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?

Green triangle has 2x area of red triangle. Heights of both triangles are equal because all sides of square are equal.  so base of green triangle = 2x base of red triangle.So  r=length of base of red triangle and 2r=base of green triangle. Let s= length of side of square also height of both triangles. for red triangle 2=1/2rs so rs=4. same relation hold for green triangle. 4=1/2(2rs) so rs=4 again. Referring to the yellow triangle bottom side=s-r and right side=s-2r. So the area of the triangle is 3= 1/2 (s-r)(s-2r) then 
6=( s-r)( s-2r)  so 6=s^2-3sr+2r^2. since rs=4. r=4/s. substitute in equation. then
6=S^2-3s(4/s)+2(16/s^2)  6=s^2-12+32/s^2 , 18=s^2+32/s^2, 18s^2=s^4+32, s^4-18s^2+32=0, Let x=s^2,  then x^2-18x+32=0, Factor (x-16) (x-2)=0  x=16, x=2 disregard x=2 x=16=s^2
then 16-4-2-3=7 area of Blue triangle
