Notes
four squares in a circle solution

Four Squares in a Circle

Solution by Properties of Chords, Similar Triangles, and Pythagoras' Theorem

In the above diagram, $O$ is the centre of the circle, $E$ is the midpoint of $B C$, and $D$ is the midpoint of $A B$ and is also the corner of the second square.

The perpendicular bisector of a chord passes through the centre of the circle. So the centre lies where the horizontal line through $E$ meets the perpendicular line to $A B$ through $D$. The triangle $D F O$ is therefore similar to triangle $D C B$, so the length of $F D$ is twice that of $F O$. Since $F D$ is half the side of a square, $F D$ has length $2$ and so $F O$ has length $1$. Therefore $O E$ has length $1 + 4 + 4 = 9$. The length of $E B$ is $2$. Let $r$ be the radius of the circle, then this is the length of $O B$. Applying Pythagoras' theorem to triangle $O E B$ shows that:

The area of the circle is therefore $85 \pi$.