# Solution to the Two Squares in a Semi-Circle Puzzle +-- {.image} [[TwoSquaresinaSemiCircle.png:pic]] > What's the total area of these two squares? =-- ## Solution by [[Pythagoras' Theorem]], [[chord|Properties of Chords]], and [[Isosceles]] [[Right-Angled Triangles]] +-- {.image} [[TwoSquaresinaSemiCircleLabelled.png:pic]] =-- In the above diagram, the point labelled $O$ is the centre of the circle and $D$ is the [[midpoint]] of the [[chord]] $A B$. Point $F$ is directly below $D$, and $E$ is where the horizontal from $A$ meets the vertical from $D$. Let $a$ be the length of a side of the blue square and $b$ of the yellow square. The horizontal distance from $A$ to $B$ is $a + b$, so the length of $A E$ is $\frac{a + b}{2}$. The height of $D$ above the diameter is the average of the heights of $A$ and $B$, so also is $\frac{a + b}{2}$. As the [[perpendicular bisector]] of a [[chord]] passes through the centre of the circle, angle $A \hat{D} O$ is a right-angle. This means that angles $A \hat{D} E$ and $F \hat{D} O$ add up to $90^\circ$ so triangles $A E D$ and $D F O$ are [[similar]]. Since $A E$ and $D F$ have the same length, they are actually [[congruent]]. This means that $A D$ and $D O$ are also the same length. Triangle $A D O$ is therefore [[isosceles]], as is triangle $B D O$, and so triangle $A O B$ is a [[right-angled triangle|right-angled]] [[isosceles]] triangle. The length of $O B$ is the radius of the semi-circle, which is $8$, so the square of the length of $A B$ is $8^2 + 8^2 = 128$. Triangle $A C B$ is also right-angled with $A C$ of length $\sqrt{2} a$ and $C B$ of length $\sqrt{2}b$. Applying [[Pythagoras' theorem]] to this triangle gives: $$ 2 a^2 + 2 b^2 = 128 $$ which gives the total area of the squares as $64$. ## Solution by [[Invariance Principle]] +-- {.image} [[TwoSquaresinaSemiCircleSpecial.png:pic]] =-- In this special case, the two squares are drawn to be the same size. The diagonal of one of the squares is a radius of the circle, so is of length $8$. The side length is therefore $\frac{8}{\sqrt{2}}$ and so each has area $32$. The total area is then $64$.