Infinite Inscription

It might help to work out the area of the first few squares and circles.

The area of the first square is clearly $100$ cm $^{2}$ . Meanwhile, the first circle's radius is equal to half the first square's side length, so its area is $π\times 5^{2}=25π$ . So the area of the first shaded region is $100-25π$ .

Using Pythagoras' Theorem yields the side length of the second square. $x^{2}=5^{2}+5^{2}=50$ so $x=\sqrt{50}$ . This yields that the area of the second square is $\sqrt{50}^{2}=50$ . It also yields that the radius of the second circle is $(\frac{\sqrt{50}}{2})^{2}$ , so the area of the second circle is $π\times (\frac{\sqrt{50}}{2})^{2}=12.25π$ . So the area of the second blue region is $50-12.25π$ .

You can repeat these calculations for the next few squares and circles if you like, but the pattern is already quite clear; the total area of the blue region will be given by:

$(100-25π)+(50-12.25π)+(25-6.25π+)...$

Rearranging yields a slightly clearer format:

$(100+50+25+...)-π(25+12.5+6.25+...)$ .

These are two geometric series whose sums to infinity can be calculated using $S_{∞}=\frac{a}{1-r}$ . Hence, the sum of the first series is $\frac{100}{1-\frac{1}{2}}=200$ and the sum of the second series is $\frac{25}{1-\frac{1}{2}}=50$ .

So the blue area is $\boxed{200-50π}$ .

QED.