Join squares 2

From the Picture , we need To calculate the Area of $\bigtriangleup{ACE}-\bigtriangleup{CHJ}\$

$Ar. of \bigtriangleup{ACE}=\dfrac{1}{2}\times9\times9 \\ Ar. of \bigtriangleup{CHJ}=\dfrac{1}{2}\times{HC}\times{HJ} \text{ , Where } HC=9-3=6 \\ \text{To calculate HJ , Let } \angle CJH =\angle CED=\theta \\ \tan{\theta}=\dfrac{9}{3}=\dfrac{HC}{HJ} \implies HJ=2 \\ \text{Answer = }\dfrac{81}{2}-\dfrac{12}{2}=\color{#3D99F6}{\boxed{34.5}}$

Consider my diagram. The area of the shaded region is area of the two squares minus area of the big triangle minus area of the red triangle.

By similar triangles, we have

$\dfrac{x}{3}=\dfrac{3-x}{6}$ $\implies$ $x=1$

The desired area therefore is

$A=9^2+3^2-\dfrac{1}{2}(9)(12)-\dfrac{1}{2}(1)(3)=\color{#D61F06}\boxed{34.5}$

Credit to Sambhrant Sachan for the image

The key is to express the black area (let's call it $\beta$ ) in two different ways. Let $HJ$ equal $x$ .

$\beta_1 = ACE - CHJ = \dfrac{AC \times CD}{2} - \dfrac{CH \times HJ}{2} = \dfrac{81 - 6x}{2}$

$\beta_2 = ABCD + DEFH - ABE - EFJ = AC^2 + DE^2 - \dfrac{AB \times (BD + DE)}{2} - \dfrac{EF \times (3 - HJ)}{2} = 36 - \dfrac{9 - 3x}{2}$

Set the two equations equal to each other.

$\beta_1 = \beta_2 \\ \dfrac{81 - 6x}{2} = 36 - \dfrac{3 - 3x}{2} \\ 81 - 6x = 72 - 9 + 3x \implies 18 = 9x \implies x = 2$

Now plug $x$ back into one of our original equations for $\beta$

$\beta = \dfrac{81 - 6 \times 2}{2} = \boxed{34.5}$

Using this image, (credits to Sambhrant Sachan), we will first find the area of $\Delta ACE$ . But this is simply $40.5$ by the area of a triangle formula. Now, we will find the area of $\Delta CHJ$ . Note that $\Delta CHJ \sim \Delta JFE$ in the ratio $2:1$ . From this, we get the length of $JH$ to be 2. From this, we get that the area of $CHJ$ is 6. Thus, the area of $ACHJE=|ACE| - |CHJ|=34.5$ . Thus, the answer is $34.5$ .

wonderful solution.