to see how it would work.
Simply put the dimensions in and use Pythagoras to solve for the values along the bottom

BTW, this is not drawn to scale 🙂

OK, now if we are using a 10 foot ladder the answer is 14.7 feet.
 
The way I understood it both ladders touch the opposing side, & the plumbline hangs from the X they create in the middle. That makes it significantly more difficult.

A right triangle with two known sides is just too easy.

Viper GTS
 
Not quite haircut. Viper is correct. The line drops from where they intersect. I wish I knew how to attach a link so I could draw a picture and show you. it is a right triangle, but only one side is known.
 
There is SO much missing information here that it is unsolvable.

Are the ladders leaning against the wall?

At what angle?

How high is the wall?

Do the ladders extend over the top of the wall?
 
Oh you are just making it too dificult mwilding. you don't need any more than you have. (Actually, you have a LOT more than you need 🙂) Ok, just because you asked though 🙂🙂, the walls are 682' tall.
 
The height of the walls is irrelevant, the only thing that matters is where the ladders contact the walls.

Knowing the angles would be nice, but again that'd make it much too easy.

Viper GTS
 
OK, you have two right triangles in the problem:
Triangle A has a hypontenuse of 6 feet and one side of 3 feet.
Triangle B has a hypotenuse of 10 feet and one side of 3 feet.

Use the pythagorean theorem to figure out the remaining sides, which should butt up against each other. Add the two sides together to get the total width.

I came up with 14.54 feet, rounding off.
 
palad

You're reading the problem wrong. The ladders don't meet at a point, they meet at an X. The drawing looks something like |x|. All you know is the lengths of the two strokes of the X.

Viper GTS
 
The answer is 2 feet wide but I can't prove it.

The 10' ladder touches the wall 9.8' high.

The 6' ladder touches the wall 5.65' high.

 
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