Notes
semi-circle in a square solution

Solution to the Semi-Circle in a Square Puzzle

Solution by Angle Between a Radius and Tangent, Isosceles Right-Angled Triangles, Area of a Circle, and Area of a Square

In the above diagram, the point labelled $O$ is the centre of the semi-circle. The point labelled $H$ is where the semi-circle touches the top edge of the square, so as the angle between a radius and tangent is $90^\circ$, $H C$ is perpendicular to $I G$. Similarly, $J E$ is perpendicular to $I A$.

The line segments $O H$ and $O J$ are radii of the circle, so are of the same length. Therefore, $O E$ and $O C$ are of the same length. Triangles $O E F$ and $B C O$ are congruent since $J E$ is parallel to $A D$ and so angles $C \hat{B} O$ and $E \hat{O} F$ are corresponding angles, and $O B$ and $O F$ have the same length. So $O E$ and $E F$ have the same length so triangle $O E F$ is isosceles.

Let $r$ be the radius of the semi-circle. Then as $O E F$ is an isosceles right-angled triangle with hypotenuse of length $r$, its shorter sides both have length $\frac{r}{\sqrt{2}}$. The side length of the square is therefore $r\left(1 + \frac{1}{\sqrt{2}}\right)$. Its area is thus

The area of the semi-circle is $\frac{1}{2} \pi r^2$.

To compare these two, consider the decimal representations of the circle area and half the area of the square, since if the blue area is larger than the pink then it will be over half the square and vice-versa:

Therefore the blue area is larger.