<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] 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 $$3k^2=\frac{15+6\sqrt{2}}{4}.$$
</s>