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!
 A: 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 :
$$area(ABCD)=height \times \frac12 (upperbase+lowerbase)= 2k \times \frac32 k = 3k^2 =3K=\frac34(5 +2 \sqrt{2})$$
A: This is my approach:

1.) Rotate $\triangle ABP$ by $90^\circ$ such that the triangle $\triangle BP'C$ formed outside the quadrilateral is congruent to $\triangle ABP$. Therefore, $BP=BP'=2$ and $AP=P'C=1$. Connect $P$ with $P'$ to form $\triangle PBP'$ and $\triangle PP'C$. Notice that $\triangle PBP'$ is an isosceles right triangle, this implies that $\angle BPP'=\angle BP'P=45^\circ$ and $PP'=2\sqrt{2}$.
2.) Note than $\triangle PP'C$ has sides that are Pythagorean triples, therefore, $\triangle PP'C$ is also a right triangle where $\angle PP'C=90^\circ$. This implies that $\angle BP'C=\angle APB=135^\circ$. This also further implies that $A$, $P$ and $P'$ are collinear.
3.) Draw a perpendicular from $B$ onto $PP'$ at point $E$. Its easy to see that $\triangle BEP$ is an isosceles right angle triangle, therefore, $BE=EP=\sqrt{2}$. Lastly, note that $\triangle AEB$ is a right angle triangle, therefore:
$$(2k)^2=(\sqrt{2})^2+(1+\sqrt{2})^2$$
$$4k^2=5+2\sqrt{2}$$
Since the area of the Quadrilateral is $3k^2$, therefore:
$$3k^2=\frac{15+6\sqrt{2}}{4}$$
A: I'm pretty bad at geometry so I just spammed a bunch of algebra.
First, note that the area is clearly $3k^2.$ Then, draw a line through $P$ perpendicular to $AD$ and $BC,$ which intersects them at $X$ and $Y,$ respectively. Let $PX=n,$ so $PY=2k-n.$ Similarly, let $AX=BY=m$ so $CY=2k-m.$ Then, note that $m^2+n^2=1.$ Also, we have $m=1-n^2=4-(2k-n)^2,$ using the second equality, we have $4k^2-4kn=3$ so $n=\frac{4k^2-3}{4k}.$ Similarly, we have $n=4-m^2=9-(2k-m)^2$ and similar to above, this yields $m=\frac{4k^2-5}{4k}.$ Therefore, we have $$\left(\frac{4k^2-3}{4k}\right)^2+\left(\frac{4k^2-5}{4k}\right)^2=1.$$ This yields $$16k^4-24k^2+9+16k^4-40k^2+25=16k^2$$ $$32k^4-80k^2+34=0$$ $$16k^4-40k^2+17=0.$$ This is a quadratic in $k^2$ and solving for $k^2$ gives $$k^2=\frac{40\pm\sqrt{1600-1088}}{32}=\frac{5\pm 2\sqrt{2}}{4}.$$ However, note that $k>1,$ so $k^2>1$
and therefore, we only take the $+$ and we have that the area is $$3k^2=\frac{15+6\sqrt{2}}{4}.$$
A: 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 $\frac34(5+2\sqrt{2}).$
