If the perimeter of an equilateral triangle and a square same and the area of equilateral triangle is P and the area of square is Q, then :

Options

A. `P lt Q`
B. `P le Q `
C. `P gt Q `
D. `P ge Q`

Answer

To solve the problem, we need to compare the areas of an equilateral triangle and a square when both have the same perimeter. Let's break it down step by step.

### Step 1: Define the perimeter of the square and equilateral triangle
Let the side length of the square be \( s \). The perimeter of the square is given by:
\[ \text{Perimeter of square} = 4s \]

Let the side length of the equilateral triangle be \( a \). The perimeter of the equilateral triangle is given by:
\[ \text{Perimeter of triangle} = 3a \]

Since the perimeters are the same, we can set them equal to each other:
\[ 4s = 3a \]

### Step 2: Express \( a \) in terms of \( s \)
From the equation \( 4s = 3a \), we can express \( a \) as:
\[ a = \frac{4s}{3} \]

### Step 3: Calculate the area of the square
The area \( Q \) of the square is given by:
\[ Q = s^2 \]

### Step 4: Calculate the area of the equilateral triangle
The area \( P \) of an equilateral triangle is given by the formula:
\[ P = \frac{\sqrt{3}}{4} a^2 \]
Substituting \( a = \frac{4s}{3} \) into the area formula:
\[ P = \frac{\sqrt{3}}{4} \left(\frac{4s}{3}\right)^2 \]
Calculating \( a^2 \):
\[ \left(\frac{4s}{3}\right)^2 = \frac{16s^2}{9} \]
Now substituting back into the area formula:
\[ P = \frac{\sqrt{3}}{4} \cdot \frac{16s^2}{9} = \frac{16\sqrt{3}}{36} s^2 = \frac{4\sqrt{3}}{9} s^2 \]

### Step 5: Compare the areas \( P \) and \( Q \)
Now we have:
- Area of the square: \( Q = s^2 \)
- Area of the equilateral triangle: \( P = \frac{4\sqrt{3}}{9} s^2 \)

To compare \( P \) and \( Q \), we can express it as:
\[ P = \frac{4\sqrt{3}}{9} Q \]

### Step 6: Determine the relationship between \( P \) and \( Q \)
Now, we need to evaluate \( \frac{4\sqrt{3}}{9} \):
- The approximate value of \( \sqrt{3} \) is about \( 1.732 \).
- Therefore, \( \frac{4\sqrt{3}}{9} \approx \frac{4 \times 1.732}{9} \approx \frac{6.928}{9} \approx 0.770 \).

Since \( 0.770 < 1 \), we conclude that:
\[ P < Q \]

### Final Conclusion
Thus, the area of the equilateral triangle \( P \) is less than the area of the square \( Q \).

### Answer
The correct relationship is \( P < Q \).

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