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The ladder and box problem - a classic challenge!

Aug 23, 2022

The ladder and box problem - a classic challenge!

watch your decisions i am preshtalwalker a

ladder

and a cube are in a room place the cube along the wall and then place the

ladder

against the wall so that it just touches the side of the cube to think precisely of the

ladder

as a line Segment If the

ladder

has length 4 and the cube has side length 1, what is the distance from the top of the cube to the top of the

ladder

given by the variable y in the following diagram? Thanks Nathaniel M Mark l and per t for suggesting the solution find an exact answer for the general case that the

ladder

has length l and the cube has side length c pause the video if you want to try this task and when you are ready, keep watching to learn how to solve this

problem

.
the ladder and box problem   a classic challenge
How can we solve this

problem

? We have a

ladder

with length 4, a cube with sides 1, and we want to solve for the distance y. It seems simple, but it's quite complicated, the solution I'm going to present comes from user blue on math stack exchange I thought it would be a very elegant solution. We start with the general case. Suppose the

ladder

has length l and the cube has side length c, which we want to solve for y in terms of l and c. To do that, we label some more parts of this diagram, the other side of the cube becomes one too have length of c and we are going to denote the horizontal distance between the

ladder

and the cube as x now we are going to do a nice little trick we are going to copy this diagram three times and construct a square that will create a clever geometric solution we are going to make this diagram now set aside let's join the four corners of these internal squares to create another square suppose this square has side lengths equal to w now let's label the area of ​​this triangle where one side length equals x as uppercase x.
the ladder and box problem   a classic challenge
We'll label the other triangle with side length y as uppercase y. We label these parts in other parts of this diagram now suppose this triangle ha s an area equal to z by symmetry these also have areas equal to z and just so we don't get lost the square made up of the four conductors we now turn blue fill in, let's calculate the area of ​​this green square well one. To calculate its area, its side length is equal to w plus c, so its area is w plus c squared. Another way is we sum the areas of the rectangles inside this square, here we have a square with an area equal to c squared, then we have two rectangles like this: This is a rectangle and this is another rectangle.
the ladder and box problem   a classic challenge
We have two x terms and two y terms, so we have double the set x plus y, and we have two z terms, so we get plus two z. We also have this square that has side lengths equal to w, so is its area w squared. We can simplify this formula a bit. Consider this rectangle area is also equal to x plus y therefore x plus y equals z we substitute tha t in this term x plus y and it becomes two times z. We then group the z terms so that we have four times z.
the ladder and box problem   a classic challenge
Simplifying this formula again, we have four times z plus w squared, which is exactly equal to the area of ​​this square, which has sides of l, so 4z plus w squared equals l squared, and we substitute that , by now having a very simple formula, we take the square root from both sides and we only take the positive square root because we consider positive lengths, then we subtract c from both sides and get an equation for w in terms of the known values from c and l. We now make an observation about w w is exactly equal to the side length of this square which is equal to the set x plus y, so if w equals x plus y and we want to solve for y to do that, we will consider these triangles, because the

ladder

has a constant slope these two right triangles will be similar hence the ratio between the horizontal leg and the vertical leg will be the same in both triangles so we have x over c is equal to c over y which means that x is equal to c squared over y.
the ladder and box problem   a classic challenge
We put that into the equation where w equals x plus y both sides of which we now multiply this equation by y and then simplify we now have a quadratic equation in y we can use brahmagupta quadratic formula to get the following to get two possible values ​​now we have a plus and a minus notice that the plus is a larger term and this corresponds to y which is the vertical longer distance, the negative term actually corresponds to x which is a horizontal distance so become we separate these two terms, so we figured out a general formula for y, how does that work in our original

problem

, c was equal to 1 and l was equal to 4.
the ladder and box problem   a classic challenge
So we plug that into the formula for w and we get that w equals minus 1 plus the square root of 17. We then plug that into our formula for y and we get the following equation. We simplify this equation a bit and we get the exact answer for y, which is about 2.76, and that's the answer to this

problem

. Thanks for making your choices one of the best channels on YouTube as always
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