Triangle, Square, or Circle

Let the area of each be $A$ . Then the perimeter of the equilateral triangle is $6\sqrt {\dfrac{A}{\sqrt 3}}\approx 4.559\sqrt A$ ,

Perimeter of the square is $4\sqrt A$ ,

Perimeter of the circle is $2\sqrt {πA}\approx 3.5449\sqrt A$

So the equilateral triangle has the longest perimeter .

Let say that all of their areas are equal to $1$ .

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Finding the perimeter of the equilateral triangle:

Using the Heron's formula and simplifying it with our variables, the area of the triangle would be equal to $\sqrt{( \frac{3x}{2} ) ( \frac{x^{3}}{8} ) }$ with $x$ being the length of the shape's side, and since the area is $1$ , squaring the formula above would still yield the same answer. $\begin{aligned} \left(\frac{3x}{2}\right) \left(\frac{x^{3}}{8}\right) = \space & 1 \\ \frac{3x^{4}}{16} = \space & 1 \\ 3x^{4} = \space & 16 \\ x^{4} = \space & \frac{16}{3} \\ x = \space & \sqrt[4]{\frac{16}{3}} \space \text{or} \space \frac{2}{\sqrt[4]{3}} \end{aligned}$ Multiplying the value of $x$ by $3$ to get the triangle's perimeter, we get $2\sqrt[4]{27}$ or $4.559$ .

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Finding the perimeter of the square:

Since the length of a square's side is the square root of its area, multiplying $1$ by $4$ would get us the perimeter which would be $4$ .

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Finding the perimeter of the circle:

$\begin{aligned} r^{2}π & = \space 1 \\ r^{2} & = \space \frac{1}{π} \\ r & = \space \sqrt{\frac{1}{π}} \space \text{or} \space \frac{1}{\sqrt{π}} \end{aligned}$ Multiplying the value of $r$ by $2π$ to get the circumference, we get $2\sqrt{π}$ or $3.545$ .

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Comparing all of the perimeters, the equilateral triangle's is the longest.