A primer for mathematics competitions - Section 1.1 - Appetizer Problem 1

Problem:

Solution:

Letting the center of the circle $O$, the square to have side length $2s$ and the circle have radius $r$.

By Pythagoras theorem, we can get this relation. (Sorry Pythagoras, I just keep forgetting how to spell your name ... )

$s^2 + (2s - r)^2 = r^2$

This can easily simplify to $5s^2 - 4sr = 0$, so we get $r = \frac{5s}{4}$.

Therefore the area ratio is $\frac{\pi r^2}{(2s)^2} = \frac{\pi (5s)^2}{4^2(2s)^2} = \frac{25\pi }{64}$.

The answer is (A).