Which has the largest perimeter?

a circle has an infinite number of sides and encloses the most area for a given perimeter the fewer sides enclosing the same area the more perimeter is required the triangle has the fewest sides and hence has the largest perimeter

Relevant wiki: Regular Polygons - Area

Consider an $n$ -sided regular polygon with radius $r$ . Then the perimeter is given by $p = 2nr \sin \dfrac \pi n$ . The area $A = \dfrac {nr^2}2\sin \dfrac {2\pi}n$ ; $\implies r = \sqrt{\dfrac {2A}{n\sin \frac {2\pi}n}}$ Therefore,

$\begin{aligned} \frac pA & = \frac {2n \sin \frac \pi n}A \sqrt{\frac {2A}{n\sin \frac {2\pi}n}} \\ & = 2\sqrt{\frac nA \color{#3D99F6}\tan \frac \pi n} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = 2 \sqrt{\frac nA\color{#3D99F6} \left(\frac \pi n + \frac {\pi^3}{3n^3} + \frac {\pi^5}{5n^5} + \cdots \right)} \\ & = 2 \sqrt{\frac nA \left(\pi + \frac {\pi^3}{3n^2} + \frac {\pi^5}{5n^4} + \cdots \right)} \end{aligned}$

Therefore, $\dfrac pA$ decreases with $n$ , therefore equilateral triangle $(n=3)$ has the largest perimeter.

Note that a circle is a regular polygon with $n\to \infty$ and it has the smallest $p-A$ ratio of $2\sqrt \pi$ .

Relevant wiki: Regular Polygons - Area

For the equilateral triangle A=s^2 (√3/4), So the perimeter p = 3 s = 4.56√A, For the circle A= πr^2, So the perimeter p = 2 πr = 2√π . √A = 3.54 √A, For the square A = (p/4)^2 , So the perimeter p = 4√A. The triangle has the largest perimeter .

For a regular N-gon, the ratio of internal area to perimeter increases with N. If all three shapes have equal areas, the one with the biggest perimeter will be the one with the fewest sides; hence, the triangle.

Area of equilateral triangle $A = \frac{\sqrt 3}{4} a^2$ $A = (9a^2)\frac{\sqrt 3}{36} =( p_1)^2 \frac{\sqrt 3}{36}$ $\implies p_1 = \frac{6\sqrt A}{\sqrt{(\sqrt 3)}}$ where a is the length of side of equilateral triangle and $p_1 = 3a$ is it's perimeter.

Area of circle $A =\pi r^2$ $A= \frac{4\pi^2r^2}{4\pi} = \frac{(p_2)^2}{4\pi}$ $\implies p_2 =2\sqrt{\pi A}$ where $r$ is the radius of circle and $p_2 = 2\pi r$ is the perimeter.

Area of square $A = l^2\ = \frac{16 l^2}{16}$ $A= \frac{(p_3)^2}{16}$ $\implies p_3 = 4\sqrt A$ where $l$ is the side of square and $p_3 = 4l$ is the perimeter.

Now dividing $\frac{p_1}{p_2} = \frac{6\sqrt A}{\sqrt{(\sqrt 3)}}.\frac{1}{2\sqrt{ \pi A}}> 1$ $\implies p_1>p_2$ Similarly dividing $\frac{p_2}{p_3} = \frac{2\sqrt {\pi A}}{4\sqrt A} < 1$ $\implies p_2 <p_3$ Also dividing $\frac{p_1}{p_3} = \frac{6\sqrt A}{3^{1/4}. 4\sqrt A}> 1$ $\implies p_1>p_3$ Therefore, we can concluded $p_1> p_3>p_2$ Thus Perimeter of equilateral triangle is largest among above three geometric figures for given identical area.

I just look into how nature usually shape things, and also if a line have infinite area, then triangle will be the closest one after triangle

easy, the less "circly" shape has the greatest perimeter, since the circle has the smallest perimeter for a given area

Formula for a triangle=(b h)/2 Square= b h If the area is the same the triangle has a bigger area since you need to divide it by 2. With the square, you don't.

If we suppose the area is 1, the perimeter of the square it's $2 \sqrt{\pi}$ , the perimeter of the triangle it's $3^{1/4}\sqrt{12}$ , and the perimeter of the square it's $2$ . So, the triangle has the greatest perimeter.

Everyone’s formulaic solutions are nice. I learn best with examples:

Take a blade of grass, bend it into a circle from tip to tip.

Now notice what it takes if you want to turn it into a circle or square, you have to push in on the sides of the shape. You can see that the area got smaller. The points bulge out a little, and the sides become straight lines. You can easily see that the amount the points go out is very small compared to how much is lost on the sides.

A circle is a shape with an uncountable number of points. A triangle is a shape with the minimum: 3 (try making a closed shape with only two lines I dare you). So it makes sense that if the circle is the biggest area for any given perimeter (or blade of grass) then the triangle would be the smallest.

to find the perimeter you must first find the area. go from the size of the square. fit the circle inside the square, then make it a bit bigger so their areas are the same. imagine you take the line around the circle and make it straight, the take the lines from the square and stack them together to form one straight line. compare the size of the 2 lines, you will notice that the squares line is longer. so trash the circle line and move on to the triangle. do the same to the triangle as you did with the circle, put it inside the square then make it a bit bigger so they have the same area. then put the triangle's sides together to form one line and compare the squares line with the triangles and you will see that the triangles line is bigger. their line is their perimeter so the triangle has the biggest perimeter. thats how i figured it out anyway. i find this alot easier to understand then $fjsfhbbxhbdddzx bd4187273471$ $\frac{numerator}{denominator}$ times 8640 873 2740e6rshj3 devided by me2 $\frac{1143}{637659}$ more fancy symbolsfffu73rxnn, i hope you feel the same way

To get rid of parameters (sides, radius), instead of comparing perimiters (p), lets compare $k(n) = p(n)^2/S(n)$ where n=3 for triangle, 4 for square, and $\infty$ for circle. This function preserves ordering. This is because S is the same in all cases, and for x>0, $x^2$ preserves ordering. $k(4)=4\cdot4$ , $k(3)=4\cdot3\cdot\sqrt{3}$ , $k(\infty)=4\pi$ .

$k(3)>k(4)>k(\infty)$ .

note that A = pi * r^2 for the circle, l^2 for the square, and (b^2)/2 for the triangle. we want to extract the value of the unknowns here (r, l, and b) in terms of A and the constants so that we may use these values in their respective perimeter formulas, that is 2 pi r, 4l, and 3b. once that is done set A=1 and examine the values of these perimeters.