# The semicircle puzzle

In a recent episode of Wrong, But Useful, I asked:

A square is inscribed within a circle of radius $r$. A second square is inscribed within a semicircle of the same radius. What is the ratio of the areas of the squares?

It’s easy enough to find the side length of the big square: the circle’s radius is $r$, and the square is made up of four right-angled isosceles triangles with the sides as hypotenuses. A bit of Pythagoras and boom: the side length is $r\sqrt 2$, and the area is $2r^2$.

The smaller square is trickier; in this case, $r$ is the distance from the middle of one side to either opposite corner. That makes another right-angled triangle, although now one of the legs (the side length) is double the length of the other, which I’ll call $x$. We’ve got $(2x)^2 + x^2 =r^2$, or $x^2 = \frac 1 5 r^2$. The area of the square is $(2x)^2 = \frac 45r^2$.

The ratio of the areas is then $2:\frac45$ or $5:2$.

Edited April 9, 2014 to correct a typo.