# Solution to the Two Squares on a Circle Puzzle +-- {.image} [[TwoSquaresonaCircle.png:pic]] > The radius of the circle is $4$. What's the combined area of the two squares? =-- ## Solution by [[Angle in a Semi-Circle]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoSquaresonaCircleLabelled.png:pic]] =-- In the above diagram, angle $C \hat{B} A$ is a [[right-angle]], so as the [[angle in a semi-circle]] is a right-angle, $A C$ must be a diameter of the circle. It therefore has length $8$. Let the squares have side lengths $a$ and $b$, with $a$ of the yellow square and $b$ of the pink. Then $A B$ has length $a\sqrt{2}$ and $B C$ has length $b\sqrt{2}$. Applying [[Pythagoras' theorem]] to triangle $A B C$ shows that: $$ (a\sqrt{2})^2 + (b\sqrt{2})^2 = 8^2 $$ So $a^2 + b^2 = 32$. Since the areas of the squares are $a^2$ and $b^2$, this shows that the combined area of the two squares is $32$.