Using the circle equation I derived roots at $(16,8)$ and $(0,0)$. In an effort to stay true to the question we decided that since the line gets a little blurry with trig and known relationships within shapes we said that just no using $\sin$, $\cos$, $\tan$ was enough for no trig requirement.
Also, I do not speak Chinese so if there is any extra info in there, let me know!
I have had several guys guess that it might not be possible, lets find out!
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Lets denote area as A; we have following areas:
$A_1 =20^2 -\frac{1}{4}. 20^2. \pi$
Big Ellipse like area: $A_2=\frac{1}{4}. 20^2.\pi-(20^2 -\frac{1}{4}. 20^2. \pi)=228$
$A_3=\frac{1}{4}. 10^2.\pi-\frac{1}{2}. 10^2=28.5; 2 A_3=57=\frac{1}{4} A_2$
Where $2 A_2$ is small ellipse like area inside the big one.
Required area $A_4$ is:
$A_4≈ 228(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}- \frac{1}{48})≈97$
I'll skip the part that you already solved. Let's start from this picture:
Triangles $\triangle DHF$ and $\triangle EJF$ are obviously similar. It means that:
$$\angle ADF=\angle BEF=\alpha$$
We want to calculate the following areas:
$$P_1=area(sector(DAF))-area(\triangle DAF)=20^2\pi\frac{\alpha}{2\pi}-\frac{20\times 16}{2}=200\alpha-160$$
$$P_2=area(sector(AEF))-area(\triangle AEF)=10^2\pi\frac{\pi-\alpha}{2\pi}-\frac{10\times 8}{2}=50(\pi-\alpha)-40$$
$$P=P_1+P_2$$
$$P=200\alpha-160 + 50(\pi-\alpha)-40$$
$$P=150\alpha + 50\pi - 200$$
However it's impossible to calculate $\alpha$ without trigonometry:
$$\tan \alpha = \frac86 = \frac43\implies \alpha \approx 0.927 \space\text{rad}, \space\space P\approx 96.174$$
If you want something really tricky try to find area of the red shape in the middle: