Triangle, Square, or Circle 2

Finding the ratio of the equilateral triangle.

The longest possible inscribed line segment in an equilateral triangle would be parallel to one of the shape's sides. The line segment would have to be as close as possible to the side. If it moves closer, it becomes longer. Thus, it can be said that the longest possible inscribed line segment in an equilateral triangle has a length approximate to the triangle's side. Since the perimeter of the said shape is three times its side length, the ratio would then be $\frac{1}{3}$ .

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Finding the ratio of the square.

The longest possible inscribed line segment in a square would be its diagonal. Let $A_{\square}$ be the area of the square. Using the Pythagorean Theorem, the diagonal's length is $\sqrt{2A_{\square}}$ . The perimeter of the square would then be $4\sqrt{A_{\square}}$ . The needed ratio would be $\frac{\sqrt{2A_{\square}}}{4\sqrt{A_{\square}}}$ . $\small \frac{\sqrt{2A_{\square}}}{4\sqrt{A_{\square}}} \implies \frac{\sqrt{2}\sqrt{A_{\square}}}{4\sqrt{A_{\square}}} \implies \frac{\sqrt{2}}{4} \approx 0.3536$

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Finding the ratio of the circle.

The longest possible inscribed line segment in a circle would be its diameter. Let $A_{\circ}$ be the area of the circle. The diameter's length would then be $\frac{2\sqrt{πA_{\circ}}}{π}$ . The circumference would be $2\sqrt{πA_{\circ}}$ . $\small \frac{\frac{2\sqrt{πA_{\circ}}}{π}}{2\sqrt{πA_{\circ}}} \implies \frac{2\sqrt{πA_{\circ}}}{2π\sqrt{πA_{\circ}}} \implies \frac{2}{2π} \implies \frac{1}{π} \approx 0.3183$

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Comparing all of the ratios, the square's has the largest value.