point in the interior of a square I've got the following problem: 
A point in the interior of a square ABCD is at distances 3, 4 and 5 meters from the vertices A, B and C, respectively. What is the area of ABCD?
I tried thinking about this three points forming a triangle and trying to solve the problem from there, but I didn't get any results.
 A: $\textbf{HINT:}$ Try to duplicate the square in this way

Then evaluate $\cos x$ and $\sin x$ using the cosine law on $PCQ$. That should be enough to find $l^2=CB^2$ using again the cosine law on $CBQ$.
$\textbf{Calculation details}$: 
by the cosine law on $CPQ$ we have
$$5^2=(4 \sqrt 2)^2 +3^2-2\cdot 3\cdot 4 \sqrt 2 \cos x$$
so $\cos x= \sqrt 2/3$. Then $\sin x=\sqrt 7/3$. Now let's consider the angle $\angle CQB=\pi/4+x$:
$$cos(\pi/4+x)=\cos x /\sqrt 2-\sin x/\sqrt 2=\frac{1}{3}-\frac{\sqrt 7}{3\sqrt 2}.$$
Let's now apply the cosine law on $CQB$:
$$CB^2=l^2=3^2+4^2-2\cdot 3 \cdot 4\cdot  cos(\pi/4+x)=17+4\sqrt {14}.$$
A: Begin by assigning coordinates: $A = (0,L)$, $B = (0,0)$, $C = (L,0)$, $P=(X,Y)$.
Now apply Pythagoras to the three triangles with corner point $A$, $B$ and $C$:
$$X^2 +(L-Y)^2 = 9$$
$$X^2 + Y^2 = 16$$
$$(L-X)^2 + Y^2 = 25$$
Subtracting the second equation from the first leads to: $Y = (L^2 + 7)/2L$.
Subtracting the second equation from the third leads to: $X = (L^2 -9)/2L$.
Substitution of these two results into the second equation gives: $L^4 -34L^2 +65=0$.
This is a quadratic equation in $L^2$. It has the solution $L^2 = 17 + 4 \sqrt14$. This the area of the square that was sought by the OP.
