P is a point inside rectangle ABCD, whose vertices are named in order. The distances are PA = 2 cm, PB = 6 cm and PC = 9 cm. What is the length of PD?

Drop perpendiculars from $P$ to the sides and write each distance with Pythagoras; the horizontal pieces cancel in pairs and so do the vertical ones, leaving $P{A}^2+P{C}^2=P{B}^2+P{D}^2$, where A, $C$ are one pair of opposite corners and $B,D$ the other. So $4+81=36+P{D}^2$, giving $P{D}^2=49$ and $PD=7$ cm. Pairing ADJACENT corners instead gives 36 + 81 − 4 = 113 and the trap √113 cm. Combinations such as 2 + 9 = 11 or 2 + 9 − 6 = 5 cannot work, because the relation holds between the SQUARES of the distances.