Square on top of another

Relevant wiki: Basic Composite Figures

When the square is rotated about the vertex, the following red and blue triangles are congruent, and so the red triangle can move to the blue triangle's position, which means the red area is equal to the blue area.

Relevant wiki: Basic Composite Figures

No matter how much you rotate the square, it will always be 1/4 of the square that is overlapped.

For me it was easier to see that the area of the square is made of 4 little blue squares but also 4 red shapes

area square = 4 x blue square

area square = 4 x red shapes

=> the area of the blue square and the red shape are equal

It is true for any regular polygon - the area of intersaction is constant for any angle of rotation.

When the vertex is fixed the area will be equal

A simple way to think about it is to rotate the second square a little further to 45 degrees to make a red triangle. You can clearly see that it still makes up a quarter of the square's area (squarea?).

Taking the side of the square = 2

x + y = 2
x = 2 - y

a = y + (1 - y)/2 + (x -1)/2

a = y + 1/2 -y/2 + x/2 - 1/2

a = y - y/2 + (2 - y)/2
a = y - y/2 + 1 - y/2
a = 1

I see it as three points. One is fixed and the others traverse on a defined locus (the perimeter of fixed square). The area swept will always be the same.

Put 4 identical squares pegged to the center of the square instead of only one. Then rotate them all the same quantity about the vertex. The overlapping are of each of them with the center square is obviously the same and their sums a the area of the wole square so the will remind the same 1/4 no matter how much you rotate them. It is another way to look at the solution.

No matter what degree will you rotate it, it will always be the same in AREA if any of the figure will rotate about a fix point.

Since squares are symmetrical, if you added a remaining three squares pinned to the centre of the centre square, no matter which way you rotated them, they could only ever have a equal portion each, which therefore must be one quarter.

Consider four overlapping squares, each with a corner at the vertex. They all overlap the entire area of the central square, with equal amounts at the corners as shown in the first picture, and also when they create four equilateral triangles when rotated through forty five degrees. To overlap a different amount at some intermediate angle of rotation would require the overlap of one square to increase or decrease, which would be opposite for an adjacent square, yet the position of each of the four squares would be identical, so the overlap can’t change and must at all positions be the same, and be quarter of the area of any overlapped square (regardless of its size).

Regardless of the rotation of the "2nd" square, it will always cover 1/4 of the area of the "1st" square

By making the square rotate we can only change the shape of overlapping surface but if the vertex of square is exactly at center of other square it doesn't affect the area..

Cover the entire square by four identical copies of the overlapping square. Then in the left picture each subsquare covers an equal amount, and will by symmetry/congruency continue to do so under arbitrary rotations of the quadruple square. Because the quadruple square always covers the entire square during a full turn, each subsquare will consistently cover 1/4 which is evidently the cover in the left picture.

Not very familiar with writing $LaTeX$ here yet, as it does not format the way I want, so I wrote my solution in overleaf instead. My over complicated solution can be viewed here Enjoy!

Basically when you rotate the square around, its just that some parts of a square are moved to the other

No matter the amount of angle rotated, it will consume an equal amount of area.

For me the easiest way to understand this is to imagine 4 squares covering the first square.

Even if you rotate them around, by symmetry, each square always covers one-quarter of the initial square.

Therefore the area is always constant.

The way I saw it very quickly was that as the overlap with the vertical (A) line shrinks, the overlap with the horizontal (B) line grows. Just like matter can't be created or destroyed, this overlap can't grow or shrink, because the length of each line consisting of the overlap isn't shrinking or growing as a whole! By "as a whole" I mean that if overlap with one line shrinks, it will grow by the same amount elsewhere!!

Simply this: If you take away a small slice by rotation, the same slice is gained on the other side. This implies the same property for all areas of rotation with similar parameters (angles of shape are all identical, shapes are identical)

The triangles 1 and 2 are congruent (identical) because (1) two angles a=b and c=d; (2) one edge m=n. Therefore, the area 1+3 = 2+3.

Red always equals blue.

If we draw a circle with half the length of second square from where the square is rotated,we get the same area in 2nd figure,as the length again remains the same.

The region that is in the rotated part, but not in the original square is equal to the part that is in the original square, but not in the rotated part. Therefore, they have equal areas.

Initial Area = l^2 Final Area = l^2 - (1/2) (l^2) tanx + (1/2) (l^2) tanx Where x is the angle by which the lower square is rotated around the centre.