# Solution to the [[Three Squares in a Circle II]] Puzzle +-- {.image} [[ThreeSquaresinaCircleII.png:pic]] > Three squares inside a circle. What’s the circle’s area? =-- ## Solution by [[Pythagoras' Theorem]], [[Difference of Two Squares]], and [[Area of a Circle]] +-- {.image} [[ThreeSquaresinaCircleIIAnnotated.png:pic]] =-- In the diagram above, $O$ is the centre of the circle and $E B$ is perpendicular to the sides of the squares at $E$ and $B$. The squares have side lengths $6$, $12$, and $18$. Therefore, $E B$ has length $36$. Let $r$ be the radius of the circle. Let $x$ be the length of $O E$ and $y$ the length of $O B$, so $x + y = 36$. Applying [[Pythagoras' theorem]] to triangle $O E D$ shows that: $$ r^2 = 3^2 + x^2 $$ and applying it to triangle $O B C$ shows that: $$ r^2 = 9^2 + y^2 $$ Subtracting these equations shows that: $$ x^2 - y^2 = 9^2 - 3^2 = 72 $$ Since $x + y = 36$, and by [[difference of two squares]] $x^2 - y^2 = (x + y)(x - y)$ this means that $x - y = 2$. So $x = 19$ and $y = 17$. Then $r^2 = 3^2 + 19^2 = 370$. The area of the circle is therefore $370\pi$.