<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] 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
</s>