**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: $$ A = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = \pi({1\over 2}AB)^2 = \pi (AB)^2/4. $$