# Four Squares in a Circle +-- {.image} [[FourSquaresinaCircle.png:pic]] > What’s the area of the circle? =-- ## Solution by [[Properties of Chords]], [[Similar Triangles]], and [[Pythagoras' Theorem]] +-- {.image} [[FourSquaresinaCircleLabelled.png:pic]] =-- 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: $$ r^2 = 9^2 + 2^2 = 85 $$ The area of the circle is therefore $85 \pi$.