<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] Not a completely geometric solution and I am assuming that only integers are allowed. 
Since area of yellow triangle is twice the red one, so $a=2d$. Now divide the square into four rectangles by drawing horizontal line from point between $a$ and $b$ and a vertical line from point between $c$ and $d$. Then $$4+8+6-a.\frac{a}{2}=y^2$$ which gives $$a^2+2y^2=36.$$ This has solution $a=2, y=4$. 
</s>