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

The correct option is: The circle has the largest area.

To understand why this statement is true, let us compare the areas of a circle, a square, and an equilateral triangle when they all have the same perimeter. Let this common perimeter be represented by P.

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

2. Area of the Square:
Let s be the side length of the square.
The perimeter of the square is:
$$4s=P$$
Solving for s:
$$s=\frac{P}{4}$$
The area of the square, A_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:
Let a be the side length of the equilateral triangle.
The perimeter of the triangle is:
$$3a=P$$
Solving for a:
$$a=\frac{P}{3}$$
The area of the equilateral triangle, A_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 the approximation $\sqrt{3}\approx 1.732$:
$$A_{\text{triangle}}\approx \frac{1.732}{36}{P}^2\approx 0.0481P^2$$

Conclusion:
Comparing the coefficients of P² for the three shapes:
$$0.0796P^2>0.0625P^2>0.0481P^2$$
This shows that:
$$A_{\text{circle}}>A_{\text{square}}>A_{\text{triangle}}$$
Therefore, for a given fixed perimeter, the circle encloses the greatest area.