An equilateral triangle, a square and a circle have equal perimeter. If T, S and C denote the area of the triangle, area of the square and area of the circle respectively, then which one of the following is correct?
T < S < C
Comparing Areas of Geometric Shapes with Equal Perimeter
The question asks us to compare the areas of three different geometric shapes – an equilateral triangle, a square, and a circle – given that they all have the same perimeter. Let's denote the common perimeter by \( P \). We need to find expressions for the area of each shape in terms of \( P \) and then compare these areas.
Step 1: Calculate Dimensions in Terms of Perimeter \( P \)
- Equilateral Triangle: Let the side length be \( a \). The perimeter is the sum of the lengths of its three equal sides.
Perimeter \( P = 3a \)
So, the side length \( a = \frac{P}{3} \). - Square: Let the side length be \( b \). The perimeter is the sum of the lengths of its four equal sides.
Perimeter \( P = 4b \)
So, the side length \( b = \frac{P}{4} \). - Circle: Let the radius be \( r \). The perimeter of a circle is its circumference.
Perimeter \( P = 2\pi r \)
So, the radius \( r = \frac{P}{2\pi} \).
Step 2: Calculate Areas in Terms of Perimeter \( P \)
Now, we will find the area of each shape using the dimensions we just found in terms of \( P \).
- Equilateral Triangle: The area \( T \) of an equilateral triangle with side length \( a \) is given by the formula \( T = \frac{\sqrt{3}}{4} a^2 \).
Substituting \( a = \frac{P}{3} \):
\( T = \frac{\sqrt{3}}{4} \left(\frac{P}{3}\right)^2 = \frac{\sqrt{3}}{4} \frac{P^2}{9} = \frac{\sqrt{3}}{36} P^2 \). - Square: The area \( S \) of a square with side length \( b \) is given by the formula \( S = b^2 \).
Substituting \( b = \frac{P}{4} \):
\( S = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16} \). - Circle: The area \( C \) of a circle with radius \( r \) is given by the formula \( C = \pi r^2 \).
Substituting \( r = \frac{P}{2\pi} \):
\( C = \pi \left(\frac{P}{2\pi}\right)^2 = \pi \frac{P^2}{4\pi^2} = \frac{P^2}{4\pi} \).
Step 3: Compare the Areas
We need to compare \( T = \frac{\sqrt{3}}{36} P^2 \), \( S = \frac{1}{16} P^2 \), and \( C = \frac{1}{4\pi} P^2 \). Since \( P^2 \) is a common positive factor (as perimeter must be positive), we can compare the coefficients of \( P^2 \): \( \frac{\sqrt{3}}{36} \), \( \frac{1}{16} \), and \( \frac{1}{4\pi} \).
Let's approximate the values of these coefficients:
- For the triangle coefficient: \( \frac{\sqrt{3}}{36} \approx \frac{1.732}{36} \approx 0.0481 \)
- For the square coefficient: \( \frac{1}{16} = 0.0625 \)
- For the circle coefficient: \( \frac{1}{4\pi} \approx \frac{1}{4 \times 3.14159} = \frac{1}{12.56636} \approx 0.0796 \)
Comparing the approximate values:
\( 0.0481 < 0.0625 < 0.0796 \)
This shows that \( \frac{\sqrt{3}}{36} < \frac{1}{16} < \frac{1}{4\pi} \). Therefore, \( \frac{\sqrt{3}}{36} P^2 < \frac{1}{16} P^2 < \frac{1}{4\pi} P^2 \), which means \( T < S < C \).
This result aligns with the general principle that, for a fixed perimeter, the shape that is closest to a circle encloses the largest area. Among a regular triangle, a square, and a circle, the circle is closest to the optimal shape for maximizing area for a given perimeter.
Based on the comparison, the correct order of areas is \( T < S < C \).
Revision Table: Area Formulas and Coefficients
| Shape | Perimeter (P) | Dimension in terms of P | Area Formula | Area in terms of P | Coefficient of \( P^2 \) | Approximate Value |
|---|---|---|---|---|---|---|
| Equilateral Triangle (T) | \( 3a \) | \( a = \frac{P}{3} \) | \( \frac{\sqrt{3}}{4} a^2 \) | \( \frac{\sqrt{3}}{36} P^2 \) | \( \frac{\sqrt{3}}{36} \) | \( \approx 0.0481 \) |
| Square (S) | \( 4b \) | \( b = \frac{P}{4} \) | \( b^2 \) | \( \frac{1}{16} P^2 \) | \( \frac{1}{16} \) | \( = 0.0625 \) |
| Circle (C) | \( 2\pi r \) | \( r = \frac{P}{2\pi} \) | \( \pi r^2 \) | \( \frac{1}{4\pi} P^2 \) | \( \frac{1}{4\pi} \) | \( \approx 0.0796 \) |
Additional Information: Isoperimetric Inequality
This problem is an example illustrating a concept in geometry known as the isoperimetric inequality. The isoperimetric inequality states that among all closed figures in a plane with a fixed perimeter, the circle encloses the maximum possible area. For other shapes like polygons, among polygons with a fixed number of sides and a fixed perimeter, the regular polygon has the largest area. As the number of sides of a regular polygon increases, its area approaches the area of a circle with the same perimeter. Therefore, it makes sense that the circle has a larger area than the square, and the square has a larger area than the equilateral triangle, when their perimeters are equal, because the square is "closer" to a circle than the triangle, and the circle is the ultimate shape for maximizing area.
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