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Is the area of a convex polygon equal to the area of a circle with the same perimeter of the polygon? I guess that it's possible, take for example an square, I guess that it's borders could be deformed to form a circle and that their areas would be the same but such condition holds only for convex polygons. I guess I've read some theorem about it in the past but I don't remember it now.

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The area of a circle is greater than the area of any polygon with the same perimeter. Equivalently, the circumference of a circle is smaller than the perimeter of any polygon with the same area. For details and references, please see the Wikipedia article on the Isoperimetric Inequality.

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Well, why don't you just try the circle and square first?

Take a circle with radius 1, which has an area of $\pi$ and perimeter of $2\pi$. A square with the same perimeter would have a side length of $\pi/2$, hence an area of $\pi^2/4$.

Since $\pi$ isn't equal to $\pi^2/4$, you already have a counterexample.

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    $\begingroup$ A circle with radius $1$ has a perimeter of $2\pi$. $\endgroup$ – JiK Aug 10 '14 at 20:38

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