Notes
chord and annulus solution

Chord and Annulus

Solution by Pythagoras' Theorem

With the points labelled as above, let $r$ be the radius of the inner circle and $R$ of the outer. The shaded region has area $\pi R^2 - \pi r^2$.

Point $A$ is the midpoint of the inner chord, so angle $O \hat{A} B$ is a right-angle. The length of $A B$ is $1$. Let $h$ be the length of $O A$. Applying Pythagoras' theorem shows that $r^2 = 1 + h^2$.

Since the dots are equally spaced, $A C$ has length $3$. Applying Pythagoras' theorem to triangle $C A O$ shows that $R^2 = 9 + h^2$. Therefore $R^2 - r^2 = 8$ and so the shaded region has area $8\pi$.

Solution by the Intersecting Chords Theorem

With the points as in the the above diagram, and $r$ and $R$ also as above, the intersecting chords theorem applies to the chords $F C$ and $D E$ of the outer circle. The relevant lengths are those of $B D$, which is $R - r$, of $B E$, which is $R + r$, of $B C$, which is $2$, and of $F B$, which is $4$. Putting these together shows that $4 \times 2 = (R + r)(R - r) = R^2 - r^2$. As above, this gives $8\pi$ for the area of the shaded region.