The fraction is 
>!$\frac{2}{3}$


Without loss of generality let the radius of the smaller circle equal 1.[![enter image description here][1]][1]


Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we have[![enter image description here][2]][2]


By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$[![enter image description here][3]][3]


Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.[![enter image description here][4]][4]
By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule
$$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$
$$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$
Combining these 2 equations,
$$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$
Solving for r produces
$$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$
We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$
which now gives 
$$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$
Squaring both sides,

>!$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$
>!$$r^2=3$$
>!$$r=\sqrt{3}$$

And the ratio of the areas is
>!$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

  [1]: https://i.sstatic.net/YvM5A.png
  [2]: https://i.sstatic.net/LOvOa.png
  [3]: https://i.sstatic.net/xP2rV.png
  [4]: https://i.sstatic.net/IM9sE.png