Odd Sangaku

First we note that $\triangle ABC$ is a $3$ - $4$ - $5$ right triangle (this makes things a lot easier).

Label the square $DEFG$ and the rectangle $GHIJ$ . Extend $FG$ to meet $BC$ at $K$ . Since the two circles are congruent and the three angles of $\triangle DGH$ and $\triangle BFK$ are the same, $\triangle DGH$ and $\triangle BFK$ are congruent (this also makes things a lot easier).

Let the side length of square $DEFG$ be $a$ , then $DG=BF=a$ , $DH=BK=\dfrac {4a}3$ , and $GH=FK= \dfrac {5a}3$ . Let $GJ=b$ . We note that $\triangle GJK$ and $\triangle BFK$ are similar (makes things easy again), then

$\begin{aligned} \frac {GJ}{GK} & = \frac {BF}{FK} = \frac 35 \\ GJ & = \frac 35 \cdot GK = \frac 35 (FK - FG) \\ \implies b & = \frac 35 \left(\frac {5a}3 - a \right) = \frac {2a}5 \end{aligned}$

We have:

$\begin{aligned} AF+FB & = AB \\ AF + \frac 35 \cdot FG + GJ & = AB \\ \frac 54a + \frac 35 a + b & = 27 \\ \frac {25+12+8}{20}a & = 27 \\ \frac {45}{20}a & = 27 \\ \implies a & = 12 \end{aligned}$

The sum of areas of the square and rectangle is $a^2 + b \cdot GH = a^2 + \dfrac {2a}5 \cdot \dfrac {5a}3 = \frac 53a^2 = \boxed{240}$

Let $\angle BAC=\alpha\implies \angle ACB=90-\alpha$ and $\angle AEF=90-\alpha$ . So, we have, $\triangle ABC\sim AFE \implies AF=27k$ and $EF=36k$ for some constant $k>0$ . Also, $AB=27,\;BC=36$ and $CA=\sqrt{AB^2+BC^2}=45$ .

Simple angle chasing gives, $\angle AEF=90-\alpha\implies \angle BEL=\alpha\implies \angle BLE=90-\alpha$ $\angle ACB=\angle GIH=90-\alpha\implies GHI=\alpha$ We conclude that, $\triangle AFE\sim \triangle HGI\sim \triangle IJC$ and $\triangle HGI\cong \triangle EBL$ , since they have the same inradius. $\begin{aligned}\triangle AFE\sim \triangle HGI&\implies \frac{AF}{HG}=\frac{FE}{GI} \\ \\ &\implies \frac{27k}{36k}=\frac{36k}{GI} \\ \\ &\implies GI=48k \\ \\ &\implies HI=\sqrt{HG^2+GI^2}=60k \\ \\ \triangle HGI\sim \triangle IJC &\implies \frac{HG}{IJ}=\frac{HI}{IC} \\ \\ &\implies \frac{36k}{IJ}=\frac{60k}{45-111k} \\ \\ &\implies IJ=\frac{3(45-111k)}{5} \\ \\ \triangle HGI\cong \triangle EBL &\implies BE=GH=36k \end{aligned}$ In right-triangle $AFE$ , $\begin{aligned} AE^2&=AF^2+EF^2 \\ \\ (27-36k)^2&=(27k)^2+(36k)^2\\ \\ (3-4k)^2&=(3k)^2+(4k)^2 \end{aligned}$ Solving the above quadratic equation, gives us $k=\dfrac{1}{3}$ . Therefore, $\begin{aligned} \text{area}[EFGH]&=(EF)^2=(36k)^2=12^2=144 \\ \\ \text{area}[HIJK]&=HI\cdot IJ=60k\cdot \frac{3(45-111k)}{5} =20\cdot \frac{24}{5}=96 \end{aligned}$ $\implies \text{area}[EFGH]+\text{area}[HIJK]=\boxed{240}$