An equilateral triangle, a square and a circle have equal perimeters. If T denotes the area of triangle, S, the area of square and C, the area of the circle, then :

Options

A. `S lt T lt C`
B. `T lt C lt S`
C. `T lt S lt C`
D. `C lt S lt T`

Answer

To solve the problem step by step, we need to find the areas of the equilateral triangle, square, and circle given that they all have equal perimeters.

### Step 1: Define the Perimeters
Let:
- \( A \) be the side length of the equilateral triangle.
- \( B \) be the side length of the square.
- \( R \) be the radius of the circle.

The perimeters of the shapes are:
- Perimeter of the equilateral triangle = \( 3A \)
- Perimeter of the square = \( 4B \)
- Perimeter of the circle = \( 2\pi R \)

Since the perimeters are equal, we can set them equal to each other:
\[ 3A = 4B = 2\pi R \]

### Step 2: Express Each Variable in Terms of a Common Variable
Let’s express \( A \), \( B \), and \( R \) in terms of a common variable \( P \), which represents the common perimeter.

From the perimeter equations:
1. \( A = \frac{P}{3} \)
2. \( B = \frac{P}{4} \)
3. \( R = \frac{P}{2\pi} \)

### Step 3: Calculate the Areas
Now we can calculate the areas of each shape using the side lengths and radius we found.

1. **Area of the Equilateral Triangle (T)**:
The area \( T \) of an equilateral triangle with side length \( A \) is given by:
\[ T = \frac{\sqrt{3}}{4} A^2 \]
Substituting \( A = \frac{P}{3} \):
\[ T = \frac{\sqrt{3}}{4} \left(\frac{P}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{P^2}{9} = \frac{\sqrt{3}P^2}{36} \]

2. **Area of the Square (S)**:
The area \( S \) of a square with side length \( B \) is given by:
\[ S = B^2 \]
Substituting \( B = \frac{P}{4} \):
\[ S = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16} \]

3. **Area of the Circle (C)**:
The area \( C \) of a circle with radius \( R \) is given by:
\[ C = \pi R^2 \]
Substituting \( R = \frac{P}{2\pi} \):
\[ C = \pi \left(\frac{P}{2\pi}\right)^2 = \pi \cdot \frac{P^2}{4\pi^2} = \frac{P^2}{4\pi} \]

### Step 4: Summary of Areas
Now we have the areas in terms of the common perimeter \( P \):
- \( T = \frac{\sqrt{3}P^2}{36} \)
- \( S = \frac{P^2}{16} \)
- \( C = \frac{P^2}{4\pi} \)

### Step 5: Conclusion
We have expressed the areas of the equilateral triangle, square, and circle in terms of their common perimeter. This allows us to compare their areas if needed.