Crescent Moon

Let $O$ be the center of the small circle, $AE=R$ and $OD=r$ . Then

$2R=9+2r$ $\color{#D61F06}\large \implies$ $R=\dfrac{9+2r}{2}$

By the pythagorean theorem , we get

$(FE)^2=r^2-(R-r)^2=2Rr-R^2$ $\color{#D61F06}\large \implies$ $FE=\sqrt{2Rr-R^2}$

However, $FE=R-5$

$FE=FE$

$R-5=\sqrt{2Rr-R^2}$

However, $R=\dfrac{9+2r}{2}$

$\dfrac{9+2r}{2}-5=\sqrt{2\left(\dfrac{9+2r}{2}\right)(r)-\left(\dfrac{9+2r}{2}\right)^2}$

$\dfrac{9+2r-10}{2}=\sqrt{9r+2r^2-\left(\dfrac{81+36r+4r^2}{4}\right)}$

$\dfrac{2r-1}{2}=\sqrt{9r+2r^2-\dfrac{81-36r-4r^2}{4}}$

Squaring both sides, we get

$\dfrac{4r^2-4r+1}{4}=9r+2r^2-\dfrac{81-36r-4r^2}{4}$

$\dfrac{4r^2-4r+1}{4}=\dfrac{36r+8r^2-81-36r-4r^2}{4}$

$4r^2-4r+1=8r^2-81-4r^2$

$-4r^2+1=-81$

$4r=81+1$

$4r=82$

$r=20.5$

It follows that,

$R=\dfrac{9+2(20.5)}{2}=25$

Therefore the area of the shaded region is,

$A=\pi(R^2-r^2)=\pi(25^2-20.5^2) \approx$ $\color{#D61F06}\boxed{\large 643}$

Let the radius of the large circle be $R$ . Using Pythagorean theorem , we have:

$\begin{aligned} {\color{#3D99F6}\overline{HF}^2} + {\color{#D61F06}\overline{FD}^2} & = \overline{HD}^2 \\ {\color{#3D99F6}\overline{HE}^2 + \overline{FE}^2} + {\color{#D61F06}\overline{FE}^2+\overline{ED}^2} & = (\overline{HE} + \overline{ED})^2 \\ {\color{#3D99F6}(R-9)^2 + (R-5)^2} + {\color{#D61F06}(R-5)^2+R^2} & = (R-9 + R)^2 \\ \implies R^4 - 38R + 131 & = R^4 - 36R + 81 \\ 2R & = 50 \\ \implies R & = 25 \end{aligned}$

Let the radius of the small circle be $r$ , then $2r = \overline{HD} = \overline{HE} + \overline{ED} = 2R-9 = 41$ , $\implies r = 20.5$ .

The area of the crescent moon $A = \pi (R^2-r^2) = \pi (25^2-20.5^2) \approx \boxed{643}$