<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] Here is an alternative solution. See the graph for labeling:

From the trapezium area formula we get:
$$\frac{a+x}{2}\cdot x=5+A \Rightarrow ax=10+2A-(9+A) \Rightarrow ax=A+1 \ \
 (1)$$
$$\frac{b+x}{2}\cdot x=7+A \Rightarrow bx=14+2A-(9+A) \Rightarrow bx=A+5 \ \
 (2)$$
$$ab=6 \ \ (3)$$
Now multiply $(1)$ and $(2)$ and substitute $(3)$:
$$abx^2=(A+1)(A+5) \Rightarrow 6(A+9)=(A+1)(A+5) \Rightarrow A^2=49 \Rightarrow A=7.$$
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