The perimeters of a circle, a square and an equilateral triangle are equal. Which one of the following statements is true?

To determine which shape has the largest area when their perimeters are equal, let us denote the common perimeter of all three shapes as $P$.

1. Area of the Circle:
The perimeter (circumference) of a circle is given by:
$P=2\pi r$
where $r$ is the radius of the circle. Solving for $r$:
$r=\frac{P}{2\pi }$
The area of the circle ($A_{\text{circle}}$) is:
$A_{\text{circle}}=\pi r^2=\pi {\left(\frac{P}{2\pi }\right)}^2=\frac{P^2}{4\pi }$
Using $\pi \approx 3.1416$:
$A_{\text{circle}}\approx \frac{P^2}{12.57}\approx 0.0796P^2$

2. Area of the Square:
The perimeter of a square with side length $s$ is:
$P=4s$
Solving for $s$:
$s=\frac{P}{4}$
The area of the square ($A_{\text{square}}$) is:
$A_{\text{square}}=s^2={\left(\frac{P}{4}\right)}^2=\frac{P^2}{16}=0.0625P^2$

3. Area of the Equilateral Triangle:
The perimeter of an equilateral triangle with side length $a$ is:
$P=3a$
Solving for $a$:
$a=\frac{P}{3}$
The area of the equilateral triangle ($A_{\text{triangle}}$) is:
$A_{\text{triangle}}=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}{\left(\frac{P}{3}\right)}^2=\frac{\sqrt{3}{P}^2}{36}$
Using $\sqrt{3}\approx 1.732$:
$A_{\text{triangle}}\approx \frac{1.732P^2}{36}\approx 0.0481P^2$

Comparing the Areas:
Comparing the coefficients of $P^2$ for each shape:
$0.0796P^2>0.0625P^2>0.0481P^2$
Therefore:
$A_{\text{circle}}>A_{\text{square}}>A_{\text{triangle}}$
This shows that for a given perimeter, the circle encloses the maximum area.