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Note by Llewellyn Sterling
2 years, 6 months ago

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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.

- 2 years, 6 months ago