<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] The problem is described as

As the title suggests, the problem in this post was meant to be a puzzle for 10th graders,

but I imagine for very mathematically talented students. I
simplify the original problem into an equivalent form.
Suppose a square $ABCE$ is such that the side lengths of $AB$
and $CE$ are equal to $b+c$ and the side lengths of $AE$ and $BC$
are equal to $a+d$.
Suppose that there is a point $P$ in the square such that the projection of $AP$ and $BP$ onto $AE$ and $BC$ has length $a$ and,
the projection of $BP$ and $CP$ onto $AB$ and $CE$ has length $c$.
Of course, since $ABCE$ is a square, its side length is $b+c=a+d$.
Finally, suppose we are given that the ratios of the segments $AP:BP:CP:EP = 1:2:3:x$ where $x$ is unknown.
We don't need to know $x$, but using the Pythagorean theorem
$$ \frac{a^2+b^2}{1^2} = 
\frac{a^2+c^2}{2^2} = 
\frac{c^2+d^2}{3^2} =
\frac{b^2+d^2}{x^2} $$
but $$  (a^2+b^2)+(c^2+d^2)=(a^2+c^2)+(b^2+d^2)$$
which implies that $\, 1^2+3^2 = 2^2+x^2 $ and $\,x=\sqrt{6}.$
Now, to solve for $\,a,b,c,d\,$ let
$$  b=a+u,\quad c=a+v,\quad d=a+u+v.$$
We know that $$ p_1 := 2^2(a^2+b^2) - 1^2(a^2+c^2) = 0, $$
and $$ p_2 := 3^2(a^2+b^2) - 1^2(c^2+d^2) = 0. $$
Compute the polynomial resultant
$$ \text{res}(p_1,p_2,v) = 8(a^2-2u^2)(2a^2+2au+u^2). $$
Solve for the positive real root of the second factor and
let $r:=u$. Then
$$a=\sqrt{2}r,\quad b=(1+\sqrt{2})r,\quad c=(4+\sqrt{2})r,
\quad d = (5+\sqrt{2})r.$$
In the original problem, we are given that $a^2+b^2=1^2,$ but
$a^2+b^2 = (5+2\sqrt{2})r^2$ which implies $r^2=1/(5+2\sqrt{2}).$
The area of the square is $(a+d)^2=(5+2\sqrt{2})^2r^2=5+2\sqrt{2}.$
In the original problem, point $D$ is the midpoint of $AE$,
and the area of the quadrilateral $ABCD$ is $3/4$ times the area of
the square $ABCE$ and so the final answer is $\frac34(5+2\sqrt{2}).$
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