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&amp;=2+3+4+x\\
&amp;\Downarrow\\
x&amp;=y^2-9\end{align}

And the following equations to do so:
\begin{align}\frac {ya}2 &amp;=4\\
\frac {bc}2 &amp;=3\\
\frac {yd}2 &amp;=2\\
a+b&amp;=y\\
c+d&amp;=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?
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.
Assume that the hypotenuses of the yellow and green triangles bisect the right side of the quadrilateral, which is assumed to be a square with side s.  Area formula is (1/2)bh where b=s and h=s/2, so the area of the yellow triangle is equal to ss/4, which is equal to 4.  ss is thus equal to 4*4=16, so s=4.  The area of the square is also equal to s*s, which is 16.  The blue triangle is thus equal to 16-4-3-2 = 7.