Pythagoras' Slanting Sangaku

Label the rectangle $ABCD$ , the right triangle $ABE$ , circles centers $O$ , $P$ , and $Q$ , $OF$ and $QG$ perpendicular to $BE$ , $PH$ and $PI$ are perpendicular to $DA$ and $AB$ , and $QJ$ perpendicular to $OF$ . Let the radii of the red circle and the blue circle be $r$ and $R$ respectively.

Then the red segment $FG = QJ = \sqrt{OQ^2-OJ^2} = \sqrt{(R+r)^2-(R-r)^2} = 2\sqrt{rR} = 12 \implies rR = 36$ . Let $\angle ABE = \theta$ . Then

$\begin{aligned} AB & = AI + IB \\ 2R & = r + r \cot \frac \theta 2 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ 2rR & = r^2 + \frac {r^2}t & \small \blue{\text{Note that }rR = 36} \\ \implies r^2 & = \frac {72t}{1+t} \end{aligned}$

And

$\begin{aligned} BE & = BG + FG + EF \\ AB \cdot \sec \theta & = r \cot \left(\frac {90^\circ -\theta}2\right) + 12 + R \cot \left(\frac {90^\circ + \theta}2\right) \\ \frac {2R}{\cos \theta} & = r \cot \left(45^\circ - \frac \theta 2\right) + 12 + R \cot \left(45^\circ + \frac \theta 2\right) \\ 2R \cdot \frac {1+t^2}{1-t^2} & = r \cdot \frac {1+t}{1-t} + 12 + R \cdot \frac {1-t}{1+t} & \small \blue{\text{Multiply both sides by }r} \\ 72 \cdot \frac {1+t^2}{1-t^2} & = \frac {1+t}{1-t} \cdot r^2 + 12r + 36 \cdot \frac {1-t}{1+t} & \small \blue{\text{Note that }rR = 36} \\ 72 \cdot \frac {1+t^2}{1-t^2} & = \frac {1+t}{1-t} \cdot \frac {72t}{1+t} + 12r + 36 \cdot \frac {1-t}{1+t} & \small \blue{\text{and }r^2 = \frac {72t}{1+t}} \\ \frac {6+6t^2}{1-t^2} & = \frac {6t}{1-t} + r + \frac {3-3t}{1+t} \\ \implies r & = \frac {3(1-t^2)}{1-t^2} = 3 \end{aligned}$

From $r^2 = \dfrac {72t}{1+t} \implies t = \dfrac 17$ , $R = \dfrac {36}r = 12$ , $BC = r \cot \left(45^\circ - \frac \theta 2\right) + 12 + R = 3 \cdot \dfrac {1+\frac 17}{1-\frac 17} + 12 + 12 = 28$ , and the area of the rectangle $AB \cdot BC = 2 \cdot 12 \cdot 28 = \boxed{672}$ .

Label the figure as follows. Denote by $R$ the radius of the big circle and by $r$ the radius of the small ones. Let $\phi =\angle HCG=\angle GCJ$ and $\theta =\angle FCK=\angle FCL$ .

By Pythagoras’s theorem on right $\triangle EFN$ ,

$FN=\sqrt{E{{F}^{2}}-E{{N}^{2}}}=\sqrt{{{\left( R+r \right)}^{2}}-{{\left( R-r \right)}^{2}}}=2\sqrt{Rr}$ Hence,

$2\sqrt{Rr}=FN=IK=HL=12\Rightarrow Rr=36 \ \ \ \ \ (1)$ $\angle EFN$ and $\angle FCL$ are corresponding angles, thus, $\angle EFN=\angle FCL=\theta$

On $\triangle ENF$ ,
$\tan \theta =\frac{EN}{NF}=\frac{R-r}{12} \ \ \ \ \ (2)$ On $\triangle GCJ$ ,
$\tan \varphi =\frac{GJ}{JC}=\frac{r}{2R-r} \ \ \ \ \ (3)$ Moreover,

$\begin{aligned} 2\theta +2\varphi =90{}^\circ & \Rightarrow \varphi =45{}^\circ -\theta \\ & \Rightarrow \tan \varphi =\tan \left( 45{}^\circ -\theta \right)=\frac{1-\tan \theta }{1+\tan \theta } \\ & \overset{\left( 2 \right)\left( 3 \right)}{\mathop{\Rightarrow }}\,\frac{r}{2R-r}=\frac{1-\frac{R-r}{12}}{1+\frac{R-r}{12}} \\ & \Leftrightarrow \left( R-12 \right)\left( R-r \right)=0 \\ & \overset{R\ne r}{\mathop{\Rightarrow }}\,R=12 \\ \end{aligned}$ From $(1)$ we get $r=3$ , hence, $\left( 2 \right)\Rightarrow \tan \theta =\frac{3}{4}$ . On $\triangle EIC$ ,

$IC=\frac{EI}{\tan \theta }=\frac{R}{\tan \theta }=\frac{12}{\frac{3}{4}}=16$ Hence, for the dimentions of $ABCD$ we have $BC=BI+IC=R+IC=12+16=28$ and $AB=2R=24$ Finally, for the area of the quadtrilateral, $\left[ ABCD \right]=AB\cdot BC=24\times 28=\boxed{672}$