Let the length of one leg be \(\sqrt{2}\). Then we can calculate the semiperimeter as \(1+\sqrt{2}\), and thus the radius of the inscribed circle as \(\frac{1}{1+\sqrt{2}}\). The area is then \(\pi(\sqrt{2}-1)\). The next two figures are quite simple, we see that in the upper-right figure, the square is half of the area of the triangle, and in the last figure, the side length of the square is \(\frac{1}{3}\) of the hypotenuse, thus the area is \(4/9\). Computing, we find that the circle contains the largest area.

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TopNewestLet the length of one leg be \(\sqrt{2}\). Then we can calculate the semiperimeter as \(1+\sqrt{2}\), and thus the radius of the inscribed circle as \(\frac{1}{1+\sqrt{2}}\). The area is then \(\pi(\sqrt{2}-1)\). The next two figures are quite simple, we see that in the upper-right figure, the square is half of the area of the triangle, and in the last figure, the side length of the square is \(\frac{1}{3}\) of the hypotenuse, thus the area is \(4/9\). Computing, we find that the circle contains the largest area.

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