Let P be a point on the circle circumscribing square ABCD that satisfies $PA.PC=56 and PB. PD=90.$ The area of the square ABCD is R, the absolute difference of the largest and smallest digit in R is _______

PAC, PBD are right angled triangles
Area of  $\Delta \mathcal{l}ePAC=\frac{1}{2}\left(PA\right)\left(PC\right)$
(PA) (PC) = 56
${\Delta }_{PAC}=28$

Area of  $\Delta \mathcal{l}ePBD=\left(\frac{1}{2}\right)\left(PB\right)\left(PD\right)$

$$\begin{gathered} \left(PB\right)\left(PD\right)=90 \\ {\Delta }_{PBD}=45 \\ {\Delta }_{PAC}=\frac{1}{2}\times \left(PM\right)\left(AC\right)\Rightarrow PM=\frac{\left(2\right)\left(28\right)}{\left(AC\right)}=\frac{56}{d}=\frac{56}{d} \\ {\Delta }_{PBD}=\left(\frac{1}{2}\right)\left(PN\right)\left(BD\right)\Rightarrow PN=\frac{\left(2\right)\left(45\right)}{\left(BD\right)}=\frac{90}{d} \end{gathered}$$

$$\begin{gathered} P{M}^2+P{N}^2=P{D}^2\Rightarrow {\left(\frac{56}{d}\right)}^2+{\left(\frac{90}{d}\right)}^2={\left(\frac{d}{2}\right)}^2 \\ \Rightarrow d^4=4[{56}^2+{90}^2]=\left(4\right)\left(4\right)\left({53}^2\right)\Rightarrow d^2=212 \end{gathered}$$

Area of square =  $\frac{d^2}{2}=106$
Largest digit – smallest digit = 6 – 0 = 6