An Old Coin In China

Let the midpoint of $CB$ = $D$

$\begin{aligned} r^2 &= (AD)^2 + (DO)^2 \\ &= (1.5)^2 + (0.5)^2 \\ &= 2.5 \\ \\ \pi r^2 &= \pi( 2.5) \\ &\approx 7.85 \end{aligned}$

Area of coin is square subtracted from circle $= 7.85 - 1 \implies \boxed{6.85}$

Let the radius of the circle be $r$ . Then

$r^2=(\frac{1}{2})^2+(1+\frac{1}{2})^2=\dfrac {5}{2}\implies πr^2\approx \dfrac {15}{2}=7.5$

$\implies$ area of the coin $\approx 7.5-1^2=\boxed {6.5}$ .

We draw a line $AE$ (the circle's diameter) , and know that $AD^2 + DE^2 = AE^2 , AD = 3 , DE = 1$ . So the radius should be $\frac{\sqrt{10}}{2}$ . The area is( $\frac{\sqrt{10}}{2}$ ) $^2$ * π - $1^2$ = 6.5