Given distances from a point to three vertices of a rectangle, find the distance to the fourth vertex

Question

Problem: $ABCD$ is a rectangle. A point $P$ is $11$ from $A$, $13$ from $B$ and $7$ from $C$. What is the length of $DP=x?$ (Note: $P$ can be inside the rectangle or outside of it.)

I drew this scenario as best as I could, but I only have triangles with two sides and no angles. How do I begin? Any weird and relatively unknown theorems I should use?

Answer 1

Hint: draw lines through $P$ parallel to the sides of rectangle. Use Pythagoras theorem for the lines $PA, PB, PC, PD$ to make up the system of four equations. Then you get the answer $1$.

Details:

$$\begin{cases} a^2+d^2=x^2 \\ a^2+b^2=11^2 \\ b^2+c^2=13^2 \\ c^2+d^2=7^2\end{cases} \stackrel{(4)-(3)+(2)}{\Rightarrow} x^2=7^2-13^2+11^2=1 \Rightarrow x=1.$$

Answer 2

Just $x^2+13^2=7^2+11^2$ by Pythagoras theorem.

Let $PK$, $PL$, $PM$ and $PN$ be altitudes of $\Delta DPA$, $\Delta DPC$, $\Delta CPB$ and $\Delta APB$ respectively. Thus, $$x^2+13^2=DK^2+KP^2+PM^2+MB^2=PN^2+KP^2+PM^2+PL^2=11^2+7^2$$