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An old one but a favourite of mine.

A 10 ft ladder is leaning on a wall. The base is touching the floor and the top is touching the wall. Under the ladder there is a box which is 3ft x 3ft x 3ft. and it is pushed hard against the wall on which the ladder is leaning. The ladder is pushed in at the base so that the ladder just touches the corner of the box.

How high does the ladder reach up the wall ?

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I don't get the deal with math problems being riddles, A squred plus B squared equals C squared

9+B^2=100 B=9.539

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I don't get the deal with math problems being riddles, A squred plus B squared equals C squared
9+B^2=100 B=9.539

Not exactly.

(x+3)^2 + (y+3)^2 = 100

The box and the ladder form 3 similar triangles that can be expressed using 3 different variables. This allows you to form 3 equations using the pythagorean theorem, but the math gets sorta hairy and nasty along the way using this approach.

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I don't get the deal with math problems being riddles, A squred plus B squared equals C squared
9+B^2=100 B=9.539

We only know side C (10). We don't know side A or side B. It also seems to me that the box doesn't have to have a full side against the wall, it just said the box is up against the wall and the ladder touches a corner of the box (it doesn't have to be an edge). So, the ladder could be touching the corner of the box 4.24 feet from the wall, couldn't it?

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Not exactly.
(x+3)^2 + (y+3)^2 = 100

The box and the ladder form 3 similar triangles that can be expressed using 3 different variables. This allows you to form 3 equations using the pythagorean theorem, but the math gets sorta hairy and nasty along the way using this approach.

I think you're making it more complicated than it needs to be....

Is the answer 8 feet?

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I just drew this out on paper, because the math involved is way beyond me. On paper with a cheap ruler done quickly it is almost 9'. Something to note, however is that no matter what the solution is there are actually 2 solutions. You can mirror the ladder so that whatever the height of the ladder on the wall, if you make it the distance from the wall on the floor, you will get a separate valid solution. basically you can lay the ladder down so that it rests on both the wall and touches the corner of the box, at a height on the wall somewhere around 4.5'. Again the math is beyond me, im going with a picture to scale.

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I think you're making it more complicated than it needs to be....

Is the answer 8 feet?

Nope

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I just drew this out on paper, because the math involved is way beyond me. On paper with a cheap ruler done quickly it is almost 9'. Something to note, however is that no matter what the solution is there are actually 2 solutions. You can mirror the ladder so that whatever the height of the ladder on the wall, if you make it the distance from the wall on the floor, you will get a separate valid solution. basically you can lay the ladder down so that it rests on both the wall and touches the corner of the box, at a height on the wall somewhere around 4.5'. Again the math is beyond me, im going with a picture to scale.

:) Excellent there are indeed two solutions. However the distance of the ladder up the wall in the second one is the same as the distance of the base of the ladder away from the wall in the first

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It can't be 8'. This would make the bottom side 6', and with a 3'x3' box you would have the lower similar triangle with two sides equal to 3', thus creating a 45º angle between the ladder and the ground. This can not be so with a triangle of two sides with different lengths.

I've managed to create 4 equations with 4 unknowns, it's just a matter of whittling it down...

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If the distance is the same on the legs then the angle is the same for the opposite corners (45 degrees) since there are 180 degrees in a triangle and the sqaure formed a 90 degrees [(180-90)/2] = 45. So the distance up the wall is the equal to the sine of 45 degrees x the hypotenuse (aka the ladder [10ft]). So the distance up the wall is 8.51ft. B))

Edited by wheezykw

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**CORRECTION**

If the sides are equidistant, then a = b. So a^2 + b^2 = c^2. Substitution applied a^2 + a^2 = c^2 ----> 2a^2 = 100ft ----> a^2 = 50ft ----> a = 7.071 ft.

Edited by wheezykw

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**CORRECTION**

If the sides are equidistant, then a = b. So a^2 + b^2 = c^2. Substitution applied a^2 + a^2 = c^2 ----> 2a^2 = 100ft ----> a^2 = 50ft ----> a = 7.071 ft.

In both your statements you have a HUGE "if." IF the sides are equidistant. IF the distance is the same on the legs. They are NOT the same. Thus, no 45 degree angle.

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Only one "if". The distant of the legs being equal defines a equidistant triangle. :P

Edited by wheezykw

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**CORRECTION**

If the sides are equidistant, then a = b. So a^2 + b^2 = c^2. Substitution applied a^2 + a^2 = c^2 ----> 2a^2 = 100ft ----> a^2 = 50ft ----> a = 7.071 ft.

This also can't be the case. In order for there to be a 45º angle, you'd have to have 6' on both sides. 36 + 36 = 72 and hence the ladder would be only 8.48'

post-6456-1213118151_thumbjpg

Pardon the quick sketch.

Edited by Jarod997

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It can't be 8'. This would make the bottom side 6', and with a 3'x3' box you would have the lower similar triangle with two sides equal to 3', thus creating a 45º angle between the ladder and the ground. This can not be so with a triangle of two sides with different lengths.

I've managed to create 4 equations with 4 unknowns, it's just a matter of whittling it down...

That's amazing. I typed every bit of that out earlier (almost word for word) but decided not to post it. I thought for a second that I accidentally clicked add reply earlier.

3/y = (x-3)/x

AND

x^2 + y^2 = 100

Where x and y are either of the legs of the triangle, depending on how you look at it.

Edited by Enlightened

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This also can't be the case. In order for there to be a 45º angle, you'd have to have 6' on both sides. 36 + 36 = 72 and hence the ladder would be only 8.48'

post-6456-1213118151_thumbjpg

Pardon the quick sketch.

a^2 + b^2 = c^2 ------> 72 =/= 100. :P

Maybe a math error?

post-7943-1213119007_thumbjpg

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a^2 + b^2 = c^2 ------> 72 =/= 100. :P

Maybe a math error?

But then your ladder doesn't touch your box. ;)

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But then your ladder doesn't touch your box. ;)

Philosophically...my ladder spoke to my box's spiritual being and reached him on a greater level then physical relation. B))

Edited by wheezykw

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Philosophically...my ladder spoke to my box's spiritual being and reached him on a greater level then physical relation. B))

lol :lol:

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I started by drawing a triangle with a 3x3 square situated in the right angle corner such that the opposite corner of the square rested on the hypotenuse.

|\

|h\ (10-y)

|---\ 10

|3_|\ y

3 .. x

Pretend the above is pretty!

I made three equations with three unknowns (using a² + b² = c²):

h² + 3² = (10-y)²

3² + x² = y²

(h+3)² + (3+x)² = 10²

And I solved. I got (it's highly possible through all the math, I messed up... it was a lot of simplifying):

x = .955461

h = 6.184

y = 3.148

Also, if you think about it, you could slide the ladder down so that it was x above the box and h out from the box and still fit the parameters.

So, the ladder could be 9.184 up the wall or 3.955461 up the wall.

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You've heard of 3-4-5 triangles right?

The 3x3 block cuts the 10' ladder in half (2*5). So you are left with an open side for both triangles... The missing 4'.

So the length up the side of the wall is a total of 3' (the box) plus the 4' so you get a total height of 7'.

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