# Solution to the Two Semi-Circles in a Square Puzzle +-- {.image} [[TwoSemiCirclesinaSquare.png:pic]] > The area of the square is $100$. What's the missing length? =-- ## Solution by [[Angle Between a Radius and Tangent]], [[Pythagoras' Theorem]], [[Angle in a Semi-Circle]], and [[Similar Triangles]] +-- {.image} [[TwoSemiCirclesinaSquareLabelled.png:pic]] =-- With the points labelled as in the above diagram, let $A C$ have length $a$, $B C$ have length $b$, and let $r$ be the radius of the smaller semi-circle. Let $c$ be the length of $A E$. As the square has area $100$, it has side length $10$. As the [[angle between a radius and tangent]] is $90^\circ$, triangle $A E D$ is [[right-angled triangle|right-angled]] so [[Pythagoras' theorem]] applies and shows that: $$ c^2 + r^2 = (a + r)^2 = a^2 + 2 a r + r^2 $$ So $c^2 = a^2 + 2 a r$. Angle $A \hat{C} B$ is $90^\circ$ as it is the [[angle in a semi-circle]], so triangles $A C B$ and $A B F$ are [[similar]]. This means that the ratio of the lengths of $A B$ to $A C$ is equal to that of $A F$ to $A B$. That is, $\frac{10}{a} = \frac{a + 2 r}{10}$. This rearranges to $10^2 = a^2 + 2 a r$ and so $c^2 = 10^2$. The missing length is therefore $10$.