Notes
four squares overlapping a circle solution

Four Squares Overlapping a Circle

Solution by the Intersecting Chords Theorem and Pythagoras' Theorem

In the above diagram, point $O$ is the centre of the circle and $E$ and $F$ are so that angles $O \hat{F} P$ and $P \hat{E} O$ are right-angles. Since $D B$ is a chord, $E$ is its midpoint and similarly $F$ is the midpoint of $C A$. From the areas of the squares, the following lengths can be deduced: $A P$ has length $10$, $B P$ has length $5$, and $C P$ has length $4$. Therefore $A F$ has length $7$.

The intersecting chords theorem says that the length of $D P$ is given by $\frac{10 \times 4}{5} = 8$. So $B D$ has length $13$ and so $E P$ has length $\frac{3}{2}$. Let $r$ be the radius of the circle. Applying Pythagoras' theorem to triangle $O F A$ shows that $r^2 = 7^2 + \left(\frac{3}{2}\right)^2 = \frac{205}{4}$. Therefore, the area of the circle is $\frac{205 \pi}{4}$.