This problem asks us to compare the areas of three different geometric shapes—a circle, a square, and an equilateral triangle—given that their perimeters are all equal. To solve this, we will express the area of each shape in terms of its perimeter and then compare the resulting expressions. The fundamental principle here is that for a fixed perimeter, the circle encloses the maximum area among all two-dimensional shapes.

Perimeter Definition

Let's assume the common perimeter for the circle, the square, and the equilateral triangle is denoted by \(P\).

Circle Area Calculation

For a circle, the perimeter is its circumference. Let \(r\) be the radius of the circle.

• The circumference (perimeter) of a circle is given by the formula: \[ C = 2\pi r \]
• Since the perimeter is \(P\), we have: \[ P = 2\pi r \]
• We can express the radius \(r\) in terms of \(P\): \[ r = \frac{P}{2\pi} \]
• The area of a circle is given by the formula: \[ A_{circle} = \pi r^2 \]
• Substituting the expression for \(r\) into the area formula: \[ A_{circle} = \pi \left( \frac{P}{2\pi} \right)^2 \] \[ A_{circle} = \pi \left( \frac{P^2}{4\pi^2} \right) \] \[ A_{circle} = \frac{P^2}{4\pi} \]
• Using the approximate value of \(\pi \approx 3.14159\): \[ A_{circle} \approx \frac{P^2}{4 \times 3.14159} \] \[ A_{circle} \approx \frac{P^2}{12.56636} \]

Square Area Calculation

For a square, let \(s\) be the length of one side.

• The perimeter of a square is given by the formula: \[ P = 4s \]
• We can express the side length \(s\) in terms of \(P\): \[ s = \frac{P}{4} \]
• The area of a square is given by the formula: \[ A_{square} = s^2 \]
• Substituting the expression for \(s\) into the area formula: \[ A_{square} = \left( \frac{P}{4} \right)^2 \] \[ A_{square} = \frac{P^2}{16} \]

Equilateral Triangle Area Calculation

For an equilateral triangle, let \(a\) be the length of one side.

• The perimeter of an equilateral triangle is given by the formula: \[ P = 3a \]
• We can express the side length \(a\) in terms of \(P\): \[ a = \frac{P}{3} \]
• The area of an equilateral triangle is given by the formula: \[ A_{triangle} = \frac{\sqrt{3}}{4} a^2 \]
• Substituting the expression for \(a\) into the area formula: \[ A_{triangle} = \frac{\sqrt{3}}{4} \left( \frac{P}{3} \right)^2 \] \[ A_{triangle} = \frac{\sqrt{3}}{4} \left( \frac{P^2}{9} \right) \] \[ A_{triangle} = \frac{\sqrt{3}P^2}{36} \]
• Using the approximate value of \(\sqrt{3} \approx 1.732\): \[ A_{triangle} \approx \frac{1.732 P^2}{36} \] \[ A_{triangle} \approx \frac{P^2}{20.784} \]

Area Comparison Summary

Now, let's compare the areas we calculated for each shape:

To compare these fractions, we look at their denominators. For fractions with the same positive numerator (\(P^2\) in this case), the smaller the denominator, the larger the value of the fraction.

• For the circle, the denominator is approximately \(12.566\).
• For the square, the denominator is \(16\).
• For the equilateral triangle, the denominator is approximately \(20.784\).

Arranging the denominators from smallest to largest:

\( 12.566 \quad < \quad 16 \quad < \quad 20.784 \)

This means that the corresponding fractions will be ordered from largest to smallest:

\( \frac{1}{12.566} \quad > \quad \frac{1}{16} \quad > \quad \frac{1}{20.784} \)

Therefore, when their perimeters are equal, the order of area from largest to smallest is:

\( A_{circle} \quad > \quad A_{square} \quad > \quad A_{triangle} \)

The circle has the largest area among the three shapes when their perimeters are equal. This result aligns with the isoperimetric inequality, which states that among all closed curves of a given length, the circle encloses the maximum possible area.