How to find the area of this square? 
In a square $ABCD$, say there is a point $P$ which lies inside
  it, the point $P$ is located at distances $x$, $y$ and $z$ meters from
  $A,B$ and $C$ respectively.

Using this information how could we compute a form for the area of the square?Please explain your approach.
 A: Using coordinate geometry. Assume that the vertices (in order) are $(0,a), (0,0), (a,0), (a,a)$ and that $P$ is $(u,v)$. Then $$
\begin{align}
u^2+(v-a)^2 &= x^2,\tag{1}
\\ u^2+v^2 &= y^2, \tag{2}
\\ (u-a)^2+v^2 &= z^2 \tag{3}.
\end{align}
$$
We can "solve" for $v$ in terms of $a$ by subtracting $(2)$ from $(1)$; similarly for $u$. Since the steps are straightforward, I am just posting the answer here:
$$
(u, v) = \left( \frac{a^2 + y^2 - z^2}{2a}, \frac{a^2 + y^2 - x^2}{2a} \right).
$$
Plugging this in $(2)$, we get:
$$
(a^2+y^2 - z^2)^2 + (a^2 + y^2 - x^2)^2 = 4a^2y^2.
$$
$$
\implies a^4 + y^4 - (a^2+y^2)(x^2+z^2)+\frac{x^4+z^4}{2}=0.
$$ 
This is a quadratic equation in $a^2$(=area of the square) and can be solved. 
Note. I do not yet understand whether both the roots of the quadratic are true solutions or not. 
A: This is just a supplement to Srivatsan's answer. I knew that this was a consequence of something but didn't quite get it but this is just Varignon's theorem. 

(I wish one day we can insert tikzpicture environment inside the answers.) I also agree with Srivatsan that there must a shorter/simpler observation and maybe someone can provide it. 
