<s>[INST] Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:

What is the blue area?
 
Hint: Think outside the box

My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:

This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
 [/INST] The hint may assume two things: to draw extra lines 1) outside; 2) inside the box so that the solution is simple to understand.
Here is yet another method: draw the lines to divide the square into four rectangles indicated by $A, B, C$ and $3+3$:

$$\begin{cases}
A+B=8 \\
A+C=4 \\
CB=6A
\end{cases} \stackrel{(1)-(2)}\Rightarrow \begin{cases}
B=C+4 \\
A=4-C \\
C(C+4)=6(4-C)
\end{cases} \Rightarrow $$
$$C^2+10C-24=0 \Rightarrow C=2, A=2, B=6$$
Hence: Blue area = $(A+B+C+6)-(2+3+4)=16-9=7.$
P.S. I wonder what brilliant method the proposer (on the brilliant.org) has.
</s>