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

Question

Prepp Team · Accepted Answer

Comparing Areas for Equal Perimeters

This problem asks us to compare the areas of a circle, a square, and an equilateral triangle when they all have the same perimeter.

Let the common perimeter be denoted by $P$.
Circle:
- Circumference $C = 2 \pi r = P$.
- Radius $r = \frac{P}{2\pi}$.
- Area $A_{circle} = \pi r^2 = \pi \left(\frac{P}{2\pi}\right)^2 = \pi \frac{P^2}{4\pi^2} = \frac{P^2}{4\pi}$.
Square:
- Perimeter $S = 4s = P$.
- Side length $s = \frac{P}{4}$.
- Area $A_{square} = s^2 = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16}$.
Equilateral Triangle:
- Perimeter $T = 3a = P$.
- Side length $a = \frac{P}{3}$.
- Area $A_{triangle} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left(\frac{P}{3}\right)^2 = \frac{\sqrt{3}}{4} \frac{P^2}{9} = \frac{\sqrt{3} P^2}{36}$.

We need to compare the areas. Since $P^2$ is a common positive factor, we compare the coefficients:

Let's approximate these values using $\pi \approx 3.14159$ and $\sqrt{3} \approx 1.732$:

$\frac{1}{4\pi} \approx \frac{1}{4 \times 3.14159} \approx \frac{1}{12.566} \approx 0.07958$
$\frac{1}{16} = 0.0625$
$\frac{\sqrt{3}}{36} \approx \frac{1.732}{36} \approx 0.0481$

Comparing the coefficients:

$0.07958 > 0.0625 > 0.0481$

Therefore:

$\frac{1}{4\pi} > \frac{1}{16} > \frac{\sqrt{3}}{36}$

This implies that $A_{circle} > A_{square} > A_{triangle}$.