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