# Square in a Semi-Circle in a Square +-- {.image} [[SquareinaSemiCircleinaSquare.png:pic]] > What fraction of the larger square is covered by the one inside the semicircle? =-- ## Solution by [[Properties of Chords]] and [[Pythagoras' Theorem]] +-- {.image} [[SquareinaSemiCircleinaSquareLabelled.png:pic]] =-- Since the top edge of the smaller square is a [[chord]] of the semi-circle, the centre of the circle lies on its [[perpendicular bisector]] and so is the [[midpoint]] of the bottom edge. This means that, in the above diagram, the length of $O A$ is half that of $A C$. Writing $x$ for the length of $O A$, this means that $O B$ is $2 x$. The area of the smaller square is then $4 x^2$. Let $y$ be the length of $O C$. Since $O A C$ is a [[right-angled triangle]], applying [[Pythagoras' theorem]] shows that $y^2 = x^2 + (2 x)^2 = 5 x^2$. As $O B$ has the same length as $O C$, the area of the outer square is then $4 y^2$. The outer square is therefore $5$ times bigger than the smaller, and so the fraction that is shaded is $\frac{1}{5}$th.