<s>[INST] Find the area of Quadrilateral $ABCD$. A puzzle for 10th graders. As the title suggests, the problem in this post was meant to be a puzzle for 10th graders, so claims the person who posted this on a language exchange platform: The problem is as follows:

Given a Quadrilateral $ABCD$ with internal point $P$, where $AP=1$, $BP=2$ and $CP=3$, and unknown sides $k$ and $2k$, compute the area of this Quadrilateral.
I first tried to inscribe this quadrilateral into a square but that approach did not turn out successful, I was thinking if there are any other ways to solve it, perhaps via setting up a coordinate system, or via a trigonometric method. I will share my own successful approach below as an answer!
 [/INST] Using $B$ as the origin and $BC$ and $BA$ as the $x$ and $y$ axis resp., we can express $C(2k,0)$ and $A(0,2k)$. Therefore, expressing in two ways the squares of lengths $AP^2,BC^2,BA^2$ we get:
$$\begin{cases}x^2+y^2&=&4\\x^2+(y-2k)^2&=&1\\(x-2k)^2+y^2&=&9\end{cases} \ \iff \ \ \begin{cases}x^2+y^2&=&4\\x^2+y^2-4ky+4k^2&=&1\\x^2+y^2-4kx+4k^2&=&9\end{cases}$$
$$\iff \ \begin{cases}x^2+y^2&=&4\\4ky-4k^2&=&3\\4kx-4k^2&=&-5\end{cases}\tag{1}$$
(Explanations: new row $R_2=$ old row $R_2$ minus row $R_1$ ; similar operation for row $R_3$).
In (1), we can extract $x=\frac{4k^2-5}{4k}$ and $y=\frac{4k^2+3}{4k}$ from the two last equations of (1) ; plugging the results into the first equation, we get:
$$\left(\frac{4k^2-5}{4k}\right)^2+\left(\frac{4k^2+3}{4k}\right)^2=4\tag{2}$$
wich is a biquadratic equation.
Setting $K:=k^2$, (2) becomes the quadratic equation:
$$(-5+4K)^2+(3+4K)^2=64K$$
which amounts to :
$$16K^2-40K+17=0$$
whose roots are $K=\frac14(5 \pm 2 \sqrt{2})$.
Only root $K=\frac14(5 +2 \sqrt{2})$ is compatible with the given lengths, giving finally :
$$area(ABCD)=height \times \frac12 (upperbase+lowerbase)= 2k \times \frac32 k = 3k^2 =3K=\frac34(5 +2 \sqrt{2})$$
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