# Solution to the Two Squares and a Semi-Circle Puzzle +-- {.image} [[TwoSquaresandaSemiCircle.png:pic]] > What’s the total area of these two squares? =-- ## Solution using [[Chords]] and [[Pythagoras' theorem]] +-- {.image} [[TwoSquaresandaSemiCircleLabelled.png:pic]] =-- The centre of the circle lies on the [[perpendicular bisectors]] of the [[chords]] $B C$ and $A B$. The perpendicular bisector of $B C$ passes through its [[midpoint]] $E$ and the corner $F$ while that of $A B$ passes through $D$ and $G$. Triangle $H O F$ is a [[right-angled triangle|right-angled]] [[isosceles]] triangle that is [[congruent]] to $F E B$. This establishes $O G$ has having the same length as the half side, $D B$. So $O D$ is three times the length of $D B$. Let $x$ be the length of $A B$, then $D B$ had length $\frac{1}{2} x$ and $O G$ has length $\frac{3}{2} x$. Since the length of $O B$ is $5$, [[Pythagoras' theorem]] applied to triangle $O B D$ establishes that: $$ 5^2 = \left(\frac{1}{2} x\right)^2 + \left(\frac{3}{2} x \right)^2 = \frac{1 + 9}{4} x^2 = \frac{5}{2} x^2 $$ The total area of the squares is then $2 x^2 = 20$.