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There is no such square whose area is equal to the difference between the areas of its circumscribed and inscribed circles. Am I right?

Imagine a square with length $L$.
This square has an area of $L^2$.
Its inscribed circle has radius $\frac{L}{2}$.
Hence, it has an area of $π \frac{L^2}{4}$.
By Pythagoras' theorem, the circumscribed circle has a radius of $\sqrt 2L$.
So it has an area of $2πL^2$.

A.T.Q. :
\begin{align} \text{Area of circumscribed circle - Area of inscribed circle} &= \text{Area of a square} \\ 2πL^2 - π\frac{L^2}{4} & \neq L^2 \end{align}

Does this complete my proof?

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It is right, I suppose? π is not equal to 4.

  • See your revised post. In it you provided a counter-example, which, indeed, proves your assertion. For rigorousness, you may want to remark that every square is completely defined by the length of its sides. – Mussé Redi Dec 28 '17 at 16:35
  • You're very welcome. – Mussé Redi Dec 28 '17 at 16:36
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    In fact I expanded on this idea: https://math.stackexchange.com/questions/2579805/no-polygon-has-the-same-area-as-the-difference-between-its-inscribed-and-circums/2580922#2580922 – For the love of maths Dec 28 '17 at 16:37