A Chordingly
Jessica Zhang of Woodbridge High School, Arwa Jamil , Year 8, the International School Brunei and Nisha Doshi and Suzanne Abbott from Year 9 the Mount School, York all sent good solutions to this problem.
Let the radius of the big circle be $R$ and the radius of the small circle be $r$. By Pythagoras theorem: $$ R^2 - r^2 = ({1\over 2}AB)^2 $$ The area of the big circle is $\pi R^2$ and the area of the small circle is $ \pi r^2 $. The area of the annulus $A$ is the area of the big circle minus the area of the small circle so: $$ A = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = \pi({1\over 2}AB)^2 = \pi (AB)^2/4. $$
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