Is the blue area greater than the red area? 
Problem:
A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.
Which area is greater?


Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.
Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.
But how can it be proven?
I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since  the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.
Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$
According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that $$\begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}$$

Assertion:
To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.

Is this true? If so, is there another way of proving the assertion?

Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.
This question is very similar to this post.

Source:
The Golden Ratio (why it is so irrational) $-$ Numberphile from $14$:$02$.
 A: The two areas are equal.  On the diagram with the red area draw the vertical and horizontal lines that define the blue area.  The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area.  Those two triangles are congruent.
A:  
The four numbered areas are congruent.

[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.

A: 
Note that for equal angles $\angle A'OB' = \angle AOB = 90^\circ$, when we subtract a common part $\angle A'OB$ from both sides, we have $\angle AOA' = \angle BOB'$, so the red and cyan triangles are congruent: $\triangle AOA' \cong \triangle BOB'$.
That implies their areas are equal, and when we add a common part $\triangle A'OB$ we get area of the $AOB$ triangle equal to the area of the $A'OB'B$ quadrilateral. Finally, the area of the two squares' common part is constant, independent on the square's rotation angle.
A: The two areas are equal.  On the diagram with the red area draw the vertical and horizontal lines that define the blue area.  The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area.  Those two triangles are congruent.
A: By pinning a square's vertex to the center of the other, you guarantee a 90 degree slice outwards. This means we could tile 4 slices perfectly. A square has rotational symmetry of n=4. Since the rotation number is an integer multiple of the slice number, the area is invariant of rotation. You can apply this generally as well. A 120 degree slice of an equilateral triangle will be invariant. A 60 degree slice of a uniform hexagon will too. 120 degrees will work for the hexagon as well since that's 3 slices on a rotation number of 6.
A: Let $f(\alpha)$ be the length of the segment from the center of the square to the outside of the square on the line at an angle of $\alpha$ degrees from the horizontal line pointing right.
Suppose that the first side of the square (in counterclockwise order) makes an angle of $\alpha$, then area you want is $\int\limits_{\alpha}^{\alpha+\frac{\pi}{2}} \frac{f(x)^2}{2} dx$ and since $f$ is periodic with period $\frac{\pi}{2}$ this is independent of $\alpha$.
A: $\hspace{5cm}$
$$b^2+b^2=(a-c)^2+c^2 \Rightarrow \frac{b^2}{2}=\frac{(a-c)^2+c^2}{4}\\
S=\frac{b^2}{2}+\frac{(a-c)c}{2}=\frac{(a-c)^2+c^2+2(a-c)c}{4}=\frac{a^2}{4}.$$
A: I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.

A: 

Solution:

Although the red area is not a triangle, the sum of its sides that do not touch the centre is equal to $1$. This can only mean that no matter how many degrees the square is rotated, no area will be greater; the red area will always be equal to the blue area, i.e. $$\frac 14$$

Credit to @dxiv who pointed this out as a hint in a comment!
A: 
If we use $\overline{FB}$ for the base of $\triangle FEB$, then its altitude is $\frac 12s$. If we use $\overline{BG}$ for the length of the base of $\triangle BEG$, then its altitude is $\frac 12s$.
So the area of $\square FBGE$ is 
$\frac 12(\frac 12s)(s-x) + \frac 12(\frac 12s)(x) = \frac 14s^2$.  Which is one-fourth of the area of the square.
The advantage of this method is that it allows you to break the square into $n$ pieces of equal area quite easily. In fact, the same method applies to an regular polygon.
